Difference between revisions of "1968 AHSME Problems/Problem 19"
m (→Solution) |
(→Solution) |
||
| (4 intermediate revisions by 3 users not shown) | |||
| Line 10: | Line 10: | ||
== Solution == | == Solution == | ||
| − | <math>\fbox{E}</math> | + | |
| + | Because we have to use at least one dime and one quarter, this part of the selection is forced. Thus, the number of ways to partition <math>\$10</math> in the desired manner is equal to the number of ways to partition <math>\$10-10\cent-25\cent=\$9.65</math> into quarters and dimes without restriction. However, <math>\$9.65</math> is not a multiple of <math>25\cent</math>, so we must use dimes until we reach a multiple of <math>25\cent</math>. We first reach such a multiple at <math>\$9.25</math>, and, because we have had no freedom up to this point, the number of ways to partition <math>\$9.25</math> into dimes and quarters will yield our desired answer. <math>\$9.25</math> can be partitioned into 37 quarters, and we can replace quarters 2 at a time with 5 dimes to partition it differently. We can replace anywhere from 0 to 18 pairs of quarters with dimes in this manner, giving us <math>\boxed{19}</math> different partitions, which is answer choice <math>\fbox{E}</math>. | ||
== See also == | == See also == | ||
| − | {{AHSME box|year=1968|num-b=18|num-a=20}} | + | {{AHSME 35p box|year=1968|num-b=18|num-a=20}} |
[[Category: Introductory Algebra Problems]] | [[Category: Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 20:12, 17 July 2024
Problem
Let 
be the number of ways 
dollars can be changed into dimes and quarters, with at least one of each coin being used. Then equals:
Solution
Because we have to use at least one dime and one quarter, this part of the selection is forced. Thus, the number of ways to partition 
in the desired manner is equal to the number of ways to partition 
into quarters and dimes without restriction. However, 
is not a multiple of 
, so we must use dimes until we reach a multiple of . We first reach such a multiple at 
, and, because we have had no freedom up to this point, the number of ways to partition into dimes and quarters will yield our desired answer. can be partitioned into 37 quarters, and we can replace quarters 2 at a time with 5 dimes to partition it differently. We can replace anywhere from 0 to 18 pairs of quarters with dimes in this manner, giving us 
different partitions, which is answer choice 
.
See also
| 1968 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
