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Difference between revisions of "2023 AMC 10B Problems/Problem 11"

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Case 1: One <math>5</math> dollar bill
 
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>5</math> Ways
+
<math>2x+10z=58</math>, we see that <math>10z</math> can be <math>0,10,20,30,40,50 </math>  or <math>6</math> ways
  
 
Case 2: Three <math>5</math> dollar bills
 
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> way.
+
<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> ways
  
−
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>
+
Now we should start to see a pattern emerge, 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>
  
 
~andyluo
 
~andyluo
Line 98: Line 98:
 
The problem is equivalent to the number of ways to make <math>\$80</math> from <math>\$2</math> bills, <math>\$5</math> bills, and <math>\$10</math> bills. We can use generating functions to find the coefficient of <math>x^{80}</math>:
 
The problem is equivalent to the number of ways to make <math>\$80</math> from <math>\$2</math> bills, <math>\$5</math> bills, and <math>\$10</math> bills. We can use generating functions to find the coefficient of <math>x^{80}</math>:
  
−
The <math>\$2</math> bills provide <math>1+x^2+x^4+x^6...+x^{78}+x^{80} = \frac{1-x^{82}}{1-x^2},</math>
+
The <math>\$2</math> bills provide <math>1+x^2+x^4+x^6 \cdots = \frac{1}{1-x^2},</math>
  
−
The <math>\$5</math> bills provide <math>1+x^5+x^{10}+x^{15}...+x^{75}+x^{80} = \frac{1-x^{85}}{1-x^5},</math>
+
The <math>\$5</math> bills provide <math>1+x^5+x^{10}+x^{15} \cdots = \frac{1}{1-x^5},</math>
  
−
The <math>\$10</math> bills provide <math>1+x^{10}+x^{20}+x^{30}...+x^{70}+x^{80} = \frac{1-x^{90}}{1-x^{10}}.</math>
+
The <math>\$10</math> bills provide <math>1+x^{10}+x^{20}+x^{30} \cdots = \frac{1}{1-x^{10}}.</math>
  
−
Multiplying, we get <math>(x^{82}-1)(x^{85}-1)(x^{90}-1)(x^2-1)^{-1}(x^5-1)^{-1}(x^{10}-1)^{-1}.</math>
+
Multiplying, we get <math>(1-x^{2})^{-1}(1-x^{5})^{-1}(1-x^{10})^{-1}.</math>
 +
 
 +
== Solution 6 (easy logic) ==
 +
 
 +
There aren't dollar signs because the <math>latex</math> thinks they're latex symbols.
 +
If you find how to override this error, please edit this.
 +
There's no <math>latex</math> here but feel free to add some!
 +
~SwordAxe
 +
 
 +
We can see in the problem that the teller gave her at least one of 20, 50, and 100. Therefore, she has 800 - 20 - 50 - 100 = 630 "left over".
 +
 
 +
Since all bills and 630 are multiples of 10, we can divide by ten.
 +
==> Question becomes: How many different collections of 2, 5, and 10 could she get if her total was 63?
 +
 
 +
We notice that because 63 is odd, we need an odd amount of 5 bills. (2 and 10 are both even, and 63 is not a multiple of 5, so we need 2 and/or 10 bills. PM SwordAxe if you don't get this.)
 +
 
 +
We can do casework.
 +
 
 +
1: She gets one 5 (50) dollar bill.
 +
She has 58 (580) left.
 +
  1) She is given only 2 dollar (20) bills => ONE COLLECTION (all 20 bills with one 50)
 +
  2) She is given one 10 dollar (100) bill
 +
      1. The rest of the money is given in 2 dollar(20) dollar bills. => ONE COLLECTION (one 100 and rest 20 with one 50)
 +
      2. She is given another 10 dollar (100) bill
 +
            I) The rest of the money is given in 2 dollar (20) dollar bills. => ONE COLLECTION (two 100, one 50, and rest 20)
 +
            II) She is given another 10 dollar (100) bill
 +
                    a) The rest of the money is given in 2 dollar (20) dollar bills. => ONE COLLECTION (three 100, one 50, and rest 20)
 +
                    b) She is given another 10 dollar (100) bill.
 +
                          1) The rest of the money is given in 2 dollar (20) dollar bills  => ONE COLLECTION (following same pattern)
 +
                          2)  AND SO ON...
 +
 
