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Difference between revisions of "2000 AMC 8 Problems/Problem 20"

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==Problem==
 
==Problem==
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<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of &#36;<math>1.02</math>, with at least one coin of each type. How many dimes must you have?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>
  
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You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of &#36;<math>1.02</math>, with at least one coin of each type. How many dimes must you have?
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<math>\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 3\qquad\text{(D)}\ 4\qquad\text{(E)}\ 5</math>
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<math> \text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 3\qquad\text{(D)}\ 4\qquad\text{(E)}\ 5 </math>
 
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==Solution==
 
  
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==Solution 1==
 
Since you have one coin of each type, <math>1 + 5 + 10 + 25 = 41</math> cents are already determined, leaving you with a total of <math>102 - 41 = 61</math> cents remaining for <math>5</math> coins.
 
Since you have one coin of each type, <math>1 + 5 + 10 + 25 = 41</math> cents are already determined, leaving you with a total of <math>102 - 41 = 61</math> cents remaining for <math>5</math> coins.
  
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There is only <math>1</math> dime in that combo, so the answer is <math>\boxed{A}</math>.
 
There is only <math>1</math> dime in that combo, so the answer is <math>\boxed{A}</math>.
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==Solution 2 (Faster)==
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We see that there must be 102 cents, so therefore there's at least 2 pennies. That leaves 7 coins. We assume that there are 3 quarters, leaving 25 cents with 4 coins left. If all 4 are nickels, that would only be 20 cents, missing 5. Therefore, one nickel must be changed into 1 dime, so the answer is <math>\boxed{A}</math>
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Solution by [[User:ILoveMath31415926535|ILoveMath31415926535]]
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==Solution 3==
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It is clear that there should only be <math>2</math> pennies; any more would take up too many coins, and the limit is <math>9</math>. Now we have &#36;<math>1</math> left, and <math>7</math> coins to use. Looking at the quarters, we can make methodical guesses. If there is <math>1</math> quarter, then we would have to make &#36;<math>0.75</math> with <math>6</math> coins. We take a few educated guesses for the nickel and dime combos, and see that <math>1</math> quarter will not work. Trying values for <math>2</math> quarters, we see that this will not work either. When we reach <math>3</math> quarters, the remaining is &#36;<math>0.25</math> made from <math>4</math> coins. We try with <math>2</math> dimes, which does not work (it only takes <math>3</math> coins) and we try with <math>1</math> dime. After trying <math>2</math> pennies, <math>3</math> quarters, <math>1</math> dime, and <math>3</math> nickels, it is evident that this combo works. Therefore, the answer is <math>\boxed{A}</math>
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*It might seem that this solution takes time, but in reality, it can be done very quickly, given ones desire to solve a problem quickly
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~mk
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== Video Solution by OmegaLearn ==
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https://youtu.be/HISL2-N5NVg?t=1409
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~ pi_is_3.14
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== Video Solution 2 ==
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https://youtu.be/Vm-proRV5wI. Soo, DRMS, NM
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https://www.youtube.com/watch?v=un-zQJRS49k    ~David
  
 
==See Also==
 
==See Also==
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{{AMC8 box|year=2000|num-b=19|num-a=21}}
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{{AMC8 box|year=2000|num-b=18|num-a=20}}
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{{MAA Notice}}

Latest revision as of 19:53, 20 November 2023

Problem

You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $$1.02$, with at least one coin of each type. How many dimes must you have?

$\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 3\qquad\text{(D)}\ 4\qquad\text{(E)}\ 5$

Solution 1

Since you have one coin of each type, $1 + 5 + 10 + 25 = 41$ cents are already determined, leaving you with a total of $102 - 41 = 61$ cents remaining for $5$ coins.

You must have $1$ more penny. If you had more than $1$ penny, you must have at least $6$ pennies to leave a multiple of $5$ for the nickels, dimes, and quarters. But you only have $5$ more coins to assign.

Now you have $61 - 1 = 60$ cents remaining for $4$ coins, which may be nickels, quarters, or dimes. If you have only one more dime, that leaves $50$ cents in $3$ nickels or quarters, which is impossible. If you have two dimes, that leaves $40$ cents for $2$ nickels or quarters, which is again impossible. If you have three dimes, that leaves $30$ cents for $1$ nickel or quarter, which is still impossible. And all four remaining coins being dimes will not be enough.

Therefore, you must have no more dimes to assign, and the $60$ cents in $4$ coins must be divided between the quarters and nickels. We quickly see that $2$ nickels and $2$ quarters work. Thus, the total count is $2$ quarters, $2$ nickels, $1$ penny, plus one more coin of each type that we originally subtracted. Double-checking, that gives a total $2 + 2 + 1 + 4 = 9$ coins, and a total of $2\cdot 25 + 2\cdot 5 + 1 + (1 + 5 + 10 + 25) = 102$ cents.

There is only $1$ dime in that combo, so the answer is $\boxed{A}$.


Solution 2 (Faster)

We see that there must be 102 cents, so therefore there's at least 2 pennies. That leaves 7 coins. We assume that there are 3 quarters, leaving 25 cents with 4 coins left. If all 4 are nickels, that would only be 20 cents, missing 5. Therefore, one nickel must be changed into 1 dime, so the answer is $\boxed{A}$

Solution by ILoveMath31415926535


Solution 3

It is clear that there should only be $2$ pennies; any more would take up too many coins, and the limit is $9$. Now we have $$1$ left, and $7$ coins to use. Looking at the quarters, we can make methodical guesses. If there is $1$ quarter, then we would have to make $$0.75$ with $6$ coins. We take a few educated guesses for the nickel and dime combos, and see that $1$ quarter will not work. Trying values for $2$ quarters, we see that this will not work either. When we reach $3$ quarters, the remaining is $$0.25$ made from $4$ coins. We try with $2$ dimes, which does not work (it only takes $3$ coins) and we try with $1$ dime. After trying $2$ pennies, $3$ quarters, $1$ dime, and $3$ nickels, it is evident that this combo works. Therefore, the answer is $\boxed{A}$

  • It might seem that this solution takes time, but in reality, it can be done very quickly, given ones desire to solve a problem quickly

~mk

Video Solution by OmegaLearn

https://youtu.be/HISL2-N5NVg?t=1409

~ pi_is_3.14

Video Solution 2

https://youtu.be/Vm-proRV5wI. Soo, DRMS, NM

https://www.youtube.com/watch?v=un-zQJRS49k ~David

See Also

2000 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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 AJHSME/AMC 8 Problems and Solutions

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