Difference between revisions of "1962 AHSME Problems/Problem 38"
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<math> \textbf{(A)}\ 3\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 17 </math> | <math> \textbf{(A)}\ 3\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 17 </math> | ||
− | ==Solution== | + | ==Solution 1== |
− | { | + | Let <math>a^2</math> <math>=</math> original population count, <math>b^2+1</math> <math>=</math> the second population count, and <math>c^2</math> <math>=</math> the third population count |
+ | We first see that <math>a^2 + 100 = b^2 + 1</math> or <math>99</math> <math>=</math> <math>b^2-a^2</math>. | ||
+ | We then factor the right side getting <math>99</math> <math>=</math> <math>(b-a)(b+a)</math>. | ||
+ | Since we can only have an nonnegative integral population, clearly <math>b+a</math> <math>></math> <math>b-a</math> and both factor <math>99</math>. | ||
+ | We factor <math>99</math> into <math>3^2 \cdot 11</math> <math>=</math> <math>(b-a)(b+a)</math> | ||
+ | There are a few cases to look at: | ||
+ | <math>1)</math> <math>b+a</math> <math>=</math> <math>11</math> and <math>b-a</math> <math>=</math> <math>9</math>. | ||
+ | Adding the two equations we get <math>2b</math> <math>=</math> <math>20</math> or <math>b</math> <math>=</math> <math>10</math>, which means <math>a</math> <math>=</math> <math>1</math>. | ||
+ | But looking at the restriction that the <math>second population</math> + <math>100</math> <math>=</math> <math>third population</math>... | ||
+ | <math>10^2</math> <math>+</math> <math>1</math> <math>+</math> <math>100</math> <math>=</math> <math>201</math> <math>\neq</math> a perfect square. | ||
+ | |||
+ | <math>2)</math> <math>b+a</math> <math>=</math> <math>33</math> and <math>b-a</math> <math>=</math> <math>3</math>. | ||
+ | Adding the two equations we get <math>2b</math> <math>=</math> <math>36</math> or <math>b</math> <math>=</math> <math>18</math>, which means <math>a</math> <math>=</math> <math>15</math>. | ||
+ | Looking at the same restriction, we get <math>18^2</math> + <math>1</math> + <math>100</math> <math>=</math> <math>324</math> + <math>101</math> <math>=</math> <math>425</math>, which is NOT a perfect square. | ||
+ | |||
+ | Finally, <math>b+a</math> <math>=</math> <math>99</math> and <math>b-a</math> <math>=</math> <math>1</math>. | ||
+ | <math>2b</math> <math>=</math> <math>100</math> or <math>b</math> <math>=</math> <math>50</math>, which means <math>a</math> <math>=</math> <math>49</math>. | ||
+ | Looking at the same restriction, we get <math>50^2</math> + <math>1</math> + <math>100</math> <math>=</math> <math>2500</math> + <math>101</math> <math>=</math> <math>2601</math> <math>=</math> <math>51^2</math>. Thus we find that the original population is <math>a^2</math> <math>=</math> <math>49^2</math> <math>=</math> <math>7^4</math>. Or <math>a^2</math> is a multiple of <math>\boxed{ (B) 7}</math> |
Revision as of 20:27, 19 February 2016
Problem
The population of Nosuch Junction at one time was a perfect square. Later, with an increase of , the population was one more than a perfect square. Now, with an additional increase of , the population is again a perfect square.
The original population is a multiple of:
Solution 1
Let original population count, the second population count, and the third population count We first see that or . We then factor the right side getting . Since we can only have an nonnegative integral population, clearly and both factor . We factor into There are a few cases to look at: and . Adding the two equations we get or , which means . But looking at the restriction that the + ... a perfect square.
and . Adding the two equations we get or , which means . Looking at the same restriction, we get + + + , which is NOT a perfect square.
Finally, and . or , which means . Looking at the same restriction, we get + + + . Thus we find that the original population is . Or is a multiple of