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Find the least number which when divided by 35, 56 and 91 leaves the same remainder 7 in each case. |
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Answer» The smallest number which is divided by 35, 56 and 91 is LCM (35, 56, 91). Write 35, 56 and 91 in the prime factorization. 35 = 5 × 7 56 = 2 × 2 × 2 × 7 = 23 × 7 91 = 7 × 13. ∴ LCM (35, 56, 91) = 5× 7 × 23 × 13 = 35 × 8 × 13 = 3640. Hence, the smallest number which is divided by 35, 56 and 91 is 3640. ∴ the smallest number that when divided by 35, 56 and 91 leaves the same remainder 7 in each case = 3640 + 7 = 3647. |
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