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

### Key Concepts

Integers

Number Theory

Algebra

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

## Arithmetic Sequence Problem | AIME I, 2012 | Question 2

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2012 based on Arithmetic Sequence.

## Arithmetic Sequence Problem – AIME 2012

The terms of an arithmetic sequence add to $715$. The first term of the sequence is increased by $1$, the second term is increased by $3$, the third term is increased by $5$, and in general, the $k$th term is increased by the $k$th odd positive integer. The terms of the new sequence add to $836$. Find the sum of the first, last, and middle terms of the original sequence.

• is 107
• is 195
• is 840
• cannot be determined from the given information

### Key Concepts

Series

Number Theory

Algebra

AIME, 2012, Question 2.

Elementary Number Theory by David Burton .

## Try with Hints

After the adding of the odd numbers, the total of the sequence increases by $836 – 715 = 121 = 11^2$.

Since the sum of the first $n$ positive odd numbers is $n^2$, there must be $11$ terms in the sequence, so the mean of the sequence is $\frac{715}{11} = 65$.

Since the first, last, and middle terms are centered around the mean, then $65 \times 3 = 195$

Hence option B correct.

Categories

## Arithmetic Mean | AIME I, 2015 | Question 12

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 based on Arithmetic Mean.

## Arithmetic Mean of Number Theory – AIME 2015

Consider all 1000-element subsets of the set {1, 2, 3, … , 2015}. From each such subset choose the least element. The arithmetic mean of all of these least elements is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p + q$.

• is 107
• is 431
• is 840
• cannot be determined from the given information

### Key Concepts

Inequalities

Algebra

Number Theory

AIME, 2015, Question 12

Elementary Number Theory by David Burton

## Try with Hints

Each 1000-element subset ${ a_1, a_2,a_3,…,a_{1000}}$ of ${1,2,3,…,2015}$ with $a_1<a_2<a_3<…<a_{1000}$ contributes $a_1$ to sum of least element of each subset and set ${a_1+1,a_2+1,a_3+1,…,a_{1000}+1}$. $a_1$ ways to choose a positive integer $k$ such that $k<a_1+1<a_2+1,a_3+1<…<a_{1000}+1$ ($k$ can be anything from $1$ to $a_1$ inclusive

Thus, the number of ways to choose the set ${k,a_1+1,a_2+1,a_3+1,…,a_{1000}+1}$ is equal to the sum. But choosing a set ${k,a_1+1,a_2+1,a_3+1,…,a_{1000}+1}$ is same as choosing a 1001-element subset from ${1,2,3,…,2016}$!

average =$\frac{2016}{1001}$=$\frac{288}{143}$. Then $p+q=288+143={431}$

Categories

## Logarithms and Equations | AIME I, 2000 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination, AIME I, 2000 based on Logarithms and Equations.

## Logarithms and Equations – AIME I 2000

$log_{10}(2000xy)-log_{10}xlog_{10}y=4$ and $log_{10}(2yz)-(log_{10}y)(log_{10}z)=1$ and $log_{10}(zx)-(log_{10}z)(log_{10}x)=0$ has two solutions $(x_{1},y_{1},z_{1}) and (x_{2},y_{2},z_{2})$ find $y_{1}+y_{2}$.

• is 905
• is 25
• is 840
• cannot be determined from the given information

### Key Concepts

Logarithms

Theory of Equations

Number Theory

AIME I, 2000, Question 9

Polynomials by Barbeau

## Try with Hints

Rearranging equations we get $-logxlogy+logx+logy-1=3-log2000$ and $-logylogz+logy+logz-1=-log2$ and $-logxlogz+logx+logz-1=-1$

taking p, q, r as logx, logy and logz, $(p-1)(q-1)=log2$ and $(q-1)(r-1)=log2$ and $(p-1)(r-1)=1$ which is first system of equations and multiplying the first three equations of the first system gives $(p-1)^{2}(q-1)^{2}(r-1)^{2}=(log 2)^{2}$ gives $(p-1)(q-1)(r-1)=+-(log2)$ which is second equation

from both equations (q-1)=+-(log2) gives (logy)+-(log2)=1 gives $y_{1}=20$,$y_{2}=5$ then $y_{1}+y_{2}=25$.

Categories

## Ordered Pairs | PRMO-2019 | Problem 18

Try this beautiful problem from PRMO, 2019, Problem 18 based on Ordered Pairs.

