Categories

## Sequence and permutations | AIME II, 2015 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME II, 2015 based on Sequence and permutations.

## Sequence and permutations – AIME II, 2015

Call a permutation $a_1,a_2,….,a_n$ of the integers 1,2,…,n quasi increasing if $a_k \leq a_{k+1} +2$ for each $1 \leq k \leq n-1$, find the number of quasi increasing permutations of the integers 1,2,….,7.

• is 107
• is 486
• is 840
• cannot be determined from the given information

Sequence

Permutations

Integers

## Check the Answer

Answer: is 486.

AIME II, 2015, Question 10

Elementary Number Theory by David Burton

## Try with Hints

First hint

While inserting n into a string with n-1 integers, integer n has 3 spots where it can be placed before n-1, before n-2, and at the end

Second Hint

Number of permutations with n elements is three times the number of permutations with n-1 elements

or, number of permutations for n elements=3 $\times$ number of permutations of (n-1) elements

or, number of permutations for n elements=$3^{2}$ number of permutations of (n-2) elements

……

or, number of permutations for n elements=$3^{n-2}$ number of permutations of {n-(n-2)} elements

or, number of permutations for n elements=2 $\times$ $3^{n-2}$

forming recurrence relation as the number of permutations =2 $\times$ $3^{n-2}$

for n=3 all six permutations taken and go up 18, 54, 162, 486

Final Step

for n=7, here $2 \times 3^{5} =486.$

as

sds

Categories

## Numbers of positive integers | AIME I, 2012 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2012 based on Numbers of positive integers.

## Numbers of positive integers – AIME 2012

Find the number of positive integers with three not necessarily distinct digits, $abc$, with $a \neq 0$ and $c \neq 0$ such that both $abc$ and $cba$ are multiples of $4$.

• is 107
• is 40
• is 840
• cannot be determined from the given information

Integers

Number Theory

Algebra

## Check the Answer

Answer: is 40.

AIME, 2012, Question 1.

Elementary Number Theory by David Burton .

## Try with Hints

Here a number divisible by 4 if a units with tens place digit is divisible by 4

Then case 1 for 10b+a and for 10b+c gives 0(mod4) with a pair of a and c for every b

[ since abc and cba divisible by 4 only when the last two digits is divisible by 4 that is 10b+c and 10b+a is divisible by 4]

and case II 2(mod4) with a pair of a and c for every b

Then combining both cases we get for every b gives a pair of a s and a pair of c s

So for 10 b’s with 2 a’s and 2 c’s for every b gives $10 \times 2 \times 2$

Then number of ways $10 \times 2 \times 2$ = 40 ways.

Categories

## Algebraic Equation | AIME I, 2000 Question 7

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2000 based on Algebraic Equation.

## Algebraic Equation – AIME 2000

Suppose that x,y and z are three positive numbers that satisfy the equation xyz=1, $x+\frac{1}{z}=5$ and $y+\frac{1}{x}=29$ then $z+\frac{1}{y}$=$\frac{m}{n}$ where m and n are relatively prime, find m+n

• is 107
• is 5
• is 840
• cannot be determined from the given information

Algebra

Equations

Integers

## Check the Answer

Answer: is 5.

AIME, 2000, Question 7

Elementary Algebra by Hall and Knight

## Try with Hints

here $x+\frac{1}{z}=5$ then1=z(5-x)=xyz putting xyz=1 gives 5-x=xy and $y=(29-\frac{1}{x}$) together gives 5-x=x$(29-\frac{1}{x}$) then x=$\frac{1}{5}$

then y=29-5=24 and z=$\frac{1}{5-x}$=$\frac{5}{24}$

$z+\frac{1}{y}$=$\frac{1}{4}$ then 1+4=5.

.

Categories

## Sequence and fraction | AIME I, 2000 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and fraction.

## Sequence and fraction – AIME I, 2000

A sequence of numbers $x_1,x_2,….,x_{100}$ has the property that, for every integer k between 1 and 100, inclusive, the number $x_k$ is k less than the sum of the other 99 numbers, given that $x_{50}=\frac{m}{n}$, where m and n are relatively prime positive integers, find m+n.

• is 107
• is 173
• is 840
• cannot be determined from the given information

Equation

Algebra

Integers

## Check the Answer

Answer: is 173.

