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1997 AHSME Problems/Problem 20: Difference between revisions

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==Solution==
==Solution==
The sum of the first <math>100</math> integers is <math>\frac{100\cdot 101}{2} = 5050</math>.
The sum of the first <math>100</math> integers is <math>\frac{100\cdot 101}{2} = 5050</math>.


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Thus, the only possible answer is <math>\boxed{A}</math>, because the last two digits are <math>50</math>.
Thus, the only possible answer is <math>\boxed{A}</math>, because the last two digits are <math>50</math>.


As an aside, if <math>5050 + 100k = 1627384950</math>, then <math>k = 16273799</math>, and the numbers added are the integers from <math>16273800</math> to <math>16273899</math>.  
As an aside, if <math>5050 + 100k = 1627384950</math>, then <math>k = 16273799</math>, and the numbers added are the integers from <math>16273800</math> to <math>16273899</math>.
 
==Solution==
Notice how the sum of 100 consecutive integers is <math>(x-49)+(x-48)+(x-47)...+x+...(x+47)+(x+48)+(x+49)+(x+50)</math>.  


Cancelling out the constants give us <math>100x + 50</math>.


==Solution==
Looking over at the list of possible values, we quickly realise that the only possible solution is <math>\boxed{A}</math>
Notice how the sum of 100 consecutive integers is <math>(x-49)+(x-48)+(x-47)...+x+...(x+47)+(x+48)+(x+49)+(x+50)</math>. cancelling out the constants give us <math>100x + 50</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}}

Latest revision as of 08:59, 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 (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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