Question

solve 1000a+100b+10c+a=a^b×c^a​

Answer 1

Let us see how this trick works.

Let the 3-digit number chosen by Minakshi be abc = 100a + 10b + c.

After reversing the order of the digits, she gets the number cba = 100c + 10b + a. On

subtraction:

• If a > c, then the difference between the numbers is

(100a + 10b + c) – (100c + 10b + a) = 100a + 10b + c – 100c – 10b – a

= 99a – 99c = 99(a – c).

• If c > a, then the difference between the numbers is

(100c + 10b + a) – (100a + 10b + c) = 99c – 99a = 99(c – a).

• And, of course, if a = c, the difference is 0.

In each case, the resulting number is divisible by 99. So the remainder is 0. Observe

that quotient is a – c or c – a. You may check the same by taking other 3-digit numbers