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Difference between revisions of "2004 AMC 12A Problems/Problem 3"
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==Problem==
 
==Problem==
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For how many ordered pairs of positive integers <math>(x,y)</math> is <math>x + 2y = 100</math>?
  
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{{empty}}
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<math>\text {(A)} 33 \qquad \text {(B)} 49 \qquad \text {(C)} 50 \qquad \text {(D)} 99 \qquad \text {(E)}100</math>
  
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==Solution==
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==Solution 1==
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Every integer value of <math>y</math> leads to an integer solution for <math>x</math>
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Since <math>y</math> must be positive, <math>y\geq 1</math>
  
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{{solution}}
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Also, <math>y = \frac{100-x}{2}</math>
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Since <math>x</math> must be positive, <math>y < 50</math>
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<math>1 \leq y < 50</math>
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This leaves <math>49</math> values for y, which mean there are <math>49</math> solutions to the equation <math>\Rightarrow \mathrm{(B)}</math>
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==Solution 2==
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If <math>x</math> and <math>2y</math> must each be positive integers, then we can say that <math>x</math> is at least 1 and <math>2y</math> is at least 1. From there, we want to find out how many ways there are to distribute the other 98 ones (the smallest positive integer addends of 100). 98 identical objects can be distributed to two distinct bins in 99 ways (think stars and bars), yet this 99 is an overcount. Because <math>y</math> must be an integer, <math>2y</math> must be even; thus only <math>\left\lfloor \frac{99}{2} \right\rfloor = \boxed{ 49 \implies B}</math> ways exist to distribute these ones.
  
 
==See Also==
 
==See Also==
  
 
{{AMC12 box|year=2004|ab=A|num-b=2|num-a=4}}
 
{{AMC12 box|year=2004|ab=A|num-b=2|num-a=4}}
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{{MAA Notice}}
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The answer is (B) since x can't be 0, which leaves only 49 different values for y.
 

Latest revision as of 00:35, 29 September 2018

Problem

For how many ordered pairs of positive integers $(x,y)$ is $x + 2y = 100$?

$\text {(A)} 33 \qquad \text {(B)} 49 \qquad \text {(C)} 50 \qquad \text {(D)} 99 \qquad \text {(E)}100$

Solution 1

Every integer value of $y$ leads to an integer solution for $x$ Since $y$ must be positive, $y\geq 1$

Also, $y = \frac{100-x}{2}$ Since $x$ must be positive, $y < 50$

$1 \leq y < 50$ This leaves $49$ values for y, which mean there are $49$ solutions to the equation $\Rightarrow \mathrm{(B)}$

Solution 2

If $x$ and $2y$ must each be positive integers, then we can say that $x$ is at least 1 and $2y$ is at least 1. From there, we want to find out how many ways there are to distribute the other 98 ones (the smallest positive integer addends of 100). 98 identical objects can be distributed to two distinct bins in 99 ways (think stars and bars), yet this 99 is an overcount. Because $y$ must be an integer, $2y$ must be even; thus only $\left\lfloor \frac{99}{2} \right\rfloor = \boxed{ 49 \implies B}$ ways exist to distribute these ones.

See Also

2004 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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
All AMC 12 Problems and Solutions

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