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 | 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 first see that <math>a^2 + 100 = b^2 + 1</math> or <math>99</math> <math>=</math> <math>b^2-a^2</math>. | ||
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<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>. | <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> | 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> | ||
+ | |||
+ | ==Solution 2== | ||
+ | Let <math>P</math> <math>=</math> original population. Translating the word problem into a system of equations, we got: | ||
+ | <cmath> | ||
+ | \begin{align} | ||
+ | P &= x^2 \\ | ||
+ | P + 100 &= y^2 + 1 \\ | ||
+ | P + 200 &= z^2 | ||
+ | \end{align} | ||
+ | </cmath> | ||
+ | for some positive integers <math>x</math>, <math>y</math> and <math>z</math>. | ||
+ | Now, by subtracting <math>(2)</math> from <math>(3)</math> (i.e. <math>(3) - (2)</math>), we got: | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | 100 &= z^2 - y^2 - 1 \\ | ||
+ | 101 &= z^2 - y^2 \\ | ||
+ | 101 &= (z - y)(z + y) | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | Since y and z are both positive integers and 101 is a prime, by factoring, the only working solution for us is <math>y = 50</math> and <math>z = 51</math>. | ||
+ | Plugging that back to <math>(2)</math> and simplify, we got <math>P = 2401 = (49)^2 = x^2</math>, a multiple of <math>7</math>. | ||
+ | Therefore, the answer is <math>\boxed{(B) 7}</math>. -nullptr07 |
Latest revision as of 15:52, 28 June 2023
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 second population + third population... 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
Solution 2
Let original population. Translating the word problem into a system of equations, we got: for some positive integers , and . Now, by subtracting from (i.e. ), we got: Since y and z are both positive integers and 101 is a prime, by factoring, the only working solution for us is and . Plugging that back to and simplify, we got , a multiple of . Therefore, the answer is . -nullptr07