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The smallest number, which when divided by 5, 10, 12 and 15, leaves remainder 2 in each case; but when divided by 7 leaves no remainder, is 1). 1892). 1823). 1754). 91 |
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Answer» Given, NUMBER is DIVISIBLE by 7 and when divided by 5, 10, 12 and 15 LEAVES remainder 2 in each case ∴ Smallest number which is divisible by 5, 10, 12 and 15 = LCM of 5, 10, 12 and 15 Finding LCM 5 = 1× 5 10 = 2 × 5 12 = 2 × 2 × 3 15 = 3 × 5 ∴ LCM = 2 × 2 × 3 × 5 = 60 ∴ The number which when divided by 5, 10, 12 and 15 leaves remainder 2 in each case is (60n + 2) where, n is an integer. Now, we have to find the smallest value of n for which (60n + 2) is divisible by 7. For, n = 1, 60n + 2 = 62, not divisible by 7 For, n = 2, 60n + 2 = 122, not divisible by 7 For, n = 3, 60n + 2 = 182, which is divisible by 7 ∴ Required number is 182 |
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