find the smallest number which when 5 is subtracted from it ,is Exactly divisible by 12 16 24and 36
Answers
We have to find the smallest number which, on being added 23 to it, is exactly divisible by 32, 36, 48 and 96.
For it first we have to find the L.C.M. of 32, 36, 48 and 96 and then subtract 23 from it. After subtracting 23, the remaining number will be the required number.
L.C.M. of 32, 36, 48 and 96
_________________
2 | 32, 36, 48, 96
|________________
2 | 16, 18, 24, 48
|________________
2 | 8, 9, 12, 24
|________________
2 | 4, 9, 6, 12
|________________
2 | 2, 9, 3, 6
|________________
3 | 1, 9, 3, 3
|________________
3 | 1, 3, 1, 1
|________________
| 1, 1, 1, 1
|
L.C.M. = 2*2*2*2*2*3*3 = 288
Now, subtracting 23 from it -
288 - 23 = 265
265 is the required smallest number which, on being added 23 to it, is exactly divisible by 32, 36, 48 and 96.
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