30. A two-digit number is obtained by either multiplying the sum of their digits by 8 and adding 1 o
Asked by poojathally | 26th Feb, 2010, 04:07: PM
Let the x and y be the ten's and unit's digit respectively,
The number is 10x+y,
8(x+y) + 1 = 10x+y
13(x-y) + 2 = 10x+y
Equating the LHS of these equations since they have the same RHS.
8(x+y) + 1 = 13(x-y) + 2
21y = 5x + 1
x,y being positive integers,
Since RHS has to be at least 21, so that we can have y = 1.
x = 4, ie. 5x4 + 1 = 21
so 41 can be the number,
Also as we can see the RHS must be a multiple of 21,
and x can't be more than 9, since it's a two digit number.
5x9 + 1 = 46 which is 21x2 = 42 and 21x3 = 63.
Hence only one number is possible and it's 41.
Answered by | 26th Feb, 2010, 04:47: PM
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