AIME I, 2000, Question 10

Elementary Number Theory by Sierpinsky

## Try with Hints

First hint

Let S be the sum of the sequence $x_k$

given that $x_k=S-x_k-k$ for any k

taking k=1,2,….,100 and adding

$100S-2(x_1+x_2+….+x_{100})=1+2+….+100$

Second Hint

$\Rightarrow 100S-2S=\frac{100 \times 101}{2}=5050$

$\Rightarrow S=\frac{2525}{49}$

for $k=50, 2x_{50}=\frac{2525}{49}-50=\frac{75}{49}$

Final Step

$\Rightarrow x_{50}=\frac{75}{98}$

$\Rightarrow m+n$=75+98

=173.

Categories

## Finding smallest positive Integer | AIME I, 1996 Problem 10

Try this beautiful problem from the American Invitational Mathematics Examination, AIME I, 1996 based on Finding the smallest positive Integer.

## Finding smallest positive Integer – AIME I, 1996

Find the smallest positive integer solution to $tan19x=\frac{cos96+sin96}{cos96-sin96}$.

• is 107
• is 159
• is 840
• cannot be determined from the given information

Functions

Trigonometry

Integers

## Check the Answer

Answer: is 159.

AIME I, 1996, Question 10

Plane Trigonometry by Loney

## Try with Hints

First hint

$\frac{cos96+sin96}{cos96-sin96}$

=$\frac{sin(90+96)+sin96}{sin(90+96)-sin96}$

=$\frac{sin186+sin96}{sin186-sin96}$

=$\frac{sin(141+45)+sin(141-45)}{sin(141+45)-sin(141-45)}$

=$\frac{2sin141cos45}{2cos141sin45}$

=tan141

Second Hint

here $tan(180+\theta)$=$tan\theta$

$\Rightarrow 19x=141+180n$ for some integer n is first equation

Final Step

multiplying equation with 19 gives

$x \equiv 141\times 19 \equiv 2679 \equiv 159(mod180)$ [since 2679 divided by 180 gives remainder 159]

$\Rightarrow x=159$.

Categories

## Amplitude and Complex numbers | AIME I, 1996 Question 11

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1996 based on Amplitude and Complex numbers.

## Amplitude and Complex numbers – AIME 1996

Let P be the product of the roots of $z^{6}+z^{4}+z^{2}+1=0$ that have a positive imaginary part and suppose that P=r(costheta+isintheta) where $0 \lt r$ and $0 \leq \theta \lt 360$ find $\theta$

• is 107
• is 276
• is 840
• cannot be determined from the given information

Equations

Complex Numbers

Integers

## Check the Answer

Answer: is 276.

AIME, 1996, Question 11

Complex Numbers from A to Z by Titu Andreescue

## Try with Hints

First hint

here$z^{6}+z^{4}+z^{2}+1$=$z^{6}-z+z^{4}+z^{2}+z+1$=$z(z^{5}-1)+\frac{(z^{5}-1)}{(z-1)}$=$\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}$ then $\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}$=0

Second Hint

gives $z^{5}=1 for z\neq 1$ gives $z=cis 72,144,216,288$ and $z^{2}-z+1=0 for z \neq 1$ gives z=$\frac{1+-(-3)^\frac{1}{2}}{2}$=$cis60,300$ where cis$\theta$=cos$\theta$+isin$\theta$

Final Step

taking $0 \lt theta \lt 180$ for positive imaginary roots gives cis72,60,144 and then P=cis(72+60+144)=cis276 that is theta=276.

.

Categories

## Triangle and integers | AIME I, 1995 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 based on Triangle and integers.

## Triangle and integers – AIME I, 1995

Triangle ABC is isosceles, with AB=AC and altitude AM=11, suppose that there is a point D on AM with AD=10 and $\angle BDC$=3$\angle BAC$. then the perimeter of $\Delta ABC$ may be written in the form $a+\sqrt{b}$ where a and b are integers, find a+b.

• is 107
• is 616
• is 840
• cannot be determined from the given information

Integers

Triangle

Trigonometry

## Check the Answer

Answer: is 616.

