Question

I am a 5 digit number between 52,000 and 53,000. The sum of my digits is 21. My ones digit is a multiple of 3. My tens digit is equal to my hundreds digit.

Answer 1

I am 52446.

Explanation:

Let the digits be a, b, c, d, and e.

Then the number is abcde.

We have five conditions:

(1) a = 5

(2) b = 2

(3) a + b + c + d + e = 21

(4) e = 3, 6, or 9

(5) c = d

From (3), 5 + 2 + c + d + e = 21

7 + c + d + e = 21

c + d + e = 14

From (5), 2c + e = 14

c = ½(14 - e)

If e = 3, c = ½(14 - 3) = 11/2. Not possible.

If e = 9, c = ½(14 - 9) = 5/2. Not possible.

If e = 6, c = ½(14 - 6) = 8/2 = 4.

The solution is

a = 5, b = 2, c = 4, d = 4, e = 6 ⟶ 52446

The number is 52446.