Question

Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468.

Answer 1

SOLUTION :  

GIVEN : Two numbers 520 & 468.

First , we have to find the LCM (520, 468) of given two numbers by Prime factorisation  and then subtract 17 from the LCM to find the smallest number .

The Prime factors of 520 & 468 are :  

520 = 2³  x 5 x 13

468 = 2 x 2 x 3 x 3 x 13 = 2² × 3² × 13

LCM (520 ,468)  = 2³  x 3² x 5 x 13  

LCM (520 ,468) = 4680

4680 is the least number which exactly divides 520 and 468  .But we want the Smallest number which when increased by 17 is exactly divided by 520 and 468.

Therefore,  4680 - 17 = 4663

Hence, 4663 is Smallest number which when increased by 17 is exactly divisible by both 520 and 468.

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

$hello \: frnd$

The Least Common Multiple of 520 and 468 are ;-

$520 = 2 \times 2 \times 2 \times 5 \times 13$

$468 = 2 \times 2 \times 3 \times 3 \times 13$

$lcm = 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 13$

Which is equal to = 4680

So The least Number Divisible by both 520 and 468 is 4680

Let the number increased by 17 gives the number divisible by 520 and 468 be x .

$x + 17 = 4680$

$x = 4663$

4663 is ur answer ✌️