Question
Thu June 28, 2012 By: Medha
 

find the number which is exactly divisible by 280 and 1245 leaving the remainder 4 and 3

Expert Reply
Thu June 28, 2012
When         number = 280

and                  remainder = 4

then              new number = 280 - 4 = 276

when                   number = 1245

and                  remainder = 3

then      new number = 1245 - 3 = 1242

Now, we find the H.C.F. of 276 are 1242

Clearly 1242 > 276

 Using Euclid's division algorithm

      1242 = 276 x 4 + 138

       276 = 138 x 2 + 0

At this stage remainder comes out zero

 H.C.F. = 138

Hence, the number which divide 280 and 1245 and leaves remainder 4 and 3 respectively is 138.

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