Find the smallest number which when increase by 17 excatly divisible b

1.	Find the smallest number which when increase by 17 excatly divisible by both 520 and 468
Answer» The smallest number divisible by 520 and 468 = LCM(520,468)Prime factors of 520 and 468 are :520 = 23{tex}\\times{/tex}\xa05 {tex}\\times{/tex}\xa013468 = 2 {tex}\\times{/tex}\xa02 {tex}\\times{/tex}\xa03 {tex}\\times{/tex}\xa03 {tex}\\times{/tex}\xa013Hence LCM(520,468) = 23{tex}\\times{/tex}\xa032 {tex}\\times{/tex}\xa05 {tex}\\times{/tex}\xa013 = 8\xa0{tex}\\times{/tex}\xa09{tex}\\times{/tex}5{tex}\\times{/tex}13 = 4680Now the smallest number which when increased by 17 is exactly divisible by both 520 and 468.=LCM(520,468)-17=4680-17=4663

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