Solving the 999 × abc = def132 Decimal Puzzle

We are given a multiplication problem in decimal notation: $\(999 \times \text{abc} = \text{def132}$\) Here, \(a, b, c, d, e,\) and \(f\) represent single digits. The notation \(abc\) represents the three-digit number \(100a + 10b + c\), and \(def132\) represents the six-digit number \(100000d + 10000e + 1000f + 132\). We need to find the values of these six digits.

The key to solving this puzzle is to understand the multiplication by 999. We can rewrite 999 as \((1000 - 1)\). So, the equation becomes: $\((1000 - 1) \times \text{abc} = \text{def132}$\) $\(1000 \times \text{abc} - 1 \times \text{abc} = \text{def132}$\) $\(1000 \times \text{abc} - \text{abc} = \text{def132}$\)

Using the Property \(1000 \times N - N\)

Let \(N\) be the three-digit number \(abc\). \(1000 \times N\) is simply the number \(N\) followed by three zeros. So, \(1000 \times \text{abc}\) is \(\text{abc}000\). The equation is thus: $\(\text{abc}000 - \text{abc} = \text{def132}$\)

We can perform this subtraction using columnar subtraction, looking at the digits from right to left.

$\( \begin{array}{@{}c@{\,}c@{}c@{}c@{}c@{}c} & a & b & c & 0 & 0 & 0 \\ - & & & & a & b & c \\ \hline & d & e & f & 1 & 3 & 2 \\ \end{array} $\)

Determining the Digits a, b, c

Let's look at the last three digits of the result, which are 1, 3, and 2. These result from subtracting \(abc\) from the trailing zeros of \(abc000\).

• Units digit (2): The rightmost subtraction is \(0 - c\). To get a result of 2, we must borrow from the left. So, \(10 - c = 2\). This gives us \(c = 10 - 2 = 8\).
• Tens digit (3): The next subtraction is from the tens place, which was originally 0. After lending 1 to the units place, it becomes 9. So, the subtraction is \(9 - b\). To get a result of 3, we have \(9 - b = 3\). This gives us \(b = 9 - 3 = 6\).
• Hundreds digit (1): The next subtraction is from the hundreds place, which was originally 0. After lending 1 to the tens place (which got it from the thousands place), it becomes 9. So, the subtraction is \(9 - a\). To get a result of 1, we have \(9 - a = 1\). This gives us \(a = 9 - 1 = 8\).

Based on the last three digits (132), the number \(abc\) must be 868. So, \(a=8\), \(b=6\), and \(c=8\).

Calculating the Full Product \(999 \times 868\)

Now that we know \(abc = 868\), we can calculate the full product \(999 \times 868\) and compare it to \(def132\). $\(999 \times 868 = (1000 - 1) \times 868$\) $\(= 1000 \times 868 - 868$\) $\(= 868000 - 868$\)

Let's perform the subtraction:

$\( \begin{array}{@{}c@{\,}c@{}c@{}c@{}c@{}c} & 8 & 6 & 8 & 0 & 0 & 0 \\ - & & & & 8 & 6 & 8 \\ \hline & 8 & 6 & 7 & 1 & 3 & 2 \\ \end{array} $\)

The result of \(999 \times 868\) is \(867132\).

Determining the Digits d, e, f

The problem states that the result is \(def132\). We calculated the result to be \(867132\). Comparing \(def132\) with \(867132\):

• The thousands digit (\(f\)) is 7.
• The ten thousands digit (\(e\)) is 6.
• The hundred thousands digit (\(d\)) is 8.

So, \(d=8\), \(e=6\), and \(f=7\).

Final Values of a, b, c, d, e, f

Combining the values we found:

• \(a = 8\)
• \(b = 6\)
• \(c = 8\)
• \(d = 8\)
• \(e = 6\)
• \(f = 7\)

The values of \(a, b, c, d, e,\) and \(f\) respectively are \(8, 6, 8, 8, 6, 7\).

Revision Table for Decimal Multiplication Puzzle

Let's quickly review the key steps and findings in this decimal puzzle involving multiplication by 999.

• The problem is \(999 \times \text{abc} = \text{def132}\).
• Using \(999 = 1000 - 1\), the equation becomes \(1000 \times \text{abc} - \text{abc} = \text{def132}\).
• This means \(\text{abc}000 - \text{abc} = \text{def132}\).
• The last three digits of the result (132) help determine \(a, b, c\) through subtraction logic (\(10-c=2, 9-b=3, 9-a=1\)).
• This yielded \(a=8, b=6, c=8\), so \(\text{abc}=868\