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ask to be
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189 because it dedicated to be
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182
Step-by-step explanation:
Given 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
First let us find LCM of 5, 10, 12 and 15. Factors of 5,10,12 and 15 are 5 x 2 x 3 x 2 = 60. Now required number is of the form 60 k + 2. We need to find the least value of k where it is exactly divisible by 7.
So 60 x 1 + 2 = 62 which is not divisible by 7
60 x 2 + 2 = 122 is not divisible by 7
60 x 3 + 2 = 182 is exactly divisible by 7
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