Find the smallest number which when increased by 17 is exactly divisib

1.	Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468.
Answer» First let’s find the smallest number which is exactly divisible by both 520 and 468. That is simply just the LCM of the two numbers. By prime factorisation, we get 520 = 2³ × 5 × 13 468 = 2² × 3² × 13 ∴ LCM (520, 468) = 2³ × 3² × 5 × 13 = 4680 Hence, 4680 is the smallest number which is exactly divisible by both 520 and 468 i.e. we will get a remainder of 0 in each case. But, we need to find the smallest number which when increased by 17 is exactly divided by 520 and 468. So that is found by, 4680 – 17 = 4663 ∴ 4663 should be the smallest number which when increased by 17 is exactly divisible by both 520 and 468.

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

Answer»

First let’s find the smallest number which is exactly divisible by both 520 and 468.

That is simply just the LCM of the two numbers.

By prime factorisation, we get

520 = 2³ × 5 × 13

468 = 2² × 3² × 13

∴ LCM (520, 468) = 2³ × 3² × 5 × 13 = 4680

Hence, 4680 is the smallest number which is exactly divisible by both 520 and 468 i.e. we will get a remainder of 0 in each case. But, we need to find the smallest number which when increased by 17 is exactly divided by 520 and 468.

So that is found by, 4680 – 17 = 4663

∴ 4663 should be the smallest number which when increased by 17 is exactly divisible by both 520 and 468.

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