1997 AHSME Problems/Problem 20: Difference between revisions
Talkinaway (talk | contribs) Created page with "==Problem== Which one of the following integers can be expressed as the sum of <math>100</math> consecutive positive integers? <math> \textbf{(A)}\ 1,\!627,\!384,\!950\qquad\t..." |
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<math> \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 </math> | <math> \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 </math> | ||
==Solution== | |||
The sum of the first <math>100</math> integers is <math>\frac{100\cdot 101}{2} = 5050</math>. | |||
If you add an integer <math>k</math> to each of the <math>100</math> numbers, you get <math>5050 + 100k</math>, which is the sum of the numbers from <math>k+1</math> to <math>k+100</math>. | |||
You're only adding multiples of <math>100</math>, so the last two digits will remain unchanged. | |||
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>. | |||
==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>. | |||
Looking over at the list of possible values, we quickly realise that the only possible solution is <math>\boxed{A}</math> | |||
== See also == | |||
{{AHSME box|year=1997|num-b=19|num-a=21}} | |||
{{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 
consecutive positive integers?
Solution
The sum of the first integers is 
.
If you add an integer 
to each of the numbers, you get 
, which is the sum of the numbers from 
to 
.
You're only adding multiples of , so the last two digits will remain unchanged.
Thus, the only possible answer is 
, because the last two digits are 
.
As an aside, if 
, then 
, and the numbers added are the integers from 
to 
.
Solution
Notice how the sum of 100 consecutive integers is 
.
Cancelling out the constants give us 
.
Looking over at the list of possible values, we quickly realise that the only possible solution is
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 | ||
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