Math  /  Numbers & Operations

Question20. How many different representations of -3 can be made with 20 or fewer counters?

Studdy Solution

STEP 1

1. We are using a system where counters can represent positive and negative values.
2. A positive counter and a negative counter can cancel each other out.
3. We need to find all combinations of positive and negative counters that result in a net value of 3-3.
4. The total number of counters used in each representation must be 20 or fewer.

STEP 2

1. Define the variables.
2. Set up an equation.
3. Determine the range of possible solutions.
4. Count the number of valid solutions.

STEP 3

Let p p represent the number of positive counters and n n represent the number of negative counters.

STEP 4

Set up the equation based on the net value of counters:
np=3 n - p = -3
Additionally, the total number of counters must satisfy:
n+p20 n + p \leq 20

STEP 5

Express n n in terms of p p using the first equation:
n=p3 n = p - 3
Substitute into the second equation:
(p3)+p20 (p - 3) + p \leq 20
Simplify:
2p320 2p - 3 \leq 20
Solve for p p :
2p23 2p \leq 23 p11.5 p \leq 11.5
Since p p must be an integer, p11 p \leq 11 .

STEP 6

Determine the range of p p and n n :
Since n=p3 n = p - 3 , and both n n and p p must be non-negative integers, we have:
p3 p \geq 3
Thus, the range for p p is 3p11 3 \leq p \leq 11 .
For each p p in this range, calculate n=p3 n = p - 3 .
Count the number of valid (p,n) (p, n) pairs:
p=3,4,5,,11 p = 3, 4, 5, \ldots, 11
This gives us 9 possible values for p p .
The number of different representations of 3-3 with 20 or fewer counters is:
9\boxed{9}

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