Difference between revisions of "2004 AMC 12A Problems/Problem 3"
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==Problem== | ==Problem== | ||
+ | For how many ordered pairs of positive integers <math>(x,y)</math> is <math>x + 2y = 100</math>? | ||
− | {{ | + | <math>\text {(A)} 33 \qquad \text {(B)} 49 \qquad \text {(C)} 50 \qquad \text {(D)} 99 \qquad \text {(E)}100</math> |
− | ==Solution== | + | ==Solution 1== |
+ | Every integer value of <math>y</math> leads to an integer solution for <math>x</math> | ||
+ | Since <math>y</math> must be positive, <math>y\geq 1</math> | ||
− | + | Also, <math>y = \frac{100-x}{2}</math> | |
+ | Since <math>x</math> must be positive, <math>y < 50</math> | ||
+ | <math>1 \leq y < 50</math> | ||
+ | This leaves <math>49</math> values for y, which mean there are <math>49</math> solutions to the equation <math>\Rightarrow \mathrm{(B)}</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | 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}} | ||
+ | {{MAA Notice}} |
Latest revision as of 01:35, 29 September 2018
Contents
Problem
For how many ordered pairs of positive integers is
?
Solution 1
Every integer value of leads to an integer solution for
Since
must be positive,
Also,
Since
must be positive,
This leaves
values for y, which mean there are
solutions to the equation
Solution 2
If and
must each be positive integers, then we can say that
is at least 1 and
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
must be an integer,
must be even; thus only
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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