1997 AHSME Problems/Problem 20

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 $16273899$ .

Solution

Notice how the sum of 100 consecutive integers is $(x-49)+(x-48)+(x-47)...+x+...(x+47)+(x+48)+(x+49)+(x+50)$ .

Cancelling out the constants give us $100x + 50$ .

Looking over at the list of possible values, we quickly realise that the only possible solution is $\boxed{A}$

1997 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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1997 AHSME Problems/Problem 20

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