AIME I, 1995, Question 9

Plane Trigonometry by Loney

## Try with Hints

First hint

Let x= $\angle CAM$

$\Rightarrow \angle CDM =3x$

$\Rightarrow \frac{tan3x}{tanx}=\frac{\frac{CM}{1}}{\frac{CM}{11}}$=11 [by trigonometry ratio property in right angled triangle]

Second Hint

$\Rightarrow \frac{3tanx-tan^{3}x}{1-3tan^{2}x}=11tanx$

solving we get, tanx=$\frac{1}{2}$

$\Rightarrow CM=\frac{11}{2}$

Final Step

$\Rightarrow 2(AC+CM)$ where $AC=\frac{11\sqrt {5}}{2}$ by Pythagoras formula

=$\sqrt{605}+11$ then a+b=605+11=616.

Categories

## Sequence and greatest integer | AIME I, 2000 | Question 11

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and greatest integer.

## Sequence and greatest integer – AIME I, 2000

Let S be the sum of all numbers of the form $\frac{a}{b}$,where a and b are relatively prime positive divisors of 1000, find greatest integer that does not exceed $\frac{S}{10}$.

• is 107
• is 248
• is 840
• cannot be determined from the given information

Equation

Algebra

Integers

## Check the Answer

Answer: is 248.

AIME I, 2000, Question 11

Elementary Number Theory by Sierpinsky

## Try with Hints

First hint

We have 1000=(2)(2)(2)(5)(5)(5) and $\frac{a}{b}=2^{x}5^{y} where -3 \leq x,y \leq 3$

Second Hint

sum of all numbers of form $\frac{a}{b}$ such that a and b are relatively prime positive divisors of 1000

=$(2^{-3}+2^{-2}+2^{-1}+2^{0}+2^{1}+2^{2}+2^{3})(5^{-3}+5^{-2}+5^{-1}+5^{0}+5^{1}+5^{2}+5^{3})$

Final Step

$\Rightarrow S= \frac{(2^{-3})(2^{7}-1)}{2-1} \times$ $\frac{(5^{-3})(5^{7}-1)}{5-1}$

=2480 + $\frac{437}{1000}$

$\Rightarrow [\frac{s}{10}]$=248.

Categories

## LCM and Integers | AIME I, 1998 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1998 based on LCM and Integers.

## Lcm and Integer – AIME I, 1998

Find the number of values of k in $12^{12}$ the lcm of the positive integers $6^{6}$, $8^{8}$ and k.

• is 107
• is 25
• is 840
• cannot be determined from the given information

Lcm

Algebra

Integers

## Check the Answer

Answer: is 25.

AIME I, 1998, Question 1

Elementary Number Theory by Sierpinsky

## Try with Hints

First hint

here $k=2^{a}3^{b}$ for integers a and b

$6^{6}=2^{6}3^{6}$

$8^{8}=2^{24}$

$12^{12}=2^{24}3^{12}$

Second Hint

lcm$(6^{6},8^{8})$=$2^{24}3^{6}$

$12^{12}=2^{24}3^{12}$=lcm of $(6^{6},8^{6})$ and k

=$(2^{24}3^{6},2^{a}3^{b})$

=$2^{max(24,a)}3^{max(6,b)}$

Final Step

$\Rightarrow b=12, 0 \leq a \leq 24$

$\Rightarrow$ number of values of k=25.

Categories

## Integers | AIME I, 1993 Problem | Question 4

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1993 based on Integers.

## Integer – AIME I, 1993

Find the number of four topics of integers (a,b,c,d) with 0<a<b<c<d<500 satisfy a+d=b+c and bc-ad=93.

• is 107
• is 870
• is 840
• cannot be determined from the given information

Integers

Digits

Algebra

## Check the Answer

Answer: is 870.

AIME I, 1993, Question 4

Elementary Algebra by Hall and Knight

## Try with Hints

First hint

Let k=a+d=b+c

or, d=k-a, b=k-c,

or, (k-c)c-a(k-a)=k(c-a)-(c-a)(c+a)

=(a-c)(a+c-k)

=(c-a)(d-c)=93

Second Hint

(c-a)(d-c)=(1,93),(3,31),(31,3),(93,1)

solving for c

(a,b,c,d)=(c-93,c-92,c,c+1),(c-31,c-28,c,c+3),(c-1,c+92,c,c+93),(c-3,c+28,c,c+31)

Final Step

taking first two solutions a<b<c<d<500

or,$1 \leq c-93, c+1 \leq 499$

or, $94 \leq c \leq 498$ gives 405 solutions

and $1 \leq c-31, c+3 \leq 499$

or, $32 \leq c \leq 496$ gives 465 solutions

or, 405+465=870 solutions.