The sum of the digits in base ten of (10(4n2 +8) +1)2, where n is a positive integer, is:
- 4
- 4n
- 2 + 2n
- 4n2
- n2 + n + 2
(10(4n2 +8) +1) will look like this 1000…00001; that is, a number with a 1 on each end and 0’s in between.
If we square it, we have
| 10…01 | ||
| x | 10…01 | |
| 10…01 | ||
| + | 10…010 … 0 | |
| 10…1..1…01 | or possibly | |
| 10…2…01 | ||
So the sum of the digits is 4, answer (a).
The 4n2 + 8 is a total red herring.
FLORIDA