May 2023
Problem of the Month

Digit Problem
By Davit Banana

A number is formed in the following manner: 123...(n-2)(n-1)n(n-1)(n-2)...321.
For example: 12345678987654321

Prove that:
The sum of the digits of the number always results in a perfect square of n.

Example 1:       12321
      1+2+3+2+1 = 9 = 3²

Example 2:       12345654321
      1+2+3+4+5+6+5+4+3+2+1 = 36 = 6²

Solution

To determine the sum of the digits of "123…(n-2)(n-1)n(n-1)(n-2)…321" recall the formula for the sum of an arithmetic sequence:


Similarly, the sum of 1 + 2 + 3 + ... + (n-1) is given by:


Therefore the sum of the digits 1+ 2 + 3 + 4 + (n-1) + n + (n-1) + ... + 3 + 2 + 1 is the sum of these two formulas:

Correctly solved by:

1. K. Sengupta	Calcutta, West Bengal, India
2. Dr. Hari Kishan	D.N. College, Meerut, Uttar Pradesh, India
3. Carolina Romero	Central High School, Grand Junction, Colorado
4. Colin (Yowie) Bowey	Beechworth, Victoria, Australia
5. Ian (Courtney Teddy's student)	Clark Elementary, Issaquah, Washington
6. Aariv (Courtney Teddy's student)	Clark Elementary, Issaquah, Washington
7. Michael Sasser	Ross S. Sterling High School, Baytown, Texas
8. Kelly Stubblefield	Mobile, Alabama
9. Seth Cohen	Concord, New Hampshire

Send any comments or questions to: David Pleacher