The	art	and	craft	of	problem	solving	solutions
                                                                                                               	
  What	is	the	problem	and	solution.	The	art	and	craft	of	problem	solving	3rd	edition	solutions.	What	are	examples	of	problem	and	solution.	What	problems	does	art	solve.	The	arts	and	crafts	of	problem	solving.	
  The	art	and	craft	of	problem	solving.	I	am	interested	in	purchasing	The	Art	and	Art	of	Problem	Solving	by	Paul	Seitz	or	The	Art	of	Problem	Solving	Volume	2	by	Ruscyzk.	I'm	leaning	more	toward	the	second	book	because	it	has	a	walkthrough	guide.	Solutions.	this	is	taken	from	Paul	Seitz's	The	Art	and	Craft	of	Problem	Solving.	This	is	version	5.5.33	on
  page	203.	here:	Find	all	ordered	pairs	of	positive	real	numbers	(x,.	The	art	and	craft	of	math	problem	solving.	539.567.721.1..	Which	math	books	should	I	buy.	The	art	of	problem	solving	math	hw	question	to	the	inequality!	help!.	The	art	and	craft	of	problem	solving.	Today	we've	seen	so	many	things	that	were	invented	many	years	ago.	Light	bulbs,
  computers,	clothes,	etc.	And	we've	also	seen	their	development	artificial	humans,	I	need	help!	I	am	in	the	Relay	for	Life	program	and	our	team	theme	this	year	is	TRANSCRIPTION	4	CHAPTER	1	LEARN	PROBLEM	SOLVING	LEARN	PROBLEM	SOLVING	To	achieve	the	perfect	size,	groups	of	two	lack	mobility	and	large	groups	often	cannot	meet	at	the
  same	time.	My	groups	are	chosen	to	work	well	I	don't	worry	about	making	groups	of	equal	strength	although	that	is	desirable	but	instead	I	try	to	Getting	people	to	interact	with	each	other,	ha	harmonious.	The	ideal	group	includes	people	I	enjoy	being	with	(no	romance	please).	People	with	compatible	schedules	(often	a	critical	trait	among	today's
  overworked	students).	Relatively	homogeneous	math	skills/skills.	Basically	a	homogenous	mix	of	introverts	and	extroverts,	unless	that	balances	out	the	mathematical	imbalance.	For	example,	a	group	of	relatively	shy	people	will	do	well,	certainly	much	better	than	a	group	with	several	shy	people	and	a	dominant	type,	unless	the	dominant	person	is
  perceived	as	one	of	the	weaker	mathematicians.	People	who	usually,	but	not	exclusively,	play	different	roles.	For	example,	it's	good	to	have	one	student	who	is	excellent	at	visualizing	and	another	who	is	prone	to	counting.It's	good	to	have	a	dreamy	creative	type,	even	if	she's	not	as	skilled/rigorous,	if	one	of	her	partners	is	a	less	imaginative	but	better
  trained	logician.	Many	successful	groups	operate	like	sports	teams,	where	novices	ask	questions,	break	rules,	and	investigate;	Power	players	who	come	up	with	the	main	theme	move	more	often	than	not;	and	loved	ones	who	can	carefully	critique,	improve,	and	finally	end	a	creative	but	delightful	discussion.	The	best	bands	I've	found	have	people	who
  mostly	play	different	roles,	but	can	also	switch	positions.	Things	to	avoid	in	groups:	Gathering	too	many	close	friends	and	avoiding	unbalanced	groups	with	a	big	extrovert	and	some	more	shy	students.	Such	groups	are	not	doomed	to	fail,	but	may	require	intervention.	Sometimes	a	broken	group	can	be	repaired	by	changing	people	between	groups.
  Sometimes	nothing	can	be	done.	But	we	teachers	are	used	to	imperfections.	In	general,	I'm	thrilled	when	4	out	of	5	groups	do	well,	and	I	know	that's	realistically	a	good	measure	of	success.	See	Appendix	B	for	the	real	story	of	how	much	a	group	can	accomplish,	which	in	part	inspired	me	to	write	TAACOPS.	Homework	is	varied:	reading	and	problems
  to	be	discussed	in	the	next	lesson,	individual	work,	group	work	(one	work	per	group),	rewriting.	For	students	(usually)	starting	out,	transcripts	are	very	important.	I	usually	give	each	issue	a	grade	(A,	B,	Cor	No	Credit)	with	a	symbol	like	/R	that	tells	me	I	have	until	next	week	to	rewrite	it,	and	then	I	change	the	grade.	It	makes	it	difficult	to	keep	track
  of,	but	it's	worth	it.	I	give	students	semi-formal	Backburner	problems	so	they	can	work	without	deadlines.	In	general,	when	students	are	busy,	it	pays	to	be	very	relaxed	about	deadlines.	I	know	my	workshop	is	a	success	when	I	find	a	group	working	on	a	blackboard	in	an	abandoned	classroomat	night,	arguing	and	laughing.	Also,	if	I	ask	about	a	specific
  problem	and	a	few	students	complain	that	they	haven't	slept	all	night,	haven't	worked	on	their	history	essay,	and	still	haven't	solved	the	problem,	I'm	in	heaven.	A	typical	homework	load	is	around	510	middle	class	assignments,	with	a	good	mix	of	easy	(secretly	designed	to	boost	confidence),	medium,	and	difficult	assignments.	Usually	at	least	half	of
  the	homework	at	the	beginning	of	the	course	is	group	work.	This	is	important	to	maintain	the	level	of	trust.	In	general,	groups	are	superior	to	individuals,	so	the	group	has	some	impressive	tasks	to	solve.	Weaker	people	take	pride	in	their	group's	accomplishments,	which	will	help	boost	their	self-confidence.	Of	course,	your	job	is	to	recognize	and
  support	these	people.	It's	important	to	start	slowly	because	it's	much	easier	to	build	trust	gradually,	and	that's	the	bottleneck.	This	is	no	different	from	exercise:	if	you	try	to	do	too	much,	you	risk	injury.	Too	many	overly	complex	tasks	can	easily	and	permanently	break	a	beginner's	ego.	A	good	way	to	avoid	this	is	to	rely	on	a	lot	of	rest	issues	in	the
  first	place.	These	problems	are	fun	and	deeper	than	they	first	appear.	If	you	have	enticed	your	students	with	the	handshake	problem	(1.1.4),	the	light	bulb	problem	(1.3.2),	the	rope	problem	(2.1.19),	or	the	professor's	problem	(2.1.21),	they	will	be	ready.	study.	great	mathematical	effort	in	solving	later	problems.	On	the	other	hand,	there	is	nothing
  wrong	with	having	serious	problems	from	day	one.	These	Backburner	issues	should	probably	be	identified	as	such.	For	example,	on	the	first	day	of	a	course,	I	often	assign	multiple	tasks,	such	as	1.3.10,	1.3.12,	and	3.3.5,	and	typically	spend	at	least	5-10	minutes	of	each	session	reporting	progress	on	these	tasks.	.	Because	they	are	considered
  particularly	harsh,	some	students	avoid	them	while	others	are	exempted	from	them.no	expectations.	I've	experimented	with	not	telling	students	a	hard	problem	is	hard,	but	with	mixed	results.	Exams	There	are	usually	3	to	4	exams	to	be	taken,	all	of	which	you	can	take	home	with	you	except	for	the	Putnam	exam	(see	below).	After	a	few	weeks	of
  working	on	the	basics	(learning	most	of	chapter	2	and	part	of	chapter	3	with	a	few	other	sections),	I'm	planning	my	first	exam2.	It	is	limited	to	about	a	week	and	takes	place	exclusively	in	groups.	The	group	submits	a	paper	and	all	members	of	the	group	receive	the	same	grade.	In	addition,	on	the	day	of	the	exam,	we	have	an	oral	exam	where	students
  are	randomly	selected	without	going	to	the	blackboard	for	notes	or	support	from	their	group	partners.	A	student	can	select	any	problem	that	has	not	been	discussed	and	submit	a	solution	in	minutes.	The	presentation	is	graded	(I	usually	invite	colleagues	from	my	department)	Olympia-style	(cards	numbered	starting	at	15),	and	polite	but	hostile
