To determine the coins that add up to 25 cents using exactly 4 coins, consider the common U.S. coin denominations: pennies (1 cent), nickels (5 cents), dimes (10 cents), and quarters (25 cents).
Let \( p \) be the number of pennies, \( n \) be the number of nickels, \( d \) be the number of dimes, and \( q \) be the number of quarters. The total number of coins is 4, and the total value is 25 cents. This gives the following equations:
\[
p + n + d + q = 4
\]
\[
1p + 5n + 10d + 25q = 25
\]
Since quarters are 25 cents each, having one quarter would already satisfy the total value, but it would only use 1 coin, not 4. Therefore, \( q = 0 \). Now, the equations simplify to:
\[
p + n + d = 4
\]
\[
1p + 5n + 10d = 25
\]
Next, consider the possible values for \( d \) (number of dimes). Since dimes are 10 cents each, the maximum number of dimes that can be used without exceeding 25 cents is 2.
Substitute \( d = 2 \) into the equations:
\[
p + n + 2 = 4 \implies p + n = 2
\]
\[
1p + 5n + 10(2) = 25 \implies p + 5n = 5
\]
Subtract the first equation from the second:
\[
(p + 5n) - (p + n) = 5 - 2 \implies 4n = 3 \implies n = 0.75
\]
Since \( n \) must be an integer, this case is invalid.
Substitute \( d = 1 \) into the equations:
\[
p + n + 1 = 4 \implies p + n = 3
\]
\[
1p + 5n + 10(1) = 25 \implies p + 5n = 15
\]
Subtract the first equation from the second:
\[
(p + 5n) - (p + n) = 15 - 3 \implies 4n = 12 \implies n = 3
\]
Then, \( p = 0 \). This gives the combination: 0 pennies, 3 nickels, and 1 dime.
Substitute \( d = 0 \) into the equations:
\[
p + n + 0 = 4 \implies p + n = 4
\]
\[
1p + 5n + 10(0) = 25 \implies p + 5n = 25
\]
Subtract the first equation from the second:
\[
(p + 5n) - (p + n) = 25 - 4 \implies 4n = 21 \implies n = 5.25
\]
Since \( n \) must be an integer, this case is invalid.
The only valid combination is 0 pennies, 3 nickels, and 1 dime. This satisfies both the total number of coins (4) and the total value (25 cents).
\(\boxed{0 \text{ pennies}, 3 \text{ nickels}, 1 \text{ dime}}\)
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