Question

find the smalleat no which when increased by 17 is exactly divisible by both 468 and 520

Answer 1

Here's the solution:

》To obtain the smallest number which when increased by 17 is exactly divisible by both 468 and 520, we need to subtract 17 with the l.c.m of 468 and 520.

=> Prime factors of 468 - 2×2×3×3×13.
=> Prime factors of 520- 2×2×2×5×13.

(Now make the pair of same term and multiply it with the left numbers):

We'll get,

LCM of 468 & 520= 2 × 2 × 2 × 3 × 3 × 5 × 13
=> 4680.

Now, the smallest number which when increased by 17 is exactly divisible by both 468 and 520 is:

》4680 - 17
》4663.

Answer 2

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