  questions	are	decided	by	the	judges	or	members	of	other	groups.	These	questions	are	intended	to	uncover	misunderstandings	etc.	The	process	is	a	little	intimidating	but	done	with	good	humor	and	structured	in	a	way	that	everyone	needs	to	understand	as	much	as	possible.	An	unbalanced	group	with	one	dominant	member	will	not	be	successful	if	that
  person	cannot	teach	others.	In	practice,	the	most	consistent	students	receive	the	highest	oral	grades.	2See	Appendix	A	for	sample	exams.	6	CHAPTER	1	PROBLEM	SOLVING	in	groups,	not	groups	of	top	mathematicians.	And	of	course	the	group	only	gets	an	oral	exam.	The	first	test	almost	always	achieves	its	goal.	The	challenges	were	simple	enough
  for	anyone	to	succeed	and	difficult	enough	for	any	group	to	develop	near-military	strength.	Almost	every	college	student	has	experienced,	or	at	least	witnessed,	an	insight,	an	apparently	sudden	revelation	that	was	actually	caused	by	hours	of	sweating.	At	this	point	the	students	have	been	converted:	almost	all	are	nowsolvers,	and	now	you	just	need	to
  practice	your	skills	and	collect	new	tactics	and	tools.	Exam	2	begins,	usually	a	month-long	individual	process.	I	meet	with	each	student	individually	to	discuss	a	set	of	problems,	a	mix	of	Putnam	problems	and	problems	selected	from	one	or	two	topics.	This	list	of	about	15	tasks	varies	greatly	from	student	to	student.	Some	people	are	ready	to	solve	the
  hardest	Putnam	problems	(say,	the	hardest	two	or	three	of	the	12	questions	on	an	exam),	while	others	will	do	well	on	the	easiest	Putnam	problems.	It	doesn't	matter	as	long	as	the	student	and	I	can	agree	on	what	the	appropriate	assignment	will	be.	While	students	work	on	this	exam,	there	are	regular	class	discussions	and	homework,	and	I	tend	to
  spend	more	lectures	at	the	end	of	this	period	than	during	the	semester	(because	students	are	too	busy	solving	their	own	problems	to	work	hard	on	group	or	public	assignments	questions).	).	The	second	exam	is	difficult	for	students.	They	are	separated	from	their	beloved	teammates	and	have	to	fight	their	weaknesses.	But	on	the	other	hand,	if	I	did	my
  homework,	at	least	some	of	the	problems	each	student	solved	were	challenging	enough	to	feel	accomplished.	Needless	to	say,	I	don't	always	succeed,	but	I	usually	make	the	mistake	of	making	the	exam	too	easy	instead	of	too	hard.	And	for	some	students	(especially	teachers	or	prospective	teachers),	I	include	problem-solving	exercises	along	with
  problem-solving	questions.	The	third	exam	is	the	Putnam	exam,	which	takes	place	on	the	first	Saturday	in	December.	My	students	need	to	pass	this	exam	(or	substitute	something	spicy	they	always	choose	for	Putnam).	I	usually	agree	to	change	my	grade	retroactively	to	A+	if	it	solves	2	or	more	tasks	(results	only	available	in	March).	For	most
  students,	Putnam	does	not	affect	their	grades.	But	this	is	our	valuable	experienceprepare	thoroughly	and	we	will	spend	most	of	the	rest	of	the	semester	discussing	problems	in	class.	I	try	not	to	encourage	exam	mania	in	my	students;	Throughout	the	course	I	emphasize	that	investigation	is	paramount	and	that	time	constraints	are	silly.	The	6-hour
  Putnam	exam	consists	of	12	problems,	which	is	a	huge	investment	of	time,	but	most	of	my	students	thoughtfully	solve	a	few	questions	during	the	exam	and	then	spend	much	more	time	studying	the	more	difficult	problems.	Teachers	of	younger	students	(for	example,	high	school	students)	may	find	that	students	like	to	compete	in	exams	more	and	less
  to	work	in	groups.	The	most	important	thing	is	to	encourage	what	makes	students	spend	a	lot	of	time	thinking	about	math.	The	final	exam	is	similar	to	the	first	exam,	it	is	a	group	meeting,	only	the	tasks	have	more	mathematical	content.	Again,	we	end	with	oral	presentations	and	often	refreshments.	The	tasks	are	usually	not	too	difficult	because	the
  students	have	barely	recovered	from	the	birdhouse.	Assessment	As	you	can	imagine,	this	is	not	the	most	pleasant	part	of	the	course	for	the	instructor.	The	main	problem,	of	course,	is	that	the	experiences	and	achievements	of	students	are	quantitatively	completely	incomparable.	I	carefully	record	who	got	which	grade	on	which	assignment,	when
  (using	a	spreadsheet),	weighted	averages,	etc.,	but	at	the	end	I	usually	just	ask:	What	did	this	person	get	out	of	the	course?	And	I'm	pretty	forgiving.	Only	if	the	student	is	not	really	trying	will	I	give	a	hard	grade	(C	or	D).	Usually	at	least	half	of	my	students	get	A's,	but	I	don't	feel	guilty	because	I	know	these	people	have	worked	like	crazy	and
  experienced	deep	disappointments	mixed	with	sporadic	triumphs.	It's	basically	a	binary,	like	a	bootcamp:	you	either	make	it	through	or	you	don't.	Most	students	do	well	with	this	course	because	most	people	like	to	be	challenged.	Chapter	2Solutions	Below	are	solutions,	solution	sketches,	or	at	least	important	hints	for	many	problems.	We	omit
  solutions	to	most	Putnam,	AIME,	USAMO,	and	IMO	problems	because	these	exams	(and	solutions)	are	widely	published	and	we	assume	(and	strongly	recommend!)	that	you	have	them	on	your	shelf.	We	also	do	not	include	the	Bay	Area	Math	Olympiad	problem	solutions	because	they	are	freely	available	at	.	In	addition,	we	do	not	provide	solutions	to
  some	routine	problems	that	are	usually	covered	by	standard	texts.	References	are	listed,	although	you	may	have	other	favorites.	All	citations	and	page	examples	are	from	TACOPS	(2nd	edition),	but	cited	references	use	the	bibliography	at	the	end	of	this	manual,	not	the	bibliography	on	TACOPS	page	3569.	A	large	number	of	problems	in	TACOPS	were
  taken	from	the	1995	Olympic	Games	from	many	countries.	Thanks	to	the	officials	from	those	countries	who	have	kindly	agreed	to	let	me	(and	others)	disseminate	my	questions,	and	thanks	to	Titus	Andreescu,	Elgin	Johnston,	and	especially	Kiran	Kedlai	for	helping	to	solve	many	of	them!	1.3	2	Leave	one	switch	on,	one	switch	off,	and	the	third	switch	on
  for	one	minute,	then	turn	it	off.	Then	check	the	three	options:	light	is	on,	light	is	off	and	cool,	light	is	off	and	warm.5	The	maximum	value	is	73.	Draw	a	picture:	the	first	equation	is	a	circle	of	radius	8	with	center	(7,3	).	The	equation	3x+4y	=	k	defines	a	family	of	parallel	lines,	all	with	slope	3/4.	Extreme	values	​​of	k	are	reached	when	the	line	touches	the
  circle.	Since	the	line	of	contact	is	perpendicular	to	the	line	drawn	from	the	radius	to	the	point	of	contact,	it	is	easy	to	calculate	the	coordinates	of	the	two	points	of	contact	and	then	substitute	them	into	3x+4y.	8	ANSWER	TO	2.1	9	11	(Outline	provided	by	Kiran	Kedlai)	ThisShow	that	cos(cos(x))	>	sin(sin(x))	for	all	x,	which	implies	that
  cos(cos(cos(cos(x)))))	>	sin(sin(cos(cos	(	x))	)	))>	sin(sin(sin(sin(x))))),	which	means	that	this	equation	has	no	solution.	Useful	facts	are	cos(cos(x))	=	sin(pi/2	cos(x)),	sin(x)+cos(x)	=	2cos(xpi/4)	and	the	monotonicity	of	sine	and	cosine	(for	angles	between	0	and	Pi/2).	An	excellent	reference	work	on	problems	of	this	kind	and	many	others	is	"Geometric
  Investigations	in	Combinatorial	Geometry"	by	Vladimir	Boltyansky	and	Alexander	der	Soifer	[3].	The	idea	is	that	triangular	packing	(so	the	centers	of	the	three	circles	are	the	vertices	of	an	equilateral	triangle)	is	a	more	efficient	use	of	space,	but	you	waste	some	space	at	the	beginning	of	the	rectangle.	So	you	need	to	catch	up	in	a	rectangle.	2.1	17	No.
