1998 JBMO Problems/Problem 1

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Problem

Prove that the number $\underbrace{111\ldots 11}_{1997}\underbrace{22\ldots 22}_{1998}5$ (which has 1997 of 1-s and 1998 of 2-s) is a perfect square.

Solution

The number $\underbrace{111\ldots 11}_{1997}\underbrace{22\ldots 22}_{1998}5$ can be rewritten as \[(10^{3995} + 10^{3994} + \cdots 10^{1999}) + 2(10^{1998} + 10^{1997} + \cdots 10) + 5.\] This number has a few geometric series and can be written as \[\frac{10^{1999}(10^{1997} - 1)}{9} + 2 \cdot \frac{10(10^{1998} - 1)}{9} + 5.\] Simplifying results in \[\frac{10^{3996} - 10^{1999} + 2 \cdot 10^{1999} - 20 + 45}{9}\] \[\frac{10^{3996} + 10^{1999} + 25}{9}.\] Notice that $10^{3996} = (10^{1998})^2$ and $10^{1999} = 10 \cdot 10^{1998}.$ That means we can "factor" the numerator, and doing so results in \[\frac{(10^{1998} + 5)^2}{9}\] \[\left(\frac{10^{1998} + 5}{3}\right)^2.\] Since $10^{1998} + 5$ is divisible by 3, we conclude that the number $\underbrace{111\ldots 11}_{1997}\underbrace{22\ldots 22}_{1998}5$ is a perfect square.

See Also

1998 JBMO (ProblemsResources)
Preceded by
First Problem
Followed by
Problem 2
1 2 3 4
All JBMO Problems and Solutions
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