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InterviewSolution
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Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468 |
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Answer» Last line isSmallest number which when increased by 17 is exactly divisible by both 520 and 468 = 4680 - 17 = 4663. The given numbers are 520 and 468.The smallest number which when increased by 17 is exactly divisible by both 520 and 468 is obtained by subtracting 17 from the LCM of 520 and 468.Prime factorisation of 520 = 2 × 2 × 2 × 5 × 13Prime factorisation of 468 = 2 × 2 × 3 × 3 × 13LCM of 520 and 468 = 2 × 2 × 2 × 3 × 3 × 5 × 13 = 4680.Smallest number which when increased by 17 is exactly divisible by both 520 and 468 = 4680 × 17 = 4663. |
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