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## 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

### Key Concepts

Lcm

Algebra

Integers

AIME I, 1998, Question 1

Elementary Number Theory by Sierpinsky

## Try with Hints

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

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

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

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

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

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

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

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

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

$\Rightarrow$ number of values of k=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 = $13 + 7=20$