2.
find the smallest number which on
added 23 to it is exactly divisible
by 32 36.45 and 96
being by vertical method
Answers
First We have to find the L.C.M of 32, 36, 48, 96 & then Subtract 23 from the L C.M & then the remaining number will be your required number.
L.C.M of 32,36,48 & 96 is 288 ( L.C.M us in the attachment)
Subtract 23 from the L.C.M
288-23= 265
_____________________________
The required smallest number is 265.
_____________________________
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.