Difference between revisions of "2008 AMC 10B Problems/Problem 8"
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− | 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 | + | 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 odd number, the product of <math>3x</math> must be even and smaller than <math>50</math>. You can try all the even substitutes 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 nine numbers in total, so the answers is <math>\boxed{9 \text{ (C)}}</math>. |
==Solution 3== | ==Solution 3== |
Revision as of 12:25, 7 June 2021
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 odd number, the product of
must be even and smaller than
. You can try all the even substitutes for
and you will end up with the numbers
,
,
,
,
,
,
,
, and
. There are nine numbers in total, so the answers 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) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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 10 Problems and Solutions |
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