Difference between revisions of "2008 AMC 10B Problems/Problem 8"
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− | ==Solution== | + | ==Solution 1== |
− | The cost of a rose is odd, hence we need an even number of roses. Let there be <math>2r</math> roses for some <math>r\geq 0</math>. Then we have <math>50-3\cdot 2r = 50-6r</math> dollars left. We can always reach the sum exactly <math>50</math> by buying <math>(50-6r)/2 = 25-3r</math> carnations. Of course, the number of roses must be such that the number of carnations is non-negative. We get the inequality <math>25-3r \geq 0</math> | + | The cost of a rose is odd, hence we need an even number of roses. Let there be <math>2r</math> roses for some <math>r\geq 0</math>. Then we have <math>50-3\cdot 2r = 50-6r</math> dollars left. We can always reach the sum exactly <math>50</math> by buying <math>(50-6r)/2 = 25-3r</math> carnations. Of course, the number of roses must be such that the number of carnations is non-negative. We get the inequality <math>25-3r \geq 0</math> which solves to <math>r\leq 8 \frac13</math>. <math>r</math> must be an integer, so there are <math>\boxed{9 \text{ (C)}}</math> possible values of <math>r</math>, and each gives us one solution. |
+ | ==Solution 2== | ||
+ | |||
+ | Let <math>x</math> and <math>y</math> be the number of roses and carnations bought. The equation should be <math>3x+2y = 50</math>. Since <math>50</math> is an even number, the product of <math>3x</math> must be even and smaller than <math>50</math>. You can try nonnegative even integers for <math>x</math> and you will end up with the numbers <math>0</math>, <math>2</math>, <math>4</math>, <math>6</math>, <math>8</math>, <math>10</math>, <math>12</math>, <math>14</math>, and <math>16</math>. There are <math>9</math> numbers in total, so the answer is <math>\boxed{9 \text{ (C)}}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | Let <math>r</math> represent the number of roses, and let <math>c</math> represent the number of carnations. Then, we get the linear Diophantine equation, | ||
+ | <math>3r+2c=50</math>. | ||
+ | Using the Euclidean algorithm, we get the initial solutions to be <math>r_0=50</math> and <math>c_0=-50</math>, meaning the complete solution will be, | ||
+ | <math>r=50+\frac{2}{\gcd(2,3)}</math> <math>k=50+2k</math>, <math>c=-50-\frac{3}{\gcd(2,3)}k=-50-3k</math> | ||
+ | |||
+ | The solution range for which both <math>r</math> and <math>c</math> are positive is <math>17</math> <math>\leq k</math> <math>\leq</math> <math>25</math>. There are <math>\boxed{9 \text{ (C)}}</math> possible values for <math>k</math>. | ||
{{AMC10 box|year=2008|ab=B|num-b=7|num-a=9}} | {{AMC10 box|year=2008|ab=B|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:32, 26 July 2022
Contents
Problem
A class collects 50 dollars to buy flowers for a classmate who is in the hospital. Roses cost 3 dollars each, and carnations cost 2 dollars each. No other flowers are to be used. How many different bouquets could be purchased for exactly 50 dollars?
Solution 1
The cost of a rose is odd, hence we need an even number of roses. Let there be roses for some
. Then we have
dollars left. We can always reach the sum exactly
by buying
carnations. Of course, the number of roses must be such that the number of carnations is non-negative. We get the inequality
which solves to
.
must be an integer, so there are
possible values of
, and each gives us one solution.
Solution 2
Let and
be the number of roses and carnations bought. The equation should be
. Since
is an even number, the product of
must be even and smaller than
. You can try nonnegative even integers for
and you will end up with the numbers
,
,
,
,
,
,
,
, and
. There are
numbers in total, so the answer is
.
Solution 3
Let represent the number of roses, and let
represent the number of carnations. Then, we get the linear Diophantine equation,
.
Using the Euclidean algorithm, we get the initial solutions to be
and
, meaning the complete solution will be,
,
The solution range for which both and
are positive is
. There are
possible values for
.
2008 AMC 10B (Problems • Answer Key • Resources) | ||
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