1998 JBMO Problems/Problem 1
Problem
Prove that the number 
(which has 1997 of 1-s and 1998 of 2-s) is a perfect square.
Solution
The number can be rewritten as
![\[(10^{3995} + 10^{3994} + \cdots 10^{1999}) + 2(10^{1998} + 10^{1997} + \cdots 10) + 5.\]](http://latex.artofproblemsolving.com/5/4/6/54696215394625b9fe9c92fea1795883671747c4.png)
This number has a few geometric series and can be written as
![\[\frac{10^{1999}(10^{1997} - 1)}{9} + 2 \cdot \frac{10(10^{1998} - 1)}{9} + 5.\]](http://latex.artofproblemsolving.com/4/6/c/46c4d6c0bde9a09cf9aade636850bc920c8c894f.png)
Simplifying results in
![\[\frac{10^{3996} - 10^{1999} + 2 \cdot 10^{1999} - 20 + 45}{9}\]](http://latex.artofproblemsolving.com/d/c/a/dcadf5d34ef6c1ef48ed7cec82691ecf6ea3ad00.png)
![\[\frac{10^{3996} + 10^{1999} + 25}{9}.\]](http://latex.artofproblemsolving.com/5/f/6/5f63da4fe953d33291b863fbca1d2b53d2a3f4da.png)
Notice that 
and 
That means we can "factor" the numerator, and doing so results in
![\[\frac{(10^{1998} + 5)^2}{9}\]](http://latex.artofproblemsolving.com/9/4/b/94b794a546fb81e9480fe21cce5215a9e4179d8d.png)
![\[\left(\frac{10^{1998} + 5}{3}\right)^2.\]](http://latex.artofproblemsolving.com/5/1/6/516d1e36bc9b273ceb2967a267e0359076964446.png)
Since 
is divisible by 3, we conclude that the number is a perfect square.
See Also
| 1998 JBMO (Problems • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 | ||
| All JBMO Problems and Solutions | ||
