Difference between revisions of "2023 AMC 12B Problems/Problem 16"
Pi is 3.14 (talk | contribs) |
Pi is 3.14 (talk | contribs) |
||
| Line 38: | Line 38: | ||
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
| − | |||
| − | |||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2023|ab=B|num-b=15|num-a=17}} | {{AMC12 box|year=2023|ab=B|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 20:38, 15 November 2023
Problem
In the state of Coinland, coins have values 
and 
cents. Suppose 
is the value in cents of the most expensive item in Coinland that cannot be purchased using these coins with exact change. What is the sum of the digits of 
Solution
This problem asks to find largest that cannot be written as
![\[ 6 a + 10 b + 15 c = x, \hspace{1cm} (1) \]](http://latex.artofproblemsolving.com/4/5/9/4599a23c5484a8b1490fa8d5b7849f3f2c7a44ef.png)
where 
.
Denote by 
the remainder of divided by 2.
Modulo 2 on Equation (1), we get
By using modulus 
on the equation above, we get
.
Following from Chicken McNugget's theorem, we have that any number that is no less than 
can be expressed in the form of 
with 
.
Therefore, all even numbers that are at least equal to 
can be written in the form of Equation (1) with .
All odd numbers that are at least equal to 
can be written in the form of Equation (1) with .
The above two cases jointly imply that all numbers that are at least 30 can be written in the form of Equation (1) with .
Next, we need to prove that 29 cannot be written in the form of Equation (1) with .
Because 29 is odd, we must have 
.
Because , we must have 
.
Plugging this into Equation (1), we get 
.
However, this equation does not have non-negative integer solutions.
All analysis above jointly imply that the largest that has no non-negative integer solution to Equation (1) is 29.
Therefore, the answer is 
.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
| 2023 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 15 |
Followed by Problem 17 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
