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 sum increased by
11 times the sum of the digits of N. Eight times the
difference of N's units and hundreds digits is 6 more
than twice its hundreds digit. Find its tens digit.​

Answer 1

Step-by-step explanation:

Let sum of the numbers excluding N be x

Let N = 100a+10b+c

Reversed N (Rev N) = 100c+10b+a

Thus we have

(x+N) + 11(a+b+c) = x + Rev N

After some evaluation you will get

10a + b - 8c = 0 ....(i)

It is also given that

8(c-a) = 6 + 2a

Which gives

10a - 8c = -6 ....(ii)

Putting the result of (ii) in (i) we get

b - 6 = 0

and so

b = 6

Hence, Tens digit of N = 6

Hope this helps...