Finding the Least Number to Subtract for Divisibility by 17

The question asks for the smallest number that needs to be subtracted from 2001 so that the resulting number is perfectly divisible by 17. To find this, we need to determine the remainder when 2001 is divided by 17.

Step-by-Step Division

We perform the division of 2001 by 17:

$$ 2001 \div 17 $$

Let's break down the division process:

• Divide 20 by 17. The quotient is 1 and the remainder is $20 - (17 \times 1) = 3$.
• Bring down the next digit (0) to make 30.
• Divide 30 by 17. The quotient is 1 and the remainder is $30 - (17 \times 1) = 13$.
• Bring down the next digit (1) to make 131.
• Divide 131 by 17. We find that $17 \times 7 = 119$. The quotient is 7 and the remainder is $131 - 119 = 12$.

So, the division can be represented as:

$$ 2001 = (17 \times 117) + 12 $$

The quotient is 117, and the remainder is 12.

Identifying the Number to Subtract

The remainder (12) is the amount by which 2001 exceeds the nearest multiple of 17 (which is $17 \times 117 = 1989$). Therefore, to get a number exactly divisible by 17, we must subtract this remainder from the original number.

The least number to be subtracted is the remainder, which is 12.

$$ 2001 - 12 = 1989 $$

Checking the result:

$$ 1989 \div 17 = 117 $$

Since 1989 is perfectly divisible by 17, subtracting 12 is the correct operation.

Conclusion

The least number that must be subtracted from 2001 to obtain a number exactly divisible by 17 is 12.

Options Analysis

The options provided were:

• 12
• 13
• 11
• 14

Our calculation shows that the remainder is 12, which is the number that needs to be subtracted. Therefore, option 1 (12) is the correct answer.