What would be the smallest natural number which when divided either by 20 or by 42 or by 76 leaves a remainder of 7 in each case?
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let x,y,z be the three integers
let us assume 20x+7,42y+7,76z+7
just find the LCM of 20(2×2×5),42(2×3×7),76(2×2×19)
2×2×5×3×7×19=7980
so the number is 7980+7=7987
hope it helps u...
let us assume 20x+7,42y+7,76z+7
just find the LCM of 20(2×2×5),42(2×3×7),76(2×2×19)
2×2×5×3×7×19=7980
so the number is 7980+7=7987
hope it helps u...
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