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

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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>.
 
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>.
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Similarly, we can simply take <math>8</math> quarters to get <math>8\cdot 1.75=14</math>.
  
 
== See also ==
 
== See also ==

Revision as of 14:29, 20 July 2019

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

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 $+20$).

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 (ProblemsAnswer KeyResources)
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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