## Orderd Pairs | PRMO | Problem-18

How many ordered pairs $(a, b)$ of positive integers with $a < b$ and $100 \leq a$, $b \leq 1000$ satisfy $gcd (a, b) : lcm (a, b) = 1 : 495$ ?

• $20$
• $91$
• $13$
• $23$

### Key Concepts

Number theory

Orderd Pair

LCM

Answer:$20$

PRMO-2019, Problem 18

Pre College Mathematics

## Try with Hints

At first we assume that $a = xp$
$b = xq$
where $p$ & $q$ are co-prime

Therefore ,

$\frac{gcd(a,b)}{LCM(a ,b)} =\frac{495}{1}$

$\Rightarrow pq=495$
Can you now finish the problem ……….

Therefore we can say that

$pq = 5 \times 9 \times 11$
$p < q$

when $5 < 99$ (for $x = 20, a = 100, b = 1980 > 100$),No solution
when $9 < 55$ $(x = 12$ to $x = 18)$,7 solution
when,$11 < 45$ $(x = 10$ to $x = 22)$,13 solution
Can you finish the problem……..

Therefore Total solutions = $13 + 7=20$

Categories

## Largest Possible Value | PRMO-2019 | Problem 17

Try this beautiful problem from PRMO, 2019 based on Largest Possible Value.

## Largest Possible Value | PRMO | Problem-17

Let a, b, c be distinct positive integers such that $b + c – a$,$c + a – b$ and $a + b – c$ are all perfect squares.
What is the largest possible value of $a + b + c$ smaller than $100$?

• $20$
• $91$
• $13$

### Key Concepts

Number theory

Perfect square

Integer

Answer:$91$

PRMO-2019, Problem 17

Pre College Mathematics

## Try with Hints

Let $b + c – a = x^2$ … (i)
$c + a – b = y^2$ … (ii)
$a + b – c = z^2$ … (iii)

Now since $a$,$b$, $c$ are distinct positive integers,
Therefore, $x$, $y$, $z$ will also be positive integers,
$a + b + c = x^2 + y^2 + z^2$
Now, we need to find largest value of $a + b + c or x^2 + y^2 + z^2$ less than $100$
Now, to get a, b, c all integers $x$,$y$, $z$ all must be of same parity, i.e. either all three are even or all three
are odd.

Can you now finish the problem ……….

Let us maximize$x^2 + y^2 + z^2$, for both cases.
If $x$, $y$, $z$are all even.
Therefore,

$b + c – a = 8^2 = 64$
$c + a – b = 42 = 16$
$a + b – c = 22 = 4$
Which on solving, give$a = 10$,$b = 34$, $c = 40$ and $a + b + c = 84$
If x, y, z are all odd
$\Rightarrow b + c – a = 92 = 81$
$c + a – b = 32 = 9$
$a + b – c = 12 = 1$
Which on solving, give $a = 5$ ,$b = 41$, $c = 45$ and$a + b + c = 91$

Can you finish the problem……..

Therefore Maximum value of $a + b + c < 100 = 91$

Categories

## Good numbers Problem | PRMO-2019 | Problem 12

Try this beautiful problem from PRMO, 2019 based on Good numbers.

## Good numbers Problem | PRMO | Problem-12

A natural number $k >$ is called good if there exist natural numbers
$a_1 < a_2 < ………. < a_k$

$\frac{1}{\sqrt a_1} +\frac{1}{\sqrt a_2}+………………. +\frac{1}{\sqrt a_k}=1$

Let $f(n)$ be the sum of the first $n$ good numbers, $n \geq 1$. Find the sum of all values of $n$ for which
$f(n + 5)/f(n)$ is an integer.

• $20$
• $18$
• $13$

### Key Concepts

Number theory

Good number

Integer

Answer:$18$

PRMO-2019, Problem 12

Pre College Mathematics

## Try with Hints

A number n is called a good number if It is a square free number.

Let $a_1 ={A_1}^2$,$a_2={A_2}^2$,………………$a_k={A_k}^2$
we have to check if it is possible for distinct natural number $A_1, A_2………….A_k$ to satisfy,
$\frac{1}{A_1}+\frac{1}{A_2}+………..+\frac{1}{A_k}=1$

Can you now finish the problem ……….