 +
This looks very tedious, but draw a simple tree diagram, and you'll see that its very easy!
 +
 
 +
If she gets one 50, there are 6 ways
 +
If she gets three 50, there are 5 ways
 +
...
 +
If she gets nine 50, there are 2 ways
 +
If she gets eleven 50, there is one way
 +
 
 +
We can add them all up, with a grand sum of 6+5+4+3+2+1 = 21.
 +
 
 +
Therefore, we get the answer (B) 21.
 +
 
 +
~SwordAxe (PM me if you have any questions! :))
  
 
==Video Solution 1 by OmegaLearn==
 
==Video Solution 1 by OmegaLearn==

Latest revision as of 12:32, 29 July 2025

Problem

Suzanne went to the bank and withdrew $$800$. The teller gave her this amount using $$20$ bills, $$50$ bills, and $$100$ bills, with at least one of each denomination. How many different collections of bills could Suzanne have received?

$\textbf{(A) } 45 \qquad \textbf{(B) } 21 \qquad \text{(C) } 36 \qquad \text{(D) } 28 \qquad \text{(E) } 32$

Solution 1

We let the number of $$20$, $$50$, and $$100$ bills be $a,b,$ and $c,$ respectively.

We are given that $20a+50b+100c=800.$ Dividing both sides by $10$, we see that $2a+5b+10c=80.$

We divide both sides of this equation by $5$: $\dfrac25a+b+2c=16.$ Since $b+2c$ and $16$ are integers, $\dfrac25a$ must also be an integer, so $a$ must be divisible by $5$. Let $a=5d,$ where $d$ is some positive integer.

We can then write $2\cdot5d+5b+10c=80.$ Dividing both sides by $5$, we have $2d+b+2c=16.$ We divide by $2$ here to get $d+\dfrac b2+c=8.$ $d+c$ and $8$ are both integers, so $\dfrac b2$ is also an integer. $b$ must be divisible by $2$, so we let $b=2e$.

We now have $2d+2e+2c=16\implies d+e+c=8$. 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 $d,e,$ and $c$ such that they add to $8$.

We still have another constraint left, that each of $d,e,$ and $c$ must be at least $1$. For $n\in\{d,e,c\}$, let $n'=n-1.$ We are now looking for how many ways we can have $d'+e'+c'=8-1-1-1=5.$

We use a classic technique for solving these sorts of problems: stars and bars. We have $5$ stars and $3$ groups, which implies $2$ bars. Thus, the total number of ways is $\dbinom{5+2}2=\dbinom72=21.$

~Technodoggo ~minor edits by lucaswujc

Solution 2

Denote by $x$, $y$, $z$ the amount of $20 bills, $50 bills and $100 bills, respectively. Thus, we need to find the number of tuples $\left( x , y, z \right)$ with $x, y, z \in \Bbb N$ that satisfy \[ 20 x + 50 y + 100 z = 800. \]

First, this equation can be simplified as \[ 2 x + 5 y + 10 z = 80. \]

Second, we must have $5 |x$. Denote $x = 5 x'$. The above equation can be converted to \[ 2 x' + y + 2 z = 16 . \]

Third, we must have $2 | y$. Denote $y = 2 y'$. The above equation can be converted to \[ x' + y' + z = 8 . \]

Denote $x'' = x' - 1$, $y'' = y' - 1$ and $z'' = z - 1$. Thus, the above equation can be written as \[ x'' + y'' + z'' = 5 . \]

Therefore, the number of non-negative integer solutions $\left( x'', y'', z'' \right)$ is $\binom{5 + 3 - 1}{3 - 1} = \boxed{\textbf{(B) 21}}$.

~stephen chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 3

To start, we simplify things by dividing everything by $10$, the resulting equation is $2x+5y+10z=80$, and since the problem states that we have at least one of each, we simplify this to $2x+5y+10z=63$. Note that since the total is odd, we need an odd number of $5$ dollar bills. We proceed using casework.

