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## 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

## GCD and Ordered pair | AIME I, 1995 | Question 8

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 based on GCD and Ordered pair.

## GCD and Ordered pair – AIME I, 1995

Find number of ordered pairs of positive integers (x,y) with $y \lt x \leq 100$ are both $\frac{x}{y}$ and $\frac{x+1}{y+1}$ integers.

• is 107
• is 85
• is 840
• cannot be determined from the given information

### Key Concepts

Integers

GCD

Ordered pair

AIME I, 1995, Question 8

Elementary Number Theory by David Burton

## Try with Hints

here y|x and (y+1)|(x+1) $\Rightarrow gcd(y,x)=y, gcd(y+1,x+1)=y+1$

$\Rightarrow gcd(y,x-y)=y, gcd(y+1,x-y)=y+1$

$\Rightarrow y,y+1|(x-y) and gcd (y,y+1)=1$

$\Rightarrow y(y+1)|(x-y)$

here number of multiples of y(y+1) from 0 to 100-y $(x \leq 100)$ are

[$\frac{100-y}{y(y+1)}$]

$\Rightarrow \displaystyle\sum_{y=1}^{99}[\frac{100-y}{y(y+1)}$]=49+16+8+4+3+2+1+1+1=85.

Categories

## GCD and Sequence | AIME I, 1985 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1985 based on GCD and Sequence.

## GCD and Sequence – AIME I, 1985

The numbers in the sequence 101, 104,109,116,…..are of the form $a_n=100+n^{2}$ where n=1,2,3,——-, for each n, let $d_n$ be the greatest common divisor of $a_n$ and $a_{n+1}$, find the maximum value of $d_n$ as n ranges through the positive integers.

• is 107
• is 401
• is 840
• cannot be determined from the given information

### Key Concepts

GCD

Sequence

Integers

AIME I, 1985, Question 13

Elementary Number Theory by David Burton

## Try with Hints

$a_n=100+n^{2}$ $a_{n+1}=100+(n+1)^{2}=100 + n^{2} +2n +1$ and $a_{n+1}-a_{n}=2n +1$

$d_{n}|(2n+1)$ and $d_{n}|(100 +n^{2})$ then $d_{n}|[(100+n^{2})-100(2n+1)]$ then $d_{n}|(n^{2}-200n)$

here $n^{2} -200n=0$ then n=200 then $d_{n}$=2n+1=2(200)+1=401.

Categories

## Problem based on LCM | AMC 8, 2016 | Problem 20

Try this beautiful problem from Algebra based on LCM from AMC-8, 2016.

## Problem based on LCM – AMC 8, 2016

The least common multiple of $a$ and $b$ is $12$, and the least common multiple of $b$ and $c$ is $15$. What is the least possible value of the least common multiple of $a$ and $c$?

• $26$
• $20$
• $28$

### Key Concepts

Algebra

Divisor

multiplication

AMC-8, 2016 problem 20

Challenges and Thrills of Pre College Mathematics

## Try with Hints

We have to find out the least common multiple of $a$ and $c$.if you know the value of $a$ and $c$ then you can easily find out the required LCM. Can you find out the value of $a$ and $c$?

Can you now finish the problem ……….

Given that the least common multiple of $a$ and $b$ is $12$, and the least common multiple of $b$ and $c$ is $15$ .then b must divide 12 and 15. There is only one possibility that b=3 which divide 12 and 15. therefore $a$=$\frac{12}{3}=4$

can you finish the problem……..

so$b$=3. Given that LCM of $b$ and $c$ is 15. Therefore c=5

Now lcm of $a$ and $c$ that is lcm of 4 and 5=20