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Difference between revisions of "2000 AMC 8 Problems/Problem 20"
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You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of <math>;1.02 </math>, with at least one coin of each type. How many dimes must you have?
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==Problem==
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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?
  
 
<math> \text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 3\qquad\text{(D)}\ 4\qquad\text{(E)}\ 5 </math>
 
<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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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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You must have <math>1</math> more penny.  If you had more than <math>1</math> penny, you must have at least <math>6</math> pennies to leave a multiple of <math>5</math> for the nickels, dimes, and quarters.  But you only have <math>5</math> more coins to assign.
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Now you have <math>61 - 1 = 60</math> cents remaining for <math>4</math> coins, which may be nickels, quarters, or dimes.  If you have only one more dime, that leaves <math>50</math> cents in <math>3</math> nickels or quarters, which is impossible.  If you have two dimes, that leaves <math>40</math> cents for <math>2</math> nickels or quarters, which is again impossible.  If you have three dimes, that leaves <math>30</math> cents for <math>1</math> nickel or quarter, which is still impossible.  And all four remaining coins being dimes will not be enough.
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Therefore, you must have no more dimes to assign, and the <math>60</math> cents in <math>4</math> coins must be divided between the quarters and nickels.  We quickly see that <math>2</math> nickels and <math>2</math> quarters work.  Thus, the total count is <math>2</math> quarters, <math>2</math> nickels, <math>1</math> penny, plus one more coin of each type that we originally subtracted.  Double-checking, that gives a total <math>2 + 2 + 1 + 4 = 9</math> coins, and a total of <math>2\cdot 25 + 2\cdot 5 + 1 + (1 + 5 + 10 + 25) = 102</math> cents.
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There is only <math>1</math> dime in that combo, so the answer is <math>\boxed{A}</math>.
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==See Also==
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{{AMC8 box|year=2000|num-b=18|num-a=20}}

Revision as of 20:27, 30 July 2011

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

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 $\boxed{A}$.

See Also

2000 AMC 8 (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
All AJHSME/AMC 8 Problems and Solutions
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