Find the least number which when divided by 35, 56 and 91 leaves the same remainder 7 in each case. I need the answer with best explanation. Who don't know answer keep away.
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The smallest number which when divided by 35, 56 and 91 = LCM of 35, 56 and 91
35 = 5 × 7
56 = 2 × 2 × 2 × 7
91 = 7 × 13
LCM = 7 × 5 × 2 × 2 × 2 × 13 = 3640
The smallest number that when divided by 35, 56, 91 leaves a remainder 7 in each case
= 3640 + 7
= 3647
Answered by
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Answer:
3647
The smallest number that when divided by 35, 56, 91 leaves a remainder 7 in each case = 3640 + 7 = 3647.
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