Find the least number that, when divided by 44 and 66, leaves the remainder 6 in each case.
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Answer:
The smallest number which when divided by 35, 56 and 91 = LCM(35,56,91)
35=5×7
56=2
3
×7
91=7×13
LCM=2
3
×5×7×13=3640
The smallest number that when divided by 35, 56, 91 leaves a remainder 7 in each case = 3640 + 7 = 3647.
Answered by
1
To find the least number that when divided by 44 and 66 leaves a remainder 6 ,we must first find the L.C.M. of 44 and 66 and then add 6 to it.
We need the least multiples of 44 and 66.
44= 2x2x11
66= 2x3x11
L.C.M of 44 and 66= 11x2x2x3
L.C.M (44 and 66) = 132
The required number = 132+6
= 138
Hence 138 is the smallest number, with a remainder of 6 in each case when divided by 44 and 66.
We need the least multiples of 44 and 66.
44= 2x2x11
66= 2x3x11
L.C.M of 44 and 66= 11x2x2x3
L.C.M (44 and 66) = 132
The required number = 132+6
= 138
Hence 138 is the smallest number, with a remainder of 6 in each case when divided by 44 and 66.
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