# how+much=power then p+o+w+e+r=?

asked Jul 19, 2014 in Infosys

How is a 3 digit number and Much is a 4 digit number and we got a 5 digit number. So left most digit is 1. (999 + 9999 = 19998).
If you add two single digit numbers we get a maximum sum of 18.  i.e., maximum carry over is 1. 1 + M = 10. So M = 9. and O = 0.
Now the remaining digits available are 2, 3, 4, 5, 6, 7, 8.
As O = 0, C + 1 = E, i.e., C and E are consecutive intezers. and E - C = 1
Also, W + H = 10 + R ------(1), and H + U = 10 + W or H - W = 10 - U ------(2)
Solving 1 and 2, we get 2H = 20 - (U - R)
Now maximum value of U - R is 9. So 2H minimum value is 12, 14 etc as 2H is even.
Say U - R = 8. only possible pairs are (9, 1), (8,0).  Both are not possible as As M = 9 and O = 0.
Say U - R = 6. Possible pairs are (9, 3), (8, 2), (7, 1), (6, 0). Only (8,2) possible.
6 0  W
9 8 C 7
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105 E 2
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As C and E are consequtive intezers, possible values for them are 3, 4. and W should take 5.
6 0  5
9 8 3  7
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10 5 4  2
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