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Difference between revisions of "2005 iTest Problems/Problem 10"

(Created page with "==Problem== How many distinct integral solutions of the form <math>(x, y)</math> exist to the equation<math> 21x + 22y = 43</math> such that <math>1 < x < 11</math> and <math>...")
 
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==Problem==
 
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How many distinct integral solutions of the form <math>(x, y)</math> exist to the equation<math> 21x + 22y = 43</math> such that <math>1 < x < 11</math> and <math>y < 22</math>?
 
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==Solution 1==
 
==Solution 1==
 
Let <math>x = \frac{43 - 22y}{21}</math> such that <math>1 < \frac{43 - 22y}{21} < 11</math>. This means that <math>-\frac{94}{11} < y < 1</math>. If <math>y \in \mathbb{Z}</math>, then <math>y ={-8,-7,-6,-5,-4,-3,-2,-1,0}</math>. However, none of these values for <math>y</math> results in a complementary integral value for <math>x</math>. Therefore, there are <math>\boxed{0}</math> integer solutions <math>x, y \in \mathbb{Z}</math> that solves <math>21x + 22y = 43</math> over <math>1 < x < 11</math> and <math>y < 22</math>.
 
Let <math>x = \frac{43 - 22y}{21}</math> such that <math>1 < \frac{43 - 22y}{21} < 11</math>. This means that <math>-\frac{94}{11} < y < 1</math>. If <math>y \in \mathbb{Z}</math>, then <math>y ={-8,-7,-6,-5,-4,-3,-2,-1,0}</math>. However, none of these values for <math>y</math> results in a complementary integral value for <math>x</math>. Therefore, there are <math>\boxed{0}</math> integer solutions <math>x, y \in \mathbb{Z}</math> that solves <math>21x + 22y = 43</math> over <math>1 < x < 11</math> and <math>y < 22</math>.

Revision as of 17:25, 13 October 2025

Solution 1

Let $x = \frac{43 - 22y}{21}$ such that $1 < \frac{43 - 22y}{21} < 11$. This means that $-\frac{94}{11} < y < 1$. If $y \in \mathbb{Z}$, then $y ={-8,-7,-6,-5,-4,-3,-2,-1,0}$. However, none of these values for $y$ results in a complementary integral value for $x$. Therefore, there are $\boxed{0}$ integer solutions $x, y \in \mathbb{Z}$ that solves $21x + 22y = 43$ over $1 < x < 11$ and $y < 22$.

See Also

2005 iTest (Problems, Answer Key)
Preceded by:
Problem 9 		Followed by:
Problem 11
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