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 the Problem:


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