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Find the smallest number which when increased by 17 is exactly divisible by both 468 and 520. |
Answer» The required number is determine with the help of LCM of 468 and 520. Using prime factorization, find the LCM: 468 = 22 × 32 × 13 520 = 23 × 5 × 13 LCM = 4680 The required number = LCM(468 and 520) – 17 = 4680 – 17 = 4663 The required number is 4663. |
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