# If you write me backward, then add me to my number, the sum is 5555. My ones digit is twice the hundreds digit. None of my digits are alike. My thousands digit is the smallest. What number am I?

1
by iyahmiles

2015-06-13T18:03:44+08:00

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The number would be abcd, then the reverse is dcba.
The number is equal to 1000a+100b+10c+d
The reverse is equal to 1000d+100c+10b+a

(1000a+100b+10c+d) + (1000d+100c+10b+a)=5555
1001(a+d) + 110(b+c) = 5555
11[91(a+d)+10(b+c)] = 5555
91(a+d) + 10(b+c) = 505
10(b+c) = 505-91(a+d)

This is a diophantine equation... Take note that the ones digit 5, will come from 91.

We can only have : a + d = 5
10(b+c) = 505 - 91(5)
10(b+c) = 505 - 455 = 50
Therefore b+c = 5

5 = 1+4 , 2+3 or 0+5
This would mean a is equal to 1 since we want the least number, and we cannot have 0 as a. This would then make d=4. We also have b=2,c=3

The number then is 1234.