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Difference between revisions of "2004 AMC 10B Problems/Problem 13"

(New page: == Problem == In the United States, coins have the following thicknesses: penny, <math>1.55</math> mm; nickel, <math>1.95</math> mm; dime, <math>1.35</math> mm; quarter, <math>1.75</math> ...)
 
m (→‎Solution 2)
 
(12 intermediate revisions by 10 users not shown)
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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 hundreds of a millimeter.
+
All numbers in this solution will be in hundredths of a millimeter.
  
 
The thinnest coin is the dime, with thickness <math>135</math>. A stack of <math>n</math> dimes has height <math>135n</math>.
 
The thinnest coin is the dime, with thickness <math>135</math>. A stack of <math>n</math> dimes has height <math>135n</math>.
Line 18: Line 18:
 
If <math>n=10</math> the possible stack heights are <math>1350,1370,1390,\dots</math>, with the remaining ones exceeding <math>1400</math>.
 
If <math>n=10</math> the possible stack heights are <math>1350,1370,1390,\dots</math>, with the remaining ones exceeding <math>1400</math>.
  
−
Therefore there are <math>\boxed{8}</math> coins in the stack.
+
Therefore there are <math>\boxed{\mathrm{(B)\ }8}</math> coins in the stack.
  
 
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>
 +
 
 +
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>
 +
 
 +
==Solution 3==
 +
 
 +
We notice that the thickness of 4 quarters is 7 mm. 7 is half of 14, so we multiply the 4 quarters by two and get <math>\boxed{(B)\,\, 8}</math>.
 +
 
 +
==Note==
 +
 
 +
We can easily add up <math>1.55\text{\ mm}</math> and <math>1.95\text{\ mm}</math> to get <math>3.50\text{\ mm}</math>. We multiply that by <math>4</math> to get <math>14\text{\ mm}</math>. Since this works and it requires 8 coins, the answer is clearly <math>\boxed{\mathrm{(B)\ }8}</math>.
 +
 
 +
Similarly, we can simply take <math>8</math> quarters to get <math>8\cdot 1.75=14</math>.
  
 
== See also ==
 
== See also ==
  
 
{{AMC10 box|year=2004|ab=B|num-b=12|num-a=14}}
 
{{AMC10 box|year=2004|ab=B|num-b=12|num-a=14}}
 +
{{MAA Notice}}

Latest revision as of 20:35, 30 September 2024

Problem

In the United States, coins have the following thicknesses: penny, $1.55$ mm; nickel, $1.95$ mm; dime, $1.35$ mm; quarter, $1.75$ mm. If a stack of these coins is exactly $14$ mm high, how many coins are in the stack?

$\mathrm{(A) \ } 7 \qquad \mathrm{(B) \ } 8 \qquad \mathrm{(C) \ } 9 \qquad \mathrm{(D) \ } 10 \qquad \mathrm{(E) \ } 11$

Solution 1

All numbers in this solution will be in hundredths of a millimeter.

The thinnest coin is the dime, with thickness $135$. A stack of $n$ dimes has height $135n$.

The other three coin types have thicknesses $135+20$, $135+40$, and $135+60$. By replacing some of the dimes in our stack by other, thicker coins, we can clearly create exactly all heights in the set $\{135n, 135n+20, 135n+40, \dots, 195n\}$.

If we take an odd $n$, then all the possible heights will be odd, and thus none of them will be $1400$. Hence $n$ is even.

If $n<8$ the stack will be too low and if $n>10$ it will be too high. Thus we are left with cases $n=8$ and $n=10$.

If $n=10$ the possible stack heights are $1350,1370,1390,\dots$, with the remaining ones exceeding $1400$.

Therefore there are $\boxed{\mathrm{(B)\ }8}$ coins in the stack.

Using the above observation we can easily construct such a stack. A stack of $8$ dimes would have height $8\cdot 135=1080$, thus we need to add $320$. This can be done for example by replacing five dimes by nickels (for $+60\cdot 5 = +300$), and one dime by a penny (for $+20$).

Solution 2

Let $p,n,d$, and $q$ be the number of pennies, nickels, dimes, and quarters used in the stack.

From the conditions above, we get the following equation:

\[155p+195n+135d+175q=1400.\]

Then we divide each side by five to get

\[31p+39n+27d+35q=280.\]

Writing both sides in terms of mod 4, we have $-p-n-d-q \equiv 0 \pmod 4$.

This means that the sum $p+n+d+q$ is divisible by 4. Therefore, the answer must be $\boxed{(B)\,\, 8}.$

Solution 3

We notice that the thickness of 4 quarters is 7 mm. 7 is half of 14, so we multiply the 4 quarters by two and get $\boxed{(B)\,\, 8}$.

Note

We can easily add up $1.55\text{\ mm}$ and $1.95\text{\ mm}$ to get $3.50\text{\ mm}$. We multiply that by $4$ to get $14\text{\ mm}$. Since this works and it requires 8 coins, the answer is clearly $\boxed{\mathrm{(B)\ }8}$.

Similarly, we can simply take $8$ quarters to get $8\cdot 1.75=14$.

See also

2004 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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All AMC 10 Problems and Solutions

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