Find 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.

TO FIND: Smallest number which when increased by 17 is exactly divisible by both 520 and 468.

L.C.M OF 520 and 468

\[520 = 2^3 \times 5 \times 13\]

`468 = 2^2xx3^2xx13`

\[\text{LCM of 520 and 468} = 2^3 \times 3^2 \times 5 \times 13\]

\[ = 4680\]

Hence 4680 is the least number which exactly divides 520 and 468 i.e. we will get a remainder of 0 in this case. But we need 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.

Concept: Euclid’s Division Lemma

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