Case 1: One $5$ dollar bill

$2x+10z=58$, we see that $10z$ can be $0,10,20,30,40,50$ or $6$ ways

Case 2: Three $5$ dollar bills

$2x+10z=48$, like before we see that $10z$ can be $0,10,20,30,40$, so $5$ ways

Now we should start to see a pattern emerge, each case there is $1$ less way to sum to $80$, so the answer is just $\frac{6(6+1)}{2}$, $21$ or $(B)$

~andyluo

Solution 4

We notice that each $100 can be split 3 ways: 5 $20 dollar bills, 2 $50 dollar bills, or 1 $100 dollar bill.

There are 8 of these $100 chunks in total--take away 3 as each split must be used at least once.

Now there are five left--so we use stars and bars.

5 chunks, 3 categories or 2 bars. This gives us $\binom{5+2}{2}=\boxed{\textbf{(B) 21}}$

~not_slay

Solution 5 (generating functions)

The problem is equivalent to the number of ways to make $$80$ from $$2$ bills, $$5$ bills, and $$10$ bills. We can use generating functions to find the coefficient of $x^{80}$:

The $$2$ bills provide $1+x^2+x^4+x^6 \cdots = \frac{1}{1-x^2},$

The $$5$ bills provide $1+x^5+x^{10}+x^{15} \cdots = \frac{1}{1-x^5},$

The $$10$ bills provide $1+x^{10}+x^{20}+x^{30} \cdots = \frac{1}{1-x^{10}}.$

Multiplying, we get $(1-x^{2})^{-1}(1-x^{5})^{-1}(1-x^{10})^{-1}.$

Solution 6 (easy logic)

There aren't dollar signs because the $latex$ thinks they're latex symbols. If you find how to override this error, please edit this. There's no $latex$ here but feel free to add some! ~SwordAxe

We can see in the problem that the teller gave her at least one of 20, 50, and 100. Therefore, she has 800 - 20 - 50 - 100 = 630 "left over".

Since all bills and 630 are multiples of 10, we can divide by ten. ==> Question becomes: How many different collections of 2, 5, and 10 could she get if her total was 63?

We notice that because 63 is odd, we need an odd amount of 5 bills. (2 and 10 are both even, and 63 is not a multiple of 5, so we need 2 and/or 10 bills. PM SwordAxe if you don't get this.)

We can do casework.

1: She gets one 5 (50) dollar bill. She has 58 (580) left. 

 1) She is given only 2 dollar (20) bills => ONE COLLECTION (all 20 bills with one 50)
 2) She is given one 10 dollar (100) bill
      1. The rest of the money is given in 2 dollar(20) dollar bills. => ONE COLLECTION (one 100 and rest 20 with one 50)
      2. She is given another 10 dollar (100) bill
            I) The rest of the money is given in 2 dollar (20) dollar bills. => ONE COLLECTION (two 100, one 50, and rest 20)
            II) She is given another 10 dollar (100) bill
                   a) The rest of the money is given in 2 dollar (20) dollar bills. => ONE COLLECTION (three 100, one 50, and rest 20)
                   b) She is given another 10 dollar (100) bill.
                         1) The rest of the money is given in 2 dollar (20) dollar bills  => ONE COLLECTION (following same pattern)
                         2)  AND SO ON...

This looks very tedious, but draw a simple tree diagram, and you'll see that its very easy!

If she gets one 50, there are 6 ways If she gets three 50, there are 5 ways ... If she gets nine 50, there are 2 ways If she gets eleven 50, there is one way

We can add them all up, with a grand sum of 6+5+4+3+2+1 = 21.

Therefore, we get the answer (B) 21.

~SwordAxe (PM me if you have any questions! :))

Video Solution 1 by OmegaLearn

https://youtu.be/UeX3eEwRS9I

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=sfZRRsTimmE

Video Solution 3 by paixiao

https://youtu.be/EvA2Nlb7gi4?si=fVLG8gMTIC5XkEwP&t=89s

Video Solution 4

https://youtu.be/D-ZvFBiZsaY

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution 5 by Lucas637

https://www.youtube.com/watch?v=kXLHjclTD44&t=27s

2023 AMC 10B (Problems • Answer Key • Resources)
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Problem 10 		Followed by
Problem 12
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All AMC 10 Problems and Solutions
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