Difference between revisions of "2004 AMC 10B Problems/Problem 13"
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<math> \mathrm{(A) \ } 7 \qquad \mathrm{(B) \ } 8 \qquad \mathrm{(C) \ } 9 \qquad \mathrm{(D) \ } 10 \qquad \mathrm{(E) \ } 11 </math> | <math> \mathrm{(A) \ } 7 \qquad \mathrm{(B) \ } 8 \qquad \mathrm{(C) \ } 9 \qquad \mathrm{(D) \ } 10 \qquad \mathrm{(E) \ } 11 </math> | ||
− | == Solution == | + | == Solution 1 == |
All numbers in this solution will be in hundredths of a millimeter. | All numbers in this solution will be in hundredths of a millimeter. | ||
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Using the above observation we can easily construct such a stack. A stack of <math>8</math> dimes would have height <math>8\cdot 135=1080</math>, thus we need to add <math>320</math>. | Using the above observation we can easily construct such a stack. A stack of <math>8</math> dimes would have height <math>8\cdot 135=1080</math>, thus we need to add <math>320</math>. | ||
This can be done for example by replacing five dimes by nickels (for <math>+60\cdot 5 = +300</math>), and one dime by a penny (for <math>+20</math>). | This can be done for example by replacing five dimes by nickels (for <math>+60\cdot 5 = +300</math>), and one dime by a penny (for <math>+20</math>). | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | Let <math>p,n,d</math>, and <math>q</math> be the number of pennies, nickels, dimes, and quarters used in the stack. | ||
+ | |||
+ | From the conditions above, we get the following equation: | ||
+ | |||
+ | <cmath>155p+195n+135d+175q=1400.</cmath> | ||
+ | |||
+ | Then we divide each side by five to get | ||
+ | |||
+ | <cmath>31p+39n+27d+35q=280.</cmath> | ||
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+ | Writing both sides in terms of mod 4, we have <math>-p-n-d-q \equiv 0 \pmod 4</math>. | ||
+ | |||
+ | This means that the sum <math>p+n+d+q</math> is divisible by 4. Therefore, the answer must be <math>\boxed{(B)\,\, 8}.</math> | ||
==Note== | ==Note== |
Latest revision as of 16:01, 25 July 2019
Contents
Problem
In the United States, coins have the following thicknesses: penny, mm; nickel, mm; dime, mm; quarter, mm. If a stack of these coins is exactly mm high, how many coins are in the stack?
Solution 1
All numbers in this solution will be in hundredths of a millimeter.
The thinnest coin is the dime, with thickness . A stack of dimes has height .
The other three coin types have thicknesses , , and . By replacing some of the dimes in our stack by other, thicker coins, we can clearly create exactly all heights in the set .
If we take an odd , then all the possible heights will be odd, and thus none of them will be . Hence is even.
If the stack will be too low and if it will be too high. Thus we are left with cases and .
If the possible stack heights are , with the remaining ones exceeding .
Therefore there are coins in the stack.
Using the above observation we can easily construct such a stack. A stack of dimes would have height , thus we need to add . This can be done for example by replacing five dimes by nickels (for ), and one dime by a penny (for ).
Solution 2
Let , and be the number of pennies, nickels, dimes, and quarters used in the stack.
From the conditions above, we get the following equation:
Then we divide each side by five to get
Writing both sides in terms of mod 4, we have .
This means that the sum is divisible by 4. Therefore, the answer must be
Note
We can easily add up and to get . We multiply that by to get . Since this works and it requires 8 coins, the answer is clearly .
Similarly, we can simply take quarters to get .
See also
2004 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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 |
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