In how many different ways can you make change for 55 cents using nothing but nickel and / or dimes

1 answer

To find the different ways to make change for 55 cents using only nickels (5 cents) and dimes (10 cents), we can set up an equation.

Let \( n \) be the number of nickels and \( d \) be the number of dimes. The value of the change can be expressed as:

\[ 5n + 10d = 55 \]

We can simplify this equation by dividing everything by 5:

\[ n + 2d = 11 \]

Now, we will express \( n \) in terms of \( d \):

\[ n = 11 - 2d \]

To ensure \( n \) is a non-negative integer, the right side of the equation \( 11 - 2d \) must be non-negative:

\[ 11 - 2d \geq 0 \]

This simplifies to:

\[ 11 \geq 2d \quad \text{or} \quad d \leq 5.5 \]

Since \( d \) must be a whole number, the maximum possible value for \( d \) is 5. Therefore, \( d \) can take any integer value from 0 to 5.

Now, let's enumerate the possible values of \( d \) and the corresponding \( n \):

  • If \( d = 0 \), then \( n = 11 - 2(0) = 11 \).
  • If \( d = 1 \), then \( n = 11 - 2(1) = 9 \).
  • If \( d = 2 \), then \( n = 11 - 2(2) = 7 \).
  • If \( d = 3 \), then \( n = 11 - 2(3) = 5 \).
  • If \( d = 4 \), then \( n = 11 - 2(4) = 3 \).
  • If \( d = 5 \), then \( n = 11 - 2(5) = 1 \).

Thus, the pairs \((n, d)\) that satisfy the equation are:

  1. \( (11, 0) \)
  2. \( (9, 1) \)
  3. \( (7, 2) \)
  4. \( (5, 3) \)
  5. \( (3, 4) \)
  6. \( (1, 5) \)

This gives us a total of 6 different combinations of nickels and dimes that can make change for 55 cents.

Thus, the final answer is:

\[ \boxed{6} \]