Math
posted by Rohan .
The sum of the digits of a three digit number is 20. The middle digit is equal to one fourth the sum of the other two. If the order of degree is reversed the number increases by 198. Find the original number

The maximum sum of two digits in the number (first and last) can be 18. Thus, the middle digit cannot be larger than 4 (since that would be larger than 1/4th the largest sum).
Since reversing the number increases it, the third digit is larger than the first. The first digit is two less than the third digit, since that condition is necessary for their difference to end with an 8.
The number is 749. 
algebra solution:
let the unit digit be z
let the tens digit be y
let the hundred digit be x
x+y+z = 20
y = (1/4)(x+z)
4y = x+z
original number is 100x + 10y + z
reversed number is 100z + 10y + x
difference of those two is 198
100x + 10y + z  100z  10y  x = 198
99x  99z = 198
x  z = 2
z = x2 , so y = (x + x2)/4
in x+y+z = 20
x + (2x2)/4 + x2 = 20
8x + 2x2  8 = 80
10x = 90
x = 9
then z = 7
and 4y = 16 > y = 4
The numbers are 947 and 749 , as Arora also found.
(I simply took the difference between the numbers to be 198, I should have reversed the order of subtraction)
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