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. | ||
Line 22: | Line 22: | ||
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>\ | ||
+ | |||
==Note== | ==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>. | 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 == |
Latest revision as of 17:27, 8 January 2024
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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