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Question
Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468.
Solution
TO FIND: Smallest number which when increased by 17 is exactly divisible by both 520 and 468.
L.C.M OF 520 and 468
`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.
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