1E. The number 2A5A0A2 is divisible by 99. What is the digit A?

The digit A that makes 2A5A0A2 to be divisible by 99 is;

A = 3

A is a single digit and as such will be between 0 and 9.

Thus, using the principle above;

2 + A + 5 + A + 0 + A + 2  = 3A + 9

3A + 9 must be divisible by 9.

Thus;

3A + 9 = 9x

A = (9x - 9)/3

When x = 1;

A = 0

When x = 2;

A = 3

When x = 3;

A = 9

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We know, a number is divisible by 9 if the sum of digits is also divisible by 9.

Also, since A is an digit, so A is in between 0 to 9.

2+A+5+A+0+A+2  = 3A + 9

3A + 9 should be divisible by 9.

3A + 9 = 9n

A = (9n -9)/3

For n= 1 :

A = 0

For n=2 :

A = 3

For n= 3 :

A = 9

After this A > 9 which is not accepted.

Therefore, the digit A can be 0, 3 and 9 .

Hence, this is the required solution.

let m = monomial

let t = trinomial

so, m times t = t.

monomial times a trinomial = another trinomial.