Understanding the Weight Measurement Problem

The goal is to find the maximum number of distinct combinations of weights (1 kg, 2 kg, 5 kg, 10 kg) that sum up to exactly 20 kg. A key constraint is that the number of 1 kg weights used must be between 8 and 11 (inclusive).

Let $n_1, n_2, n_5, n_{10}$ represent the number of 1 kg, 2 kg, 5 kg, and 10 kg weights used, respectively. The total weight equation is:

$1 \cdot n_1 + 2 \cdot n_2 + 5 \cdot n_5 + 10 \cdot n_{10} = 20$

The constraint on the 1 kg weights is $8 \le n_1 \le 11$. We examine each possible value for $n_1$.

Calculating Ways for Each 1 kg Weight Count

Case 1: Using 8 weights of 1 kg ($n_1 = 8$)

The remaining weight needed is $20 \text{ kg} - 8 \times 1 \text{ kg} = 12 \text{ kg}$.

We need to find combinations for $2n_2 + 5n_5 + 10n_{10} = 12$.

• If $n_{10} = 1$: $2n_2 + 5n_5 = 12 - 10 = 2$. The only solution is $n_5=0, n_2=1$. Combination: (1, 0, 1).
• If $n_{10} = 0$: $2n_2 + 5n_5 = 12$.
  • If $n_5=0$, $2n_2=12 \implies n_2=6$. Combination: (6, 0, 0).
  • If $n_5=1$, $2n_2=7$. No integer solution.
  • If $n_5=2$, $2n_2=2 \implies n_2=1$. Combination: (1, 2, 0).

Total ways for $n_1=8$: 3

Case 2: Using 9 weights of 1 kg ($n_1 = 9$)

The remaining weight needed is $20 \text{ kg} - 9 \times 1 \text{ kg} = 11 \text{ kg}$.

We need combinations for $2n_2 + 5n_5 + 10n_{10} = 11$.

• If $n_{10} = 1$: $2n_2 + 5n_5 = 1$. No non-negative integer solution.
• If $n_{10} = 0$: $2n_2 + 5n_5 = 11$.
  • If $n_5=0$, $2n_2=11$. No integer solution.
  • If $n_5=1$, $2n_2=6 \implies n_2=3$. Combination: (3, 1, 0).
  • If $n_5=2$, $2n_2=1$. No integer solution.

Total ways for $n_1=9$: 1

Case 3: Using 10 weights of 1 kg ($n_1 = 10$)

The remaining weight needed is $20 \text{ kg} - 10 \times 1 \text{ kg} = 10 \text{ kg}$.

We need combinations for $2n_2 + 5n_5 + 10n_{10} = 10$.

• If $n_{10} = 1$: $2n_2 + 5n_5 = 0$. The only solution is $n_2=0, n_5=0$. Combination: (0, 0, 1).
• If $n_{10} = 0$: $2n_2 + 5n_5 = 10$.
  • If $n_5=0$, $2n_2=10 \implies n_2=5$. Combination: (5, 0, 0).
  • If $n_5=1$, $2n_2=5$. No integer solution.
  • If $n_5=2$, $2n_2=0 \implies n_2=0$. Combination: (0, 2, 0).

Total ways for $n_1=10$: 3

Case 4: Using 11 weights of 1 kg ($n_1 = 11$)

The remaining weight needed is $20 \text{ kg} - 11 \times 1 \text{ kg} = 9 \text{ kg}$.

We need combinations for $2n_2 + 5n_5 + 10n_{10} = 9$.

• If $n_{10} = 1$: $2n_2 + 5n_5 = -1$. No non-negative integer solution.
• If $n_{10} = 0$: $2n_2 + 5n_5 = 9$.
  • If $n_5=0$, $2n_2=9$. No integer solution.
  • If $n_5=1$, $2n_2=4 \implies n_2=2$. Combination: (2, 1, 0).

Total ways for $n_1=11$: 1

Total Maximum Number of Ways

Summing the ways from each case:

Total Ways = (Ways for $n_1=8$) + (Ways for $n_1=9$) + (Ways for $n_1=10$) + (Ways for $n_1=11$)

Total Ways = $3 + 1 + 3 + 1 = 8$

Therefore, the maximum number of different ways to measure 20 kg under the given conditions is 8.