  Color	the	vertices	of	the	square	black,	white,	black,	white18.	The	first	set	contains	only	numbers	whose	digits	are	drawn	exclusively	with	curves;	the	second	contains	linear	digits;	the	third	contains	mixed	numbers.19	The	main	idea	is	to	cut	one	of	the	ropes	under	the	ceiling	into	a	small	loop.21	First	the	professor	writes	down	a	random	number,	shows
  it	to	another	professor,	who	then	writes	down	her	salary	and	the	first	number	her	note.	This	process	of	subtotaling	continues	until	the	last	professor	shows	their	total	to	the	first	professor,	who	then	subtracts	their	random	number.	The	sum	(and	thus	the	mean)	is	now	known22.	Both	have	the	same	share.	This	is	easy	to	calculate	with	algebra,	but	can
  be	seen	by	imagining	that	the	bottles	don't	contain	liquid,	but	colorful	ping-pong	balls.	Then	remember:	Mass	is	conserved!23	Basic	idea:	Two	slow	people	have	to	move	together	at	some	point.24	See	exercise	6.4.12.25.	Study	the	formulas	in	any	standard	documentation.26	See	the	discussion	of	Euler	paths	in	Section	4.1.27	(	and	)	30	years.	(b)	1	(a
  book	can	have	all	color	illustrations).	c)	5	times	(results	0,1,2,3,4,5	goals).	10	CHAPTER	2	IMPORTANT	DECISIONS	(d)	Answer5.	The	original	pyramid	has	5	sides.	Adding	a	tetrahedron	does	not	add	new	faces!	The	reason:	when	gluing	the	tetrahedron	to	the	pyramid,	the	two	sides	of	the	tetrahedron	are	in	one	plane	with	the	faces	of	the	pyramid!	Of
  the	other	two	faces	of	the	tetrahedron,	one	is	glued	to	the	face	of	the	pyramid	(so	these	two	faces	are	now	inside	the	new	body)	and	the	other	becomes	one	of	the	faces	of	the	new	body.	The	big	question	is	how	do	you	show	that	faces	are	broke?	One	way	is	to	do	some	pretty	heavy	3D	vector	calculations	with	a	lot	of	trigonometry.	It's	a	good	exercise
  and	worth	doing.	An	ingenious	way	to	see	this	at	first	glance	is	to	imagine	two	pyramids	placed	side	by	side	with	their	square	faces	touching	each	other.	It	is	easy	to	see	that	the	gap	between	them	is	exactly	the	size	and	shape	of	a	regular	tetrahedron	(draw	a	picture	or	better	yet	build	some	models!).	So	if	you	put	a	regular	tetrahedron	in	that	slot,	you
  get	a	nice	prism.	In	other	words,	two	sides	of	a	tetrahedron	coincide	with	two	sides	of	a	pyramid.	(e)	All	n.	Just	draw	parallel	lines!	2.2	9	Check	that	f	3(x)	=	x,10	Tn	=	n!+2n.11	f(n)	is	equal	to	the	number	of	ones	in	the	representation	of	base	2	n.12	(binary)	See	Euler's	formula	(	3.4.3	,	3.4.	40).13	If	f(n)	is	the	number	of	regions	for	n	lines,	then	the
  (n+1)th	line	intersects	each	of	the	n	lines,	each	time	creating	a	new	region	plus	one	more	region	(the	initial	direction).	So	f(n+1)	=	f(n)+n+1.	The	result	is	an	algorithm	and	it	is	possible	to	use	the	arithmetic	series	formulas	(see	section	5.3)	to	obtain	the	formula	f(n)	=	(n2	+	n+2)/2.14	Using	the	same	methods	as	in	2.2.13	above,	the	answer	will
  essentially	be	double:	n2n	+2.15	For	example	1/8+1/56	=	pattern	1/7?16	Motivated	by	examples	1/2+	1/3+	1/6	=	1	and	1/2+	1/3+	1/7	+1/42	=	1,	it	is	easy	to	guess	that	the	next	example	is	1	/2+1/3+1/7+1/43+1/(2	3	7	43)	=	1,	now	it	is	easy	to	generate	any	solution:	let	x1	=	2,	x2	=	3adefine	xm	=	x1	x2	xm1	+1	until	xn	is	defined.17	See	2.3.38.18	If	n
  is	a	perfect	square.19	As	long	as	n	60	=	(71)(111)	we	can	always	do	this.20	The	sum	stays	the	same	!	To	prove	this	we	have	to	do	a	rather	tricky	inductive	proof	(see	Section	2.3).21	(a)	n(n+1)/2	b)	n2(n+1)2/4	c)	There	is	no	simple	formula,	but	see	5.3	for	ideas	Example	2	(you	need	a	binomial	theorem).	ANSWERS	TO	2.3	11	22	s(n)	=	2n1.23	Use	the
  fact	that	(u2)2	+(2u)2	+22	=	(u2	+2)2.25	Let's	work	out	the	first	few	product	terms.	We	get	(1+	x3)(1+2x9)(1+3x27)(1+4x81)	=	=	1+	x3	+2x9	+2x12	+3x27	+3x30	+6x36	+6x39	+4x81	+	.	What	are	the	(positive)	exponents	of	ki?	All	integers	of	the	form	3u1	+3u2	+	+3ur	,	where	integers	u	j	1	satisfy	u1	<	u2	<	<	ur.	In	other	words,	they	are	numbers
  that,	when	written	in	base	3,	contain	only	ones	and	zeros	and	end	in	zero.	As	a	result,	the	first	few	exponents	(written	in	base	3)	are	10,100,110,1000,1010,1100,1110,10000,	.	.	.	.Of	course	these	numbers	are	just	a	basic	representation	of	the	sequence	2,4,6,8,2.	.	..	Specifically,	to	find	k1996,	we	just	write	2	1996	=	3992	in	base-2:	3992	=
  2048+1024+512+256+128+16+8,	so	the	base-2	representation	of	3994	is	equal	to	111010011.	111110011000	(3rd	Base).	In	other	words,	k1996	=	33	+34	+37	+38	+39	+310	+311.	26	Solved	in	example	2.4.3.27.	See	[9]	for	a	very	nice	treatment	of	this	and	related	problems.28	g(n)	=	2r	where	r	is	the	number	of	1's	in	the	base-2	representation	of
  n.32.	Don't	let	m	be	a	perfect	square.	We	can	write	m	=	n2+b,	where	1	b	2n.	Since	(n+	1/2)2	=	n2	+	n+	1/4	we	see	that	if	b	n	then	{m}	=	n	and	if	b	>	n	then	{	m	}	=	n+	1.	So	in	the	first	case	we	have	f(m)	=	m+	n	=n2	+	b+	n	<	n2	+	2n+	1.	In	the	second	case	we	have	f(m)	=	m+	n+	1	=n2.	+b+n+1	>	n2	+2n+1.	In	no	case	will	f(m)	be	a	perfect
  square.	Sum	up	things	by	noting	that	if	m	=	n2	then	f(m)	=	n2	+	n,	which	is	not	a	perfect	square.35	a)	No;	assume	that	n	=	41.	See	also	7.5.1.	Example.	b)	Correct	formula	is(n)1+(n2)+(n4).	See	the	link	provided	in	the	problem	for	the	derivation	and	further	discussion.	2.3	21	See	p.	223.22	Assume	that	neither	U	nor	T	is	closed	under	multiplication.
