Difference between revisions of "2023 AMC 10B Problems/Problem 11"
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | == Solution 4 == | ||
+ | To start, we simplify things by dividing everything by <math>10</math>, the resulting equation is <math>2x+5y+10z=80</math>, and since the problem states that we have at least one of each, we simplify this to <math>2x+5y+10z=63</math>. Note that since the total is odd, we need an odd number of <math>5</math> dollar bills. We proceed using casework. | ||
+ | |||
+ | Case 1: One <math>5</math> dollar bill | ||
+ | |||
+ | <math>2x+10z=58</math>, we see that <math>10z</math> can be <math>10,20,30,40,50 </math> or <math>0</math>. | ||
+ | <math>6</math> Ways | ||
+ | |||
+ | Case 2: Three <math>5</math> dollar bills | ||
+ | |||
+ | <math>2x+10z=48</math>, like before we see that <math>10z</math> can be <math>0,10,20,30,40</math>, so <math>5</math> cases. | ||
+ | |||
+ | Now we should start to see a pattern emerges, each case there is <math>1</math> less way to sum to <math>80</math>, so the answer is just <math>\frac{6(6+1)}{2}</math>, <math>21</math> or <math>(B)</math> | ||
==Video Solution 1 by OmegaLearn== | ==Video Solution 1 by OmegaLearn== |
Revision as of 10:09, 16 November 2023
Contents
Problem
Suzanne went to the bank and withdrew . The teller gave her this ammount using
bills,
bills, and
bills, with at least one of each denomination. How many different collections of bills could Suzanne have received?
Solution 1
We let the number of ,
, and
bills be
and
respectively.
We are given that Dividing both sides by
, we see that
We divide both sides of this equation by :
Since
and
are integers,
must also be an integer, so
must be divisible by
. Let
where
is some positive integer.
We can then write Dividing both sides by
, we have
We divide by
here to get
and
are both integers, so
is also an integer.
must be divisible by
, so we let
.
We now have . Every substitution we made is part of a bijection (i.e. our choices were one-to-one); thus, the problem is now reduced to counting how many ways we can have
and
such that they add to
.
We still have another constraint left, that each of and
must be at least
. For
, let
We are now looking for how many ways we can have
We use a classic technique for solving these sorts of problems: stars and bars. We have stars and
groups, which implies
bars. Thus, the total number of ways is
~Technodoggo ~minor edits by lucaswujc
Solution 2
First, we note that there can only be an even number of dollar bills.
Next, since there is at least one of each bill, we find that the amount of dollar bills is between
and
. Doing some casework, we find that the amount of
dollar bills forms an arithmetic sequence:
+
+
+
+
+
.
Adding these up, we get .
~yourmomisalosinggame (a.k.a. Aaron)
Solution 3
Denote by ,
,
the amount of $20 bills, $50 bills and $100 bills, respectively.
Thus, we need to find the number of tuples
with
that satisfy
First, this equation can be simplified as
Second, we must have . Denote
.
The above equation can be converted to
Third, we must have . Denote
.
The above equation can be converted to
Denote ,
and
.
Thus, the above equation can be written as
Therefore, the number of non-negative integer solutions is
.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 4
To start, we simplify things by dividing everything by , the resulting equation is
, and since the problem states that we have at least one of each, we simplify this to
. Note that since the total is odd, we need an odd number of
dollar bills. We proceed using casework.
Case 1: One dollar bill
, we see that
can be
or
.
Ways
Case 2: Three dollar bills
, like before we see that
can be
, so
cases.
Now we should start to see a pattern emerges, each case there is less way to sum to
, so the answer is just
,
or
Video Solution 1 by OmegaLearn
See also
2023 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.