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

(Solution)
(Solution)
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Cancelling out the constants give us <math>100x + 50</math>.  
 
Cancelling out the constants give us <math>100x + 50</math>.  
  
Looking over at the solutions, we quickly realise that the only possible solution is <math>\boxed{A}</math>
+
Looking over at the list of possible values, we quickly realise that the only possible solution is <math>\boxed{A}</math>
  
 
== See also ==
 
== See also ==
 
{{AHSME box|year=1997|num-b=19|num-a=21}}
 
{{AHSME box|year=1997|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 04:34, 19 August 2017

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 $\boxed{A}$

See also

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