  Then	there	exist	t1,	t2	T	and	u1,u2	U	such	that	t1t2	6	T	and	u1u2	6U	.	But	then	nt1t2	U	and	u1u2	T	,	because	if	something	is	not	a	member	of	T,	it	must	be	a	member	of	U	and	vice	versa	(since	all	these	products	are	in	S,	which	is	closed	under	multiplication).	Now	consider	t1t2u1u2.	It	is	the	three-element	product	T	of	t1,	t2	T	and	u1u2	T	.	So	by
  assumption	t1t2u1u2	is	T.	But	t1t2u1u2	is	also	the	product	of	three	elements	of	U,	namely	t1t2,u1	and	u2,	and	so	by	assumption	it	must	be	a	member	of	U.	But	this	is	a	contradiction:	t1t2u1u2	cannot	belong	to	T	and	U	because	these	two	sets	do	not	intersect.	Thus,	our	original	assumption	that	neither	T	nor	U	is	closed	cannot	be	true.	We	conclude
  that	at	least	one	of	these	sets	is	closed.	23	No,	that	is	not	possible.	WLOG,	circle	3	(odd)	numbers;	name	them	a,	b,	c.	Now	consider	the	relations	a/b,b/c,c/a.	Each	of	these	is	either	a	prime	number	or	the	reciprocal	of	a	prime	number	and	the	product	is	1.	A	product	is	a	fraction	whose	numerator	(before	simplification)	is	the	product	of	prime	numbers
  and	the	denominator	is	the	product	of	prime	numbers.	Since	there	are	only	3	primes	in	the	numerator	and	denominator,	we	have	a	contradiction:	the	numerator	and	denominator	end	up	with	a	different	number	of	primes,	and	there	is	no	way	for	the	two	to	be	equal	(so	the	fraction	is	1).	As	long	as	an	odd	number	of	numbers	are	placed	on	the	ring,	this
  is	not	possible.	However,	with	an	even	number	it	is	easy.32	Since	xn	+1/xn	is	an	integer,	we	show	that	xn+1	+1/xn+1	is	also	an	integer.	Assuming	x+1/x	is	an	integer,	so	the	product	is	(xn	+	xn)(x+	x1)	=	xn+1	+2+	x(n+1).	So	xn+1	+1/xn+1	is	an	integer40.	Key	Idea:	Look	at	this	picture!	2.4	7	MonOn	the	distance-time	plot,	it's	easy	to	see	that	Pat
  walks	55	minutes.8	Draw	a	picture	with	the	points	arranged	in	a	grid.	ANSWERS	TO	QUESTION	2.4	13	9	Sunrise	was	at	4:30	am.	This	is	quite	a	difficult	problem	to	solve	with	algebra;	Instead	we	draw	a	time-distance	plot	and	do	some	geometry.	In	the	graph	below,	which	is	not	to	scale,	the	horizontal	axis	is	time	since	sunrise	and	the	vertical	axis	is
  the	distance	along	the	road	from	point	A	to	point	B.	Points	F,	G,	and	H	are	at	noon,	5:00	p.m.	,	and	23:15,	respectively,	and	segments	AD	and	BH	represent	the	temporal	distances	of	Pat	and	Dana	meeting	at	C.	We	can	determine	the	time	of	sunrise	after	calculating	the	length	of	AF.	B	A	C	F	E	D	G	H	The	triangles	ACH	and	DCB	are	similar,	so	BEED	=
  FHAF	.	Since	ED=	5,	FH	=	11.25	and	BE	=	AF	we	have	AF2	=	4545	and	AF	=	15	2.	10	The	error	should	move	along	two	lines:	first	from	(7.11)	to	O	=	(0,	0)	.	and	then	from	O	to	(17.3).	This	is	because	of	the	following	principle:	the	error	must	bypass	quadrant	II	entirely,	even	if	the	direct	path	from	(7.11)	to	(17.3)	passes	through	quadrant	II.	To	see
  why,	let	a	and	b	be	any	positive	numbers	and	consider	a	path	starting	at	A	=	(0,a)	and	ending	at	B	=	(b,0).	Certainly	the	fastest	route	in	quadrant	II	is	segment	AB	and	the	length	of	this	route	is	a2+b2.	Now	consider	the	alternate	path	AO	followed	by	OB.	This	path	is	outside	of	quadrant	II	(because	quadrant	II	does	not	contain	an	x	​​or	y	axis)	and	has	a
  total	length	of	a	+	b.	Compare	the	two	lengths.	By	the	arithmetic-geometric	inequality	we	have	a2	+	b2	2ab,	which	means	that	2a2	+	2b2	a2	+	2ab	+	b2	=	(a+b)2.	So	a+b	2	a2+b2.	We	conclude	that	as	long	as	the	speed	in	quadrant	II	is	less	than	12,	any	path	from	A	to	B	passing	through	quadrant	II	takes	longer	than	the	shortest	path	outside
  quadrant	II	(along	the	y	and	x	axes)	.	Since	12	<	12,	our	error	saves	time	by	avoiding	quarter	II.11.Between	midnight	and	noon,	the	hands	start	and	end	in	alignment,	and	they	line	up	eleven	more	times:	immediately	after	1,	then	after	2,	etc.	(point	of	view	of	the	ant	living	on	the	clock	hand),	so	the	intervals	occur	every	12/	11	hours.	So	the	answer	is	1
  hour	plus	1/11	hour	or	1:05:27	and	3/11	sec.15.	Repeat	the	grading	task.	You	have	to	characterize	the	trapezoidal	numbers.17	The	answer	is	8.	Notice	that	the	minimized	value	is	simply	the	square	of	the	distance	between	the	two	points	(u,	2u2)	and	(v,	9/v).	This	is	the	distance	between	any	point	on	the	circle	x2	+	y2	=	2	and	the	point	in	the	upper	half
  of	the	hyperbola	y	=	9/x.	Therefore,	we	need	to	find	the	minimum	distance	between	these	two	curves.	It	is	easy	to	see	(draw)	that	this	is	achieved	by	choosing	a	point	(1,1)	on	the	circle	and	a	point	(3,3)	on	the	hyperbola.18	Expand	the	cube;	the	shortest	distance	is	a	straight	line;	Answer:	5.	19	Let	R	be	the	interior	of	a	rectangle	with	vertices	(0,0),
  (b,0),(b,a),(0,a).	The	line	y	=	ax/b	does	not	intersect	the	lattice	nodes	in	R	(it	passes	through	(0,0)	and	(b,a),	but	these	points	are	not	in	R	and	there	are	no	other	lattice	nodes	on	the	line	because	a	and	b	have	no	common	divisors)	.	Note	that	bai/bc	is	exactly	the	number	of	grid	points	below	this	line	in	R	for	x	=	i.	So	the	sum	on	the	left	is	just	the	number
  of	grid	points	under	the	line	in	R.	Similarly,	the	sum	on	the	right	is	equal	to	the	number	of	grid	points	lying	in	across	the	line.	The	total	value	should	be	half	the	total	number	of	grid	points,	which	is	obviously	(a1)(b1).	Note	that	at	least	one	of	a	and	b	must	be	odd,	otherwise	the	two	numbers	will	share	a	common	divisor	(namely	2).	So	(a1)(b1)	is	even
  and	divisible	by	2.20.	See	example	8.2.1.	The	only	difference	is	that	we	now	use	the	y	=	plotx	instead	of	a	hyperbola.	21	Answer:	104	+172.	Each	term	(2k1)2	+	a2k	can	be	interpreted	as	the	length	of	a	vector	with	a	horizontal	offset	(2k	1)	and	a	vertical	offset	ak.	If	we	place	these	vectors	end	to	end,	then	the	sum	n	k=1	(2k1)2	+a2k	will	be	the	total
  length	of	a	potentially	jagged	path	with	a	net	horizontal	offset	of	1+3+(2	101)=102	and	a	net	vertical	offset	of	a1	+a2	+	+a10	=	17.	If	we	change	ai	so	that	the	length	of	the	path	is	as	short	as	possible,	the	path	will	obviously	be	a	straight	line.	The	length	of	this	line,	of	course,	is	only	104	+172.	ANSWERS	FOR	EXERCISE	3.1	15	23	Answer:	1235.
