Question

Find the smallest number which, on being added 23 to it, is exactly divisible by 32, 36, 48 and 96.

Answer 1

Solution :-

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.

Answer.

Answer 2

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
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The required smallest number is 265.
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