Question

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

Answer 1

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.