Difference between revisions of "1973 AHSME Problems/Problem 21"

(Solution to Problem 21)
 
(Solution)
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<cmath>2a+n-1=\frac{200}{n}</cmath>
 
<cmath>2a+n-1=\frac{200}{n}</cmath>
 
<cmath>2a = 1-n + \frac{200}{n}</cmath>
 
<cmath>2a = 1-n + \frac{200}{n}</cmath>
We know that <math>n</math> and <math>a</math> are positive integers, so we check values of <math>n</math> that are a factor of <math>200</math>.  Of these values, the only ones that result in a positive integer <math>a</math> is when <math>n = 5</math> or when <math>n = 8</math>, so there are <math>\boxed{\textbf{(B)}\ 2}</math> sets of integers whose sum is <math>100</math>.
+
We know that <math>n</math> and <math>a</math> are positive integers, so we check values of <math>n</math> that are a factor of <math>200</math>.  Of these values, the only ones that result in a positive integer <math>a</math> is when <math>n = 5</math> or when <math>n = 8</math>, so there are <math>\boxed{\textbf{(B)}\ 2}</math> sets of two or more consecutive positive integers whose sum is <math>100</math>.
  
 
==See Also==
 
==See Also==

Revision as of 17:29, 5 July 2018

Problem

The number of sets of two or more consecutive positive integers whose sum is 100 is

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

Solution

If the first number of a group of $n$ consecutive numbers is $a$, the $n^\text{th}$ number is $a+n-1$. We know that the sum of the group of numbers is $100$, so \[\frac{n(2a+n-1)}{2} = 100\] \[2a+n-1=\frac{200}{n}\] \[2a = 1-n + \frac{200}{n}\] We know that $n$ and $a$ are positive integers, so we check values of $n$ that are a factor of $200$. Of these values, the only ones that result in a positive integer $a$ is when $n = 5$ or when $n = 8$, so there are $\boxed{\textbf{(B)}\ 2}$ sets of two or more consecutive positive integers whose sum is $100$.

See Also

1973 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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 31 32 33 34 35
All AHSME Problems and Solutions