  Change	the	problem	to	one	that	counts	grid	points	in	triangular	and	rectangular	areas;	2.4.19.	the	task	is	somewhat	similar.	Consider	the	general	problem	of	n	balls	m0,m1.	.	.	,mn1	with	arbitrary	starting	positions.	Each	ball	has	a	spirit	path,	the	path	it	would	take	if	it	didn't	bounce	off	its	neighbors,	but	instead	passed	through	them.	Whenever	the
  balls	bounce,	the	path	of	the	real	ball	is	the	same	as	the	path	of	the	other	spirit	ball.	After	a	minute,	each	ghost	path	returned	to	the	original	location	of	each	ball.	Thus,	after	one	minute,	the	actual	position	of	the	balls	is	the	reverse	of	the	original	position.	Also,	this	permutation	must	be	a	cyclic	permutation	because	the	balls	cannot	pass	through	each
  other.	We	say	that	the	permutation	goes	from	m0	to	md,	where	d	is	counterclockwise,	that	is,	the	difference	modulo	n	between	the	number	of	balls	counterclockwise	and	the	number	of	balls	clockwise.	To	see	this,	let	vi(t)	be	a	function	of	the	ball's	speed	mi,	where	+1.1	ve	is	counterclockwise	and	clockwise,	respectively.	Note	that	at	any	time	t	n1i=0
  vi(t)	=	d,	since	the	number	of	balls	rotating	clockwise	and	counterclockwise	never	changes	(even	if	the	balls	collide).	There	will	be	many	bounces	and	any	time	interval	between	bounces	of	any	speedis	a	constant.	Let	t1,	t2,	.	.	.	,	tk	are	the	times	within	each	interval,	and	let	each	interval	be	of	length	`i.	For	each	mile	of	marble,	denote	the	distance
  traveled	counterclockwise	from	t	=	0	to	t	=	1	as	si	=	vi(t1)`1	+	vi(t2)`2	+	vi(tk)`k.	Summing	it	over	all	balls,	we	get	n1i=0	si	=	d(`1	+	`2	+	+	`k)	=	d1	=	d.	The	only	cyclic	permutation	associated	with	this	sum	of	net	distance	traveled	is	the	one	that	takes	m0	in	mcr.	3.1	13	Answer	170.	Let	the	path	go	from	(3.5)	through	(0,a)	to	(b,0)	and	(8.2).	This
  path	has	the	same	length	as	the	path	(draw	a	picture!)	from	(3,5)	to	(0,a)	to	(b,0)	to	(8,2),	and	this	path	is	explicitly	minimized	when	it	is	direct	from	(3.5)	to	(8.2).	15	The	product	of	n	divisors	is	nd(n).	16	CHAPTER	2	CHAPTER	16	SOLUTIONS	The	answer	is	229.	Note	that	the	sum	of	all	elements	of	the	set	S	=	{1,2,3,	.	.	.	.30}	is	1+	2+	3+	+	30	=	30
  31/2	=	465.	Let	A	be	a	subset	of	S,	and	let	Ac	denote	the	complement	of	A	(the	elements	of	S	that	are	notes	of	A.	The	sum	of	the	elements	of	A	plus	the	sum	of	the	elements	of	Ac	must	equal	465.	Since	465	=	232	+	233,	if	the	sum	of	the	elements	of	A	is	greater	than	232,	the	sum	of	the	elements	of	Ac	must	be	less	than	232.	In	other	words,	there	is	a
  one-to-one	correspondence	between	subsets	whose	sum	of	elements	is	greater	than	232	and	subsets	whose	sum	of	elements	is	not	equal	to	(in	particular,	A	and	Ac),	Thus,	the	number	of	subsets	whose	sum	of	elements	is	greater	than	232	is	exactly	half	of	the	total	number	of	subsets	of	S,	and	the	number	of	subsets	of	S	is	230.19	There	will	be	an
  infinite	number	of	rotations.	Consider	the	situation	after	rotating	the	errors	by	1	degree.	Now	it	still	lies	on	four	vertices	of	the	square,	with	exactly	the	same	situation	as	at	the	beginning	in	terms	of	angular	relations.	It	would	be	impossible	to	tell	if	this	were	not	the	beginning	of	the	g	configuration.	So	the	beetles	turn	another	hundred	trunk	and	so	on
  and	so	forth...20	The	first	player	can	always	winfirst	place	the	penny	perpendicular	to	the	center	of	the	table	and	then	mirror	its	opponents	moves	relative	to	the	center21.	Mark	the	points	where	the	ball	bounces,	C,	C1,	C2,	.	.	..	Draw	segment	BD1	such	that	BD1	=	BC	and	CBD1	=	.	Now	we	mirror	the	trajectory	of	the	billiard	ball	in	ABC	around	the
  axis	BC.	Let	E1	be	the	reflection	of	C1.	For	calculation	purposes,	it	does	not	matter	if	we	consider	the	real	path	CC1C2.	.	.	or	a	reflected	path	starting	from	CE1,	so	let's	assume	that	the	billiard	ball	actually	begins	its	path	by	moving	from	C	to	E1.	Redraw	segment	BD2	using	D1BD2	=	and	BD2	=	BD1.	Rebounce	the	new	ball	path	starting	at	E1	and	let
  the	bounce	of	C2	be	E2.	Since	CE1D1	=	C2E1B	=	BE1E2,	the	path	of	CE1E2	must	be	a	straight	line.	Now	we	repeat	the	process	and	construct	the	segments	BD2,	BD3,	.	.	.z	DiBDi+1	=	i	BDi	=	BDi+1	and	mirror	the	path	until	we	get	a	new	path	that	has	a	radius	of	CE1.	Each	reflection	Ci	of	the	original	path	corresponds	to	the	intersection	of	the	ray
  with	BDi.	To	count	the	number	of	intersections,	draw	a	circle	with	center	B	and	radius	AB.	The	path	we	are	interested	in	is	the	segment	CE	(when	the	ray	leaves	the	circle,	there	are	no	more	reflections	on	the	original	path),	where	BCE	=	.	To	count	the	number	of	intersections	with	BDi,	we	need	to	find	CBE.	Since	BE	=	BC,	BEC	=	BC,	so	CBE	=	180	2	.
