Difference between revisions of "1963 AHSME Problems/Problem 31"
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Solving for <math>x</math> in the equation yields <math>x =rfthe meaning of theta 0</math>. Solving the inequality results in <math>y \le 254 \frac{1}{3}</math>. From the two conditions, <math>y</math> can be an odd number from <math>1</math> to <math>253</math>, so there are <math>127</math> solutions where <math>x</math> and <math>y</math> are integers. The answer is <math>\boxed{\textbf{(D)}}</math>. | Solving for <math>x</math> in the equation yields <math>x =rfthe meaning of theta 0</math>. Solving the inequality results in <math>y \le 254 \frac{1}{3}</math>. From the two conditions, <math>y</math> can be an odd number from <math>1</math> to <math>253</math>, so there are <math>127</math> solutions where <math>x</math> and <math>y</math> are integers. The answer is <math>\boxed{\textbf{(D)}}</math>. | ||
==Solution 2== | ==Solution 2== | ||
| − | We will prove that <math>y</math> is an odd number by contradiction. If <math>y</math> is even, then we know that <math>y = 2m</math> where <math>m</math> is some integer. However, this immediately assumes that <math>\text{ even } + \text{ even } = \text{ odd }</math> which is impossible. therefore <math>y</math> must ben odd. | + | We will prove that <math>y</math> is an odd number by contradiction. If <math>y</math> is even, then we know that <math>y = 2m</math> where <math>m</math> is some integer. However, this immediately assumes that <math>\text{ even } + \text{ even } = \text{ odd }</math> which is impossible. therefore <math>y</math> must ben odd. then we can easily prove that <math>x</math> ..... |
| − | then we can | + | ==Solution 3== |
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| + | We can solve for first solution by applying extended euclids division algorithm or we can apply hit and trial for the first solution to get <math>x_0</math> = <math>380</math> and <math>y_0</math> =<math>1</math> . then the general solution of the given diophanitine equation will be <math>x</math> = <math>x_0</math> +<math>3t</math> and <math>y</math> = <math>y_0</math> - <math>2t</math>. Since we need only positive integer solutions So we solve <math>380</math> + <math>3t</math> <math>></math> <math>0</math> and <math>1</math>-<math>2t</math> <math>></math> <math>0</math> to get <math>t</math> <math>></math> <math>0</math> (applying Greatest integer function) also we can clearly see that <math>t_{(min)}</math> <math>=</math> <math>0</math> so,t <math><</math> <math>GIF</math>(<math>383</math>/<math>3</math>). That implies <math>t</math> ranges from <math>0</math> to <math>127</math>. Hence,the correct answer is <math>127</math>, <math>\boxed{\textbf{(D)}}</math>. | ||
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| + | ~Geometry-Wizard | ||
==See Also== | ==See Also== | ||
Latest revision as of 13:07, 14 January 2024
Problem
The number of solutions in positive integers of 
is:
Solution 1
Solving for 
in the equation yields 
. Solving the inequality results in 
. From the two conditions, 
can be an odd number from 
to 
, so there are 
solutions where and are integers. The answer is 
.
Solution 2
We will prove that is an odd number by contradiction. If is even, then we know that 
where 
is some integer. However, this immediately assumes that 
which is impossible. therefore must ben odd. then we can easily prove that .....
Solution 3
We can solve for first solution by applying extended euclids division algorithm or we can apply hit and trial for the first solution to get 
= 
and 
= . then the general solution of the given diophanitine equation will be = +
and = - 
. Since we need only positive integer solutions So we solve + 
and - to get 
(applying Greatest integer function) also we can clearly see that 
so,t 
(
/
). That implies ranges from to . Hence,the correct answer is , .
~Geometry-Wizard
See Also
| 1963 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 30 |
Followed by Problem 32 | |
| 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 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
