Question

Five three digit numbers including n were to be added. While adding, the reverse of n was added by mistake instead of n. Hence the summ increased by 11 times the summ of the digits of n.Eight times the difference of n's units and hundreds digit is 6 more than twice its hundreds digit. Find its tens digit.

Answer 1

$<B>if N's digits are a,b,c, then we know that c>a, since reversing the digits increased the sum. cba - abc = 99(c-a) = 11(a+b+c) Since 99 is a multiple of both 9 and 11, a+b+c must be a multiple of 9. So, a+b+c is 9 or 18, since 7+8+9=24 is the maximum sum of digits. Now, we also know that 8(c-a) = 2c+6 6c-8a=6 3c-4a=3 4a = 3(c-1) That means that c-1 is a multiple of 4, so c is either 5 or 9, making a either 3 or 6.$

The difference c-a is either 2 or 3.
8*2 = 10+6, making c=5
8*3 = 18+6, making c=9
No news there.

That means N' is either N+198 or N+297.
198 = 11*18 = 11(3+10+5)
297 = 11*27 = 11(6+12+9)

And now we have a problem, since b must be less than 10.