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.

Read the clues. Find the number

Answer 1

Explanation:

5 digit number

5abcd

a = 2 or a = 3. but a = 3 means the upper limit of 53,000. and the sum of digits is 8 and not 21.

so, a = 2.

d = 3n

c = b

5 + a + b + c + d = 21

5 + a + 2b + 3n = 21

a + 2b + 3n = 16

2 + 2b + 3n = 16

2b + 3n = 14

so, n must be an even number to allow an even result (14). therefore, n = 0 or n = 2 (any larger even number n like 4 would not create a single digit result)

if n = 0, then 2b = 14, b = 7

if n = 2, then 2b + 6 = 14, 2b = 8, b = 4

so, we have the options

52770

52446

I suspect that 0 as 0×3 is not considered a multiple of 3 by your teacher, so only n = 2 (or d = 6) is a desired solution.

therefore, I would use

52446

as answer.