Difference between revisions of "1997 AHSME Problems/Problem 20"

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== See also ==
 
== See also ==
 
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{{AHSME box|year=1997|num-b=19|num-a=21}}
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Revision as of 13:13, 5 July 2013

Problem

Which one of the following integers can be expressed as the sum of $100$ consecutive positive integers?

$\textbf{(A)}\ 1,\!627,\!384,\!950\qquad\textbf{(B)}\ 2,\!345,\!678,\!910\qquad\textbf{(C)}\ 3,\!579,\!111,\!300\qquad\textbf{(D)}\ 4,\!692,\!581,\!470\qquad\textbf{(E)}\ 5,\!815,\!937,\!260$

Solution

The sum of the first $100$ integers is $\frac{100\cdot 101}{2} = 5050$.

If you add an integer $k$ to each of the $100$ numbers, you get $5050 + 100k$, which is the sum of the numbers from $k+1$ to $k+100$.

You're only adding multiples of $100$, so the last two digits will remain unchanged.

Thus, the only possible answer is $\boxed{A}$, because the last two digits are $50$.

As an aside, if $5050 + 100k = 1627384950$, then $k = 16273799$, and the numbers added are the integers from $16273800$ to $16273899$.

See also

1997 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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