Question
Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468.
Answer: Option A
:
A
Let the required number is x.
⇒(x+17) is the smallest number divisible by 520 and 468.
⇒(x+17)=LCM(520,468)
Prime factorisation of 520 and 468:
520=23×5×13
468=22×32×13
⇒LCM=23×32×5×13=4680
Thus, x+17=4680.
⇒x=4663
Therefore, the required number is 4663.
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:
A
Let the required number is x.
⇒(x+17) is the smallest number divisible by 520 and 468.
⇒(x+17)=LCM(520,468)
Prime factorisation of 520 and 468:
520=23×5×13
468=22×32×13
⇒LCM=23×32×5×13=4680
Thus, x+17=4680.
⇒x=4663
Therefore, the required number is 4663.
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