Difference between revisions of "1973 AHSME Problems/Problem 21"
Rockmanex3 (talk | contribs) (Solution to Problem 21) |
Rockmanex3 (talk | contribs) (→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
Solution
If the first number of a group of consecutive numbers is
, the
number is
. We know that the sum of the group of numbers is
, so
We know that
and
are positive integers, so we check values of
that are a factor of
. Of these values, the only ones that result in a positive integer
is when
or when
, so there are
sets of two or more consecutive positive integers whose sum is
.
See Also
1973 AHSC (Problems • Answer Key • Resources) | ||
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 |