A machine dispenses four kinds of tokens. Red tokens are worth 25 points, blue
tokens are worth 10 points, green tokens are worth 5 points, and yellow tokens
are worth 1 point. The machine never runs out of tokens to dispense.
To get tokens from the machine, people enter a total number of points, and the
machine will dispense the fewest number of tokens that total that number of
points. For example, if someone enters 30 points, the machine will dispense 1 red
token and 1 green token, because their total value is 25 + 5 = 30 points, and this
total cannot be obtained using fewer tokens.
For how many different total numbers of points will the machine dispense exactly
4 tokens?

To solve this problem, we need to determine for how many different total numbers of points the machine will dispense exactly 4 tokens, using red tokens (worth 25 points), blue tokens (worth 10 points), green tokens (worth 5 points), and yellow tokens (worth 1 point).

### Approach:
We need to minimize the number of tokens while achieving a specific number of points. Given that exactly 4 tokens are dispensed, we need to account for all possible combinations of these tokens, and calculate the total points for each combination. The task is to find the different total numbers of points the machine can dispense using exactly 4 tokens.

Let the number of red tokens be denoted as \( r \), blue tokens as \( b \), green tokens as \( g \), and yellow tokens as \( y \). The total number of tokens is constrained by:

\[
r + b + g + y = 4
\]

The total points are given by:

\[
25r + 10b + 5g + 1y
\]

We now consider different combinations of \( r \), \( b \), \( g \), and \( y \) that satisfy \( r + b + g + y = 4 \), and calculate the corresponding total points.

### Case 1: \( r = 0 \)
In this case, we need to satisfy \( b + g + y = 4 \).
- If \( b = 0 \), we have \( g + y = 4 \). The total points are \( 5g + 1y \):
  - \( g = 0, y = 4 \) → total points = \( 5(0) + 1(4) = 4 \)
  - \( g = 1, y = 3 \) → total points = \( 5(1) + 1(3) = 8 \)
  - \( g = 2, y = 2 \) → total points = \( 5(2) + 1(2) = 12 \)
  - \( g = 3, y = 1 \) → total points = \( 5(3) + 1(1) = 16 \)
  - \( g = 4, y = 0 \) → total points = \( 5(4) + 1(0) = 20 \)
  
- If \( b = 1 \), we have \( g + y = 3 \). The total points are \( 10(1) + 5g + 1y \):
  - \( g = 0, y = 3 \) → total points = \( 10(1) + 5(0) + 1(3) = 13 \)
  - \( g = 1, y = 2 \) → total points = \( 10(1) + 5(1) + 1(2) = 17 \)
  - \( g = 2, y = 1 \) → total points = \( 10(1) + 5(2) + 1(1) = 21 \)
  - \( g = 3, y = 0 \) → total points = \( 10(1) + 5(3) + 1(0) = 25 \)

- If \( b = 2 \), we have \( g + y = 2 \). The total points are \( 10(2) + 5g + 1y \):
  - \( g = 0, y = 2 \) → total points = \( 10(2) + 5(0) + 1(2) = 22 \)
  - \( g = 1, y = 1 \) → total points = \( 10(2) + 5(1) + 1(1) = 26 \)
  - \( g = 2, y = 0 \) → total points = \( 10(2) + 5(2) + 1(0) = 30 \)

- If \( b = 3 \), we have \( g + y = 1 \). The total points are \( 10(3) + 5g + 1y \):
  - \( g = 0, y = 1 \) → total points = \( 10(3) + 5(0) + 1(1) = 31 \)
  - \( g = 1, y = 0 \) → total points = \( 10(3) + 5(1) + 1(0) = 35 \)

- If \( b = 4 \), we have \( g + y = 0 \). The total points are \( 10(4) + 5(0) + 1(0) = 40 \).

### Case 2: \( r = 1 \)
In this case, we need to satisfy \( b + g + y = 3 \). The total points are \( 25(1) + 10b + 5g + 1y \):
- If \( b = 0 \), we have \( g + y = 3 \). The total points are \( 25 + 5g + 1y \):
  - \( g = 0, y = 3 \) → total points = \( 25 + 5(0) + 1(3) = 28 \)
  - \( g = 1, y = 2 \) → total points = \( 25 + 5(1) + 1(2) = 32 \)
  - \( g = 2, y = 1 \) → total points = \( 25 + 5(2) + 1(1) = 36 \)
 

A machine dispenses four kinds of tokens. Red tokens are worth 25 points, blue tokens are worth 10 points, green tokens are worth 5 points, and yellow tokens are worth 1 point. The machine never runs out of tokens to dispense. To get tokens from the machine, people enter a total number of points, and the machine will dispense the fewest number of tokens that total that number of points. For example, if someone enters 30 points, the machine will dispense 1 red token and 1 green token, because their total value is 25 + 5 = 30 points, and this total cannot be obtained using fewer tokens. For how many different total numbers of points will the machine dispense exactly 4 tokens?

This problem explores how to dispense the fewest number of tokens from a machine that offers red, blue, green, and yellow tokens, worth 25, 10, 5, and 1 point respectively. We calculate for how many different total points the machine will dispense exactly 4 tokens using combinatorics and optimization techniques. Suitable for students in Grades 8-10.

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