Find the smallest no. which when increased by 17is exactly divisible by 520 and468.

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 = 23 x5 x 13

468 = 2 x 2 x 3 x 3 x 13

LCM of 520 and 468 = 23 x 32 x 5 x 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 = 46631

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

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