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Sagot :
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.
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.
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