Find the least number which when divided by 35,56 and 91 leaves the same remainders 7 in each case
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The least number which can be divided by 35,56 and 91 to leave no remainder is the l.c.m of these numbers.
Hence, to leave a remainder of 7 in each case, the number should be 'l.c.m' + 7
l.c.m(35,56,91)
35 = 5*7
56 = 2^3*7
91 = 7*13
l.c.m = 2^3 * 5 * 7 * 13 = 3640
The required number = 3640 + 7 = 3647
Hence, the least number which when divided by 35,56 and 91 to leave a remainder of 7 in each case is '3647'
Hope it helps.
Hence, to leave a remainder of 7 in each case, the number should be 'l.c.m' + 7
l.c.m(35,56,91)
35 = 5*7
56 = 2^3*7
91 = 7*13
l.c.m = 2^3 * 5 * 7 * 13 = 3640
The required number = 3640 + 7 = 3647
Hence, the least number which when divided by 35,56 and 91 to leave a remainder of 7 in each case is '3647'
Hope it helps.
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