For $k = 2$; it is obvious that there do not exist distinct$A_1, A_2$, such that $\frac{1}{A_1}+\frac{1}{A_2}=1 \Rightarrow 2$ is not a good number

For $k = 3$; we have $\frac{1}{2} +\frac{1}{3}+\frac{1}{6}=1 \Rightarrow 3$ is a good number.

$\frac{1}{2}+\frac{1}{2}\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$ $\Rightarrow 4$ is a good number

Let $k$ wil be a good numbers for all $k \geq 3$

$f(n) = 3 + 4 +… n$ terms =$\frac{n(n + 5)}{2}$
$f(n + 5) =\frac{(n + 5)(n +10)}{2}$

$\frac{f(n+5}{f(n)}=\frac{n+10}{n}=1+\frac{10}{n}$

Can you finish the problem……..

Therefore the integer for n = $1$, $2$, $5$ and $10$. so sum=$1 + 2 + 5 + 10 = 18$.

Categories

## Integer Problem | AMC 10A, 2020 | Problem 17

Try this beautiful problem from Number theory based on Integer.

## Integer Problem – AMC-10A, 2020- Problem 17

Define $P(x)=(x-1^2)(x-2^2)……(x-{100}^2)$

How many integers $n$ are there such that $P(n) \geq 0$?

• $4900$
• $4950$
• $5000$
• $5050$
• $5100$

### Key Concepts

Number system

Probability

divisibility

Answer: $5100$

AMC-10A (2020) Problem 17

Pre College Mathematics

## Try with Hints

Given $P(x)=(x-1^2)(x-2^2)……(x-{100}^2)$. at first we notice that $P(x)$ is a product of of $100$ terms…..now clearly $P(x)$ will be negetive ,for there to be an odd number of negetive factors ,n must be lie between an odd number squared and even number squared.

can you finish the problem……..

$P(x)$ is nonpositive when $x$ is between $100^2$ and $99^2$, $98^2$ and $97^2 \ldots$ , $2^2$ and $1^2$

can you finish the problem……..

Therefore, $(100+99)(100-99)+((98+97)(98-97)+1)$+….

.+$((2+1)(2-1)+1)$=$(200+196+192+…..+4)$ =$4(1+2+…..+50)=4 \frac{50 \times 51}{2}=5100$

Categories

## Divisibility Problem from AMC 10A, 2003 | Problem 25

Try this beautiful problem from Number theory based on divisibility from AMC 10A, 2003.

## Number theory in Divisibility – AMC-10A, 2003- Problem 25

Let $n$ be a $5$-digit number, and let $q$ and $r$ be the quotient and the remainder, respectively, when $n$ is divided by $100$. For how many values of $n$ is $q+r$ divisible by $11$?

• $8180$
• $8181$
• $8182$
• $9190$
• $9000$

### Key Concepts

Number system

Probability

divisibility

Answer: $8181$

AMC-10A (2003) Problem 25

Pre College Mathematics

## Try with Hints

Since $11$ divides $q+r$ so may say that $11$ divides $100 q+r$. Since $n$ is a $5$ digit number …soTherefore, $q$ can be any integer from $100$ to $999$ inclusive, and $r$ can be any integer from $0$ to $99$ inclusive.

can you finish the problem……..

Since $n$ is a five digit number then and $11 | 100q+r$ then $n$ must start from $10010$ and count up to $99990$

can you finish the problem……..

Therefore, the number of possible values of $n$ such that $900 \times 9 +81 \times 1=8181$

Categories

## Theory of Equations | AIME I, 2015 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2015 based on Theory of Equations.

## Theory of Equations – AIME I, 2015

The expressions A=$1\times2+3\times4+5\times6+…+37\times38+39$and B=$1+2\times3+4\times5+…+36\times37+38\times39$ are obtained by writing multiplication and addition operators in an alternating pattern between successive integers.Find the positive difference between integers A and B.

• is 722
• is 250
• is 840
• cannot be determined from the given information

### Key Concepts

Series

Equations

Number Theory

AIME I, 2015, Question 1

Elementary Number Theory by Sierpinsky

## Try with Hints

A = $(1\times2)+(3\times4)$

$+(5\times6)+…+(35\times36)+(37\times38)+39$

B=$1+(2\times3)+(4\times5)$

$+(6\times7)+…+(36\times37)+(38\times39)$

B-A=$-38+(2\times2)+(2\times4)$

$+(2\times6)+…+(2\times36)+(2\times38)$

=722.