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Question
What is the smallest number that when divided by 35, 56 and 91 leaves remainders of 7 in each case?
What is the smallest number that when divided by 35, 56 and 91 leaves remainders of 7 in each case?
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Solution
The smallest number which when divided by 35, 56 and 91 = LCM of 35, 56 and 91
35 = 5 x 7
56 = 2 x 2 x 2 x 7
91 = 7 x 13
LCM = 7 x 5 x 2 x 2 x 2 x 13 = 3640
The smallest number that when divided by 35, 56, 91 leaves a remainder 7 in each case = 3640 + 7 = 3647.
35 = 5 x 7
56 = 2 x 2 x 2 x 7
91 = 7 x 13
LCM = 7 x 5 x 2 x 2 x 2 x 13 = 3640
The smallest number that when divided by 35, 56, 91 leaves a remainder 7 in each case = 3640 + 7 = 3647.
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