Find the smallest number which when increased by 17 is exactly di… - Vidyadip

Question

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

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Answer

given that the smallest member which when increased by $17$ is (exactly) by both 520 and 468.
Prime factors of 520 and 468 are
$520 = 2^3 \times 5 \times 13$
$468 = 2^2 \times 3^2 \times 13$
$L.C.M (520, 468) = 2^3 \times 3^2 \times 5 \times 13$
$= 8 \times 9 \times 65$
$= 4680$
Number is increased by $17$ so, smallest number will be
$= L.C.M (520, 468) - 17$
$= 4680 - 17$
$= 4663$
Thus, the required number is $4663.$

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