Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468. 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. Concept: Euclid’s Division Lemma Is there an error in this question or solution? Uh-Oh! That’s all you get for now. We would love to personalise your learning journey. Sign Up to explore more. Sign Up or Login Skip for now Uh-Oh! That’s all you get for now. We would love to personalise your learning journey. Sign Up to explore more. Sign Up or Login Skip for now |