  So	the	number	of	intersections	is	(1802)	+1.	25	The	answer	is	1600/3.	The	temperature	function	is	given	by	the	relation	T(x,y,z)	=	(x+y)2+(yz)2.	Now	consider	two	similar	functions:	U(x,y,z)=T(y,z,x)	and	V(x,y,z)=T(z,x,y).	In	other	words,	U	and	V	are	obtained	from	T	by	circular	permutation	of	the	variables.	The	average	values	​​of	T,	U,	and	V	collected
  on	the	surface	of	the	sphere	are	the	same	because	of	symmetry.	Let's	call	this	average	value	A.	Then	the	average	value	of	T	+	U	+	V	is	equal	to	3A.	But	T	(x,y,z)+U(x,y,z)+V	(x,y,z)	=(x+	y)2	+(y	z)2	+(y+	z)2	+(z	x)2	+	(z+x)2y)2=	4x2	+4y2	+4z2.	However,	on	the	surface	of	the	planet	it	is	4x2	+4y2	+4z2	=	4,202	=	1600,	a	constant!	Therefore	the	mean
  of	T	+	U	+	V	is	1600	=	3A.26	Let	f	(x)	:=	1/(1+	tgx)	2	and	let	g(x)	:=	1/(1+	cotx)	2.	Since	cotx	=	tan(	pi/2	x),	the	graphs	of	the	two	functions	from	x	=	0	to	x	=	pi/2	are	simply	mirror	images	(reflected	at	x	=	pi/4).	Therefore	pi/20	f	(x)dx	=	pi/2	0g(x)dx.	On	the	other	hand,	cotx	=	1/tanx	and	simple	algebra	gives	f	(x)	+	g(x)	=	1.	So	pi/2	0f	(x)dx	=	1	2	pi/20
  [	f	(x)+g	(	x)	]dx	=	pi/4.	30a	See	Frederick	Mosteller's	excellent	book	Fifty	Hard	Problems	in	Probability	with	Solutions	[19].	It	is	only	88	pages	long,	but	contains	many	wonderful	insights	into	probability,	a	subject	unfortunately	ignored	in	TAA-COPS.30b	(Jim	Propp).	Above.	For	1k	52,	let	Ak	denote	the	number	of	possibilities	where	the	kth	card
  remains	the	kth	from	the	top.	For	k	=	26,	Ak	is	the	number	of	options	where	all	cards	1	through	k	are	selected,	so	Ak	=	(52k26k).	By	symmetry,	Ak	=	A53k	for	27k	52,	so	52k=1	Ak	=	226k=1	(52k26k	).	The	second	sum	is	(52	27)	(see	stick	identity,	example	6.2.5	on	page	218).	Multiplying	by	2	and	dividing	by	(5226)	gives	52/27.30c.	Consider	a
  sequence	of	n	distinct	integers.	The	ith	number	(i=	2,3,...,n)	is	swapped	if	and	only	if	it	was	the	smallest	of	the	first	i	elements.	The	probability	of	this	is	1/i.	So	the	average	number	of	swaps	is	only	12	+	13	+	+	1n.	18	CHAPTER	2	KEY	DECISIONS	3.2	7	Assume	that	not	all	values	​​are	equal.	Let	a	>	0	be	the	smallest	value	on	the	board.	There	must	be	an
  east-adjacent	square	(WLOG)	of	a	square	with	a	value	of	b	strictly	greater	than	a.	But	then	a	is	equal	to	the	average	of	4	numbers	not	less	than	a,	one	of	which	is	strictly	greater	than	a.	This	is	a	contradiction.10	coins	withdiameter	cannot	touch	more	than	5	others12.	Of	course,	if	n	is	even,	this	is	not	true:	imagine	a	group	of	pairs	of	people	standing	a
  few	inches	apart,	each	pair	quite	far	from	all	other	pairs.	Now,	if	n	is	odd,	first	eliminate	all	pairs,	as	in	the	case	above	where	two	people	shoot	each	other.	Since	n	is	odd,	some	dry	people	remain.	Now	consider	the	person	whose	nearest	neighbor	is	maximal	(there	may	be	connections).	This	man	will	stay	dry	because	the	only	way	he	could	be	shot	is	if
  someone	else	was	as	close	to	him	as	he	is	to	his	next	door	neighbor.	But	this	is	contradicted	by	the	fact	that	distances	are	different	for	each	person.	We	will	show	that	no	palindrome	can	exist	on	the	basis	of	a	contradiction.	Suppose	the	concatenation	of	numbers	from	1	to	n	is	a	palindrome	P	:=	1234567891011	4321.	Consider	the	longest	consecutive
  zero	sequence	P;	note	that	it	exists	because	it	must	be	greater	than	10.	There	can	be	multiple	consecutive	zeros	of	the	same	length;	select	the	last	one	(on	the	right).	Note	that	there	is	one	digit	to	the	left	of	this	line,	and	that	digit	plus	zeros	makes	up	one	of	the	numbers	from	1	to	n.	To	be	more	specific,	suppose	the	longest	string	of	zeros	is	0000.
  Then	such	a	string	on	the	right	consists	of	the	last	digits	of	one	of	the	numbers	from	1	to	n,	and	not	the	1	in	the	middle,	and	is	not	covered	(for	example,	400005,	then	to	the	left	of	it	the	number	400000	will	appear,	which	contradicts	the	fact	that	0000	Similarly,	a	number	ending	in	0000	would	have	to	start	with	one	digit,	because	if,	say,	the	number
  were	7310000,	then	the	number	7000000	would	be	to	the	left	of	it.	So,	let's	say	that	the	rightmost	line	0000	is	the	last	digit	of	the	number	70000.	Then	if	we	write	down	the	numbers	of	the	predecessor	and	successor,	these	four	zeros	will	be	inserted	into	the	line	699997000070001.	one	more	lineQ.	Since	P	is	a	palindrome,	the	first	0000s	must	be
  embedded	in	the	string	100070000799996.	But	that	doesn't	make	sense,	since	the	first	occurrence	of	0000	is	the	last	digit	of	10000.	So	the	only	possibility	is	that	there	is	only	one	string.	0000,	which	definitely	finds,	is	exactly	in	the	middle	of	P	and	is	the	last	four	digits	of	10000.	By	writing	an	ancestor	and	a	descendant	and	letting	|	Mark	exactly	the
  center	of	P,	in	the	middle	should	be	the	following	line:	9999100|0010001	But	it	is	not	symmetrical	(9	6	=	0),	so	we	come	to	a	contradiction.	ANSWERS	TO	3.3	19	15	Consider	the	shortest	path	from	1	to	n2,	where	the	path	passes	through	neighboring	cells.	In	the	worst	case,	the	length	of	the	path	is	n	(if	1	and	n2	are	in	opposite	corners	of	the
  diagonal).	In	any	case,	the	path	elements	will	have	no	more	than	n	different	numbers	from	1	to	n2	inclusive.	If	their	successive	differences	were	less	than	or	equal	to	n,	then	there	is	1	successive	difference	that	fills	the	gap	from	1	to	n2.	Since	n21	=(n1)(n+1),	the	largest	difference	must	be	at	least	n+1.16	See	4.1.18.	Subsection	This	is	Eisenstein's
  irreducibility	criterion,	which	is	discussed	and	proved	in	most	modern	texts	on	algebra,	such	as	[12].	3.3	11	Suppose	there	are	n	people.	Possible	dates	are	from	0	to	n	1	inclusive.	If	no	one	knows	zero	people,	we	can	pack	n	1	numbers	from	1	to	n	1.	If	at	least	one	person	knows	zero	people,	then	the	maximum	number	of	friends	is	n	2,	so	the	box	will
  work	again.	area	into	50	rectangles	of	size	0.02.	Pigeonhole	places	three	points	in	one	of	these	rectangles.	(Actually,	it's	not	entirely	trivial	to	say	that	the	area	of	​​a	triangle	inside	a	rectangle	is	at	most	half	the	area	of	​​the	rectangle.	Try	to	find	a	convincing	argument!)	18	The	following	numbers	are	coprime.	19	Let	there	be	n	people.	Each	person	sits	at
  a	distance	d	from	him	in	the	right	place,	where	0	<	d	<	nmeasured	counterclockwise.	There	are	n	people	but	1	different	values	​​of	d.	Therefore,	at	least	two	people	are	equidistant.20	Main	idea:	squares	are	products	of	numbers	raised	to	even	powers,	and	even	powers	are	obtained	by	dividing	two	numbers	by	exponents	of	the	same	parity.	To
  understand	the	idea,	let's	solve	this	problem	for	a	specific	case:	n	=	3.	Let's	take	n	different	numbers	a,	b,	c	and	let	the	sequence	we're	studying	equal	to	u1,	u2,	.	.	.	,uN	,	where	N	23	=	8.	Now	we	define	successive	multiplications	pi	:=	u1u2	ui	for	i	=	1,2,	.	.	.	,N.	Each	of	these	N	numbers	can	be	written	as	akb`cm,	where	k,	`,m	are	nonnegative
  integers.	We	are	only	interested	in	the	parity	of	the	exponents:	there	are	23	different	possibilities	for	the	parity	of	the	elements	of	the	ordered	triple	(k,	`,	m).	For	example,	one	possibility	(even,	even,	odd),	another	(even,	odd,	even).	Now	if	one	of	the	pis	has	a	shape	(straight,	straight,	straight)	we're	done	because	it's	going	to	be	a	perfect	square.
  Otherwise	we	use	at	most	231	different	parities	N	for	23	different	pi.	Therefore,	two	of	them,	like	pi	and	pj,	have	the	same	parity	values	​​for	their	exponents.	Suppose	both	numbers	are	(odd,	odd,	even).	Then	the	quotient	(dividing	the	largest	by	the	smallest!)	is	a	perfect	square	because	when	we	divide	powers	we	subtract	powers,	so	all	three	powers
  become	even!	This	method	generalizes	unambiguously;	the	tricky	part	is	the	notation,	which	also	obscures	the	ideas.21	Define	f	(x,	y)	=	x	+	y2	and	let	S	=	{	f	(a,	b)|a,	b,	integers	with	0	a,	b	m}.	irrationally,	S	has	(m+1)2	distinct	elements,	the	largest	of	which	is	m(1+2).	We	divide	the	interval	[0,m(1	+	2)]	into	(1	+	2)/(m	+	2)	subintervals	of	length	m2
  +	2m.	According	to	the	sorting	principle,	there	are	two	different	f(a1,b1)	>	f(a2,b2)	in	a	subinterval.	Without	loss	of	generality,	f	(a1,b1)	>	f	(a2,b2)	>	0.	It	is	easy	to	verify	that	a	=	a1a2	and	b	=	b1b2	fulfill	all	requirements22.(n+1)-end	sequence	7,77,777,.	.	..	When	dividing	by	n,	there	are	at	most	n	distinct	residues,	so	that	after	classification	at	least
  two	members	of	the	sequence	have	the	same	residue.	Their	difference	is	therefore	a	multiple	of	n	and	contains	only	the	numbers	7	and	0.24.	Let	td	be	the	total	number	of	games	played	at	the	end	of	the	d-th	day.	Members	of	sequences	t1,	t2.	.	.	,	t56	are	different	and	vary	from	1	to	88	inclusive.	From	the	same	reasoning	as	in	3.3.22,	division	by	23
  gives	23	possible	residues,	so	according	to	the	classification	there	are	at	least	d56/23e=3	members	of	the	sequence	with	the	same	residue,	e.g.	ta,	tb,	tc,	where	a	1.	But	the	condition	of	the	problem	gives	(	an,	au)	=	(n,	u)	=	(n,	rn)	=	n.	But	an	=	u	and	au	is	a	multiple	of	u,	so	(an,au)	=	u	>	n,	a	contradiction.	7.2	6	All	perfect	squares	correspond	to	0	or
  +1	modulo	3;	this	follows	by	simply	raising	only	three	possibilities	mod	3:	02	0.12	1.22	1.	So	a2	and	b2	cannot	both	be	1	mod	3,	because	that	would	make	c2	equal	to	2,	which	is	impossible.	The	only	possibility	is	that	one	or	both	of	a	is	0,	which	makes	ab	a	multiple	of	3.7.	One	dice	game	modulo	7	turns	out	to	be	0.1.1.9.	Again	we	use	mod	7.10	N	=
  381654729.15.	.	Then	ar	=	2r	p	+	2r	1.	According	to	Fermat's	theorem,	2p1	1	(mod	p),	so	ap1	will	be	a	multiple	of	p.	=	f	(1)	f	(1),	which	means	that	f	(1)	is	either	0	or	1.	The	first	possibility	is	ruled	out	because	the	range	of	f	is	N.6	Let	d|n	where	n=	ab	with	ab	.	According	to	FTA	we	can	write	d	=with	u|and	v|b.	The	difficulty	is	in	showing	that	this	is	a
  one-on-one	match.	It	is	enough	to	show	that	if	u,	u	and	v,	v	are	divisors	of	a	and	the	bands	uv	=	uv,	respectively,	then	u	=	u,	v	=	v.	This	is	done	by	following	the	last	steps	of	the	FTA	proof	for	p)	=	14	(if	n	=	2rs,	then	(	n)	=	(2r)(s),	which	is	a	multiple	of	4	because	(2r)	=	2r1	and	(s	)	is	even).	If	n	is	odd	then	its	PPF	cannot	contain	more	than	one	prime
  because	if	we	write	n	=	ab	where	a,b	>	1	and	a	b	we	get	(n)	=	(a)(b)	which	is	the	sum	of	the	multiplication	of	two	even	numbers.	.	The	only	possibility	is	that	n	=	pr	for	some	odd	prime	number	p.	But	then	(n)	=	pr1(p	1)	=	14,	which	has	no	solution	(since	p	=	3	and	p	=	7	are	the	only	possibilities)22.	F	is	a	function	of	the	perfect	square	index,	i.e.	F(n)	=
  1	if	n	is	a	perfect	square	and	F(n)	=	0	otherwise24.	Of	course,	if	n	is	prime	then	(n)+(n)	=	(n	1)+(n	+	1)	=	2n.	To	prove	the	opposite,	we	will	show	that	if	n	is	composite,	then	(n)+(n)>2n.	Suppose	n	is	not	prime	but	has	only	one	prime	factor	p.	Then	(n)	=	nn/p	and	(n)	n+n/p+1,	so	(n)+(n)>2n.	Then	let	n	be	two	different	prime	factors,	p	and	q.	Then	we
  have	(n)	=	n	np	n	q+	n	pq	and	(n)	n+	n	p+	n	q+	n	pq	and	again	(n)+(n)	>	2n.	The	scheme	is	now	clear:	if	n	is	not	prime,	(n)	will	be	a	sum	of	variables	starting	from	n,	and	(n)	contains	at	least	all	(n)	terms,	but	all	positive	ones,	so	when	(n)	is	added	to	(n	),	some	terms	will	be	reduced,	but	not	all,	leaving	2n	plus	some	extra	elements26.	(Sketch)	We
  argue	by	contradiction.	Suppose	there	is	a	strictly	increasing	multiplicative	function	f(n)	where	f(n)	>	n	starts	at	some	n	=	n0.	Our	strategy	is	to	obtain	some	upper	bounds	on	f(n)	and	then	disprove	them.	We	use	the	fact	that	f	(2m)	=	2	f	(m)	for	odd	m.	For	any	k	we	have	f	(2k)<	f	(2k	+2)=	2	f	(2k1	+1)<	2	f	(2k1	+2)	52	CHAPTER	2SOLUTION	=	22	f
  (2d2	+1).	.	.	<	2k1fa	(3).	So	f(2k)/2k	<	f(3),	which	is	a	constant.	Since	k	is	arbitrary,	this	means	that	the	function	grows	essentially	linearly;	it	can't	get	out	of	control	too	violently	when	climbing.	We	obtain	a	contradiction	by	showing	that	we	can	find	n	such	that	f(n)/n	can	be	arbitrarily	large.	But	it's	simple:	let	p1	be	the	first	prime	for	which	f	(p1)	>
  p1.	Then	we	have	f(p1)p1+1	and	indeed	f(k)k+1	for	all	k	p1.	Since	there	are	infinitely	many	primes,	we	can	assume	that	n	=	p1	p2	pr,	where	pi	are	consecutive	primes	and	r	is	arbitrarily	large.	Then	f	(n)n	(	p1	+1p1	)(p2	+1	p2	)	(	pr	+1pr	)	and	this	expression	is	differentiated	as	r	(see	Example	9.4.7)	to	get	the	desired	contradiction.27	This	can	be
  done	in	many	ways	to	achieve	One	method	is	to	write	999	=	(101)	99	and	use	the	binomial	theorem	with	most	terms	being	multiples	of	100.	Another	method	is	to	use	Euler's	extension	of	Fermat's	little	theorem	(Task	7.3.25).	Since	(100)	=	40,	we	have	980	1	(mod	100),	so	the	problem	boils	down	to	finding	919	modulo	100.	Divide	100	into	relatively
  prime	factors	of	4	25.	It	is	easy	to	see	that	9r	1	(mod	4)	for	any	r,	and	the	expansion	Eulers	is	again	920	1	(mod	25).	So	920	1	(mod	100).	So	for	919	x	(mod	100)	we	have	9x	1	(mod	100),	which	corresponds	to	the	linear	Diophantine	equation	9x100y	=	1.	This	is	easily	solved	with	7.1.13.	methods	that	give	x	=	89.	The	second	method	is	obviously	more
  difficult,	but	more	informative	and	probably	more	useful	(we	were	lucky	that	9	=	101).	the	clarity	of	Vanden	Eydens'	approach	to	Mobia.	version	formula	[25].	This	is	the	perfect	resource	for	beginners.	On	a	more	sophisticated	level,	Wilfa	2.6	generating	functionality.	contains	a	fascinating	discussion	that	is	also	a	good	addition	to	TACOPS	9.4.	Section.
  7.4	6	The	first	equation	can	be	solved	by	factoringThis	is	equivalent	to	5n	=	m2	2500	=	(m+	50)(m	50)	and	the	only	solution	is	n	=	5	(since	there	is	only	one	pair	of	5	powers	that	differ	by	100.	The	second	equation	has	a	similar	solution;	n	=	8.	ANSWERS	TO	7.4	53	7	Assume	without	loss	of	generality	a	b	c	Note	that	we	must	have	2	(1+1/c)	3	If	c	=	1
  then	(1+1/a)(1+	1/b)	=	1,	which	is	obviously	impossible,	c	=	2	leads	to	(1+1/a)(1+1/b)	=	4	/3,	which	means	4/3	(1+1/b)2,	which	gives	b	<	7.	Since	(1+1/a)	>	1,	we	must	also	have	b	>	3.	Adding	values	​​gives	solutions	of	(	7,6,2),	(	9,5,2),(15	,4,2)	c	=	3	gives	(1+	1/a)(1+	1/b)	=	3/2	A	similar	analysis	gives	b	2	,	then	2+	3c	<	4c,	which	means	ab	<	4.	So
  there	are	few	cases:	1.	c	2:	this	only	leads	to	the	solution	(2,2,2)	2.	c	>	2	and	ab	=	1	:	this	implies	1	c	=	1+1	+	c+2,	impossible	3.	c	>	2	and	ab	=	2:	this	leads	to	(1,2,5)	4.	c	>	2	and	ab	=	3:	this	leads	to	(1	,3,3).	They	are	only	ro	constraints	(for	permuth	ations).12	Hint:	Show	that	(x+1)3	<	x3+8x26x+8<	(x+3)3.	The	only	solutions	are	(0,9),(9,11).13
  (sketch)	There	are	no	other	solutions	than	x	=	y	=	1.	When	y	2,	in	the	analysis	modulo	9,	the	forces	x	are	equal.	Then	we	can	write	x	=	2m	and	use	the	factorization	tool	to	factor	3y	=	72m4	=	(7m	+2)(7m2).	54	CHAPTER	2	AGREEMENTS	CHAPTER	2.	15	See	3.4.31.22.	Note	that	8	and	9	do	the	math.	Inspired	by	this,	we	just	need	to	find	solutions	for
  8x2	+1	=	9y2.	Replacing	u	=	2x,	v	=	3y	converts	this	to	v22u2	=	1,	which	was	solved	earlier	in	7.4.19.	7.5	13	Hint:	Let	g	=	gcd(m,n)	and	write	m	=	ag,n	=	bg,	where	a	b.	Then	the	equation	becomes	g(abab+1)	=	g(a1)(b1)	=	0.14	(sketch	by	Kiran	Kedlaya)	No	geometric	sequence	of	integers	from	1	to	100	can	be	longer	than	7	terms.	To	see	this,	let	m/n
  be	a	common	ratio	where	m	is	n	and	look	at	prime	factors	and	inequalities.	Thus,	12	progressions,	each	with	a	maximum	of	7	words,	can	only	cover	84	or	fewer	numbers	from	1	toAlternative	solution:	Use	the	fact	that	there	are	25	prime	numbers	between	1	and	100,	in	which	case	modulo	4	analysis	leads	to	a	contradiction.	Therefore,	all	three	are
  equal.	We	can	write	x	=	2a,y	=	2b,z	=	2c,	where	a,b,c	Z.	Substituting	and	simplifying,	we	get	a	new	equation	a2	+	b2	+	c2	=	4abc	and	a	similar	mod	4	analysis	power	a,b	,c	to	be	one.	Writing	them	as	a	=	2u,b	=	2v,c	=	2w	gives	u2	+	v2	+	w2	=	8uvw.	This	process	can	be	repeated	indefinitely.	This	means	that	x,y,z	can	be	divided	by	2	any	number	of
  times	and	the	result	is	always	an	integer.	The	only	numbers	that	satisfy	this	are	x	=	y	=	z	=	0.	This	solution	method	is	called	infinite	descent	and	was	discovered	by	Fermat.	An	excellent	discussion	can	be	found	in	[21].19	There	are	many	interesting	approaches	to	this	problem.	One	idea	is	to	use	the	formula	developed	in	Solution	7.1.22	for	the	largest
  power	of	two	that	divides	n!	and	use	it	to	calculate	the	highest	power	of	2	that	divides	(no).	Another	idea	is	to	use	(strong)	induction	by	writing	Pascal's	triangle	modulo2.	A	third,	more	complicated	approach	is	to	generate	functional	considerations	of	(1+	x)n	modulo	2	using	the	fact	that	(1+	x)2	1+	x2	(mod	2).21	If	p	6	=	2,	then	d(8p	)	=	8	If	p	=	2,	then
  d(6p)	=	6.	If	p	6	=	3,	then	d(9p2)	=	9.	If	p	=	3,	then	d(12p2)	=	12.24	See	7.3.23.	task.	ANSWERS	TO	7.5	55	25	Yes.	Just	pick	relatively	prime	squares	from	1997	like	22,32,52,72,112,	.	.	..	The	Chinese	remainder	theorem	(7.2.16)	guarantees	that	x	can	be	found	such	that	x	0	(mod	22),	x	1	(mod	32),	x	2	(mod	52),	.	.	.	Then	(x)	=	(x+1)	=	(x+2)	=	=	0.27
  (draft)	If	m	is	odd,	then	(2m)	=	(2)	(m)	=	(m).	Thus,	the	inverted	image	sets	consist	of	pairs.	And	notice	that	(2m)	=	(m).28	Yes,	there	are	infinitely	many	such	lines.	For	example,	(20368)=	2(20367)	and	(20385)=2(20383).	There	are	infinitely	many	rows	of	twoElements	in	a	1:2	ratio,	if	2(	n	k	)=(	n	k+1	),	reduce	to	2(k+1)	=	n	k	or	n	=	3k+2.	So,	as	long
  as	n	is	2	(mod	3),	the	ratio	of	two	neighboring	elements	is	1:2.	Next,	we	look	for	nonadjacent	duplicates.	The	next	simplest	case	is	2(n	k	)=(n	k+2	),	which	reduces	to	2(k+2)(k+1)	=	(n	k)(n	k1	).	Substitute	u	=	n	k	and	v	=	k	+	2;	our	equation	becomes	2(v2	v)	=	u2u.	Multiplying	both	sides	by	4	and	completing	the	square	gives	2(4v24v+1)	=
  4u24u+1+1,	so	substituting	x	=	2v1,y	=	2u1	reduces	the	original	equation	to	2x2	y2	=	1.	This	is	the	Pell	equation	with	infinitely	many	solutions	that	can	be	obtain	in	the	standard	way	(see	7.4.177.4.20),	or	just	note	that	(1,1),	(5,7)	aresol	aresol