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.

Question

How to answer this question? Helpp me

Answer 1

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.

Time for other eyes. Have I messed up somewhere in all that?

Answer 2

U r taking a very long approach.
The difference is between the 2 n is 99(a-c)
99(c-a)=9(a+b+c)
9(c-a)=a+b+c
8c-10a=b - i equation
Also
8(c-a)= 6+2a
8c-10a=6 -ii equation
From i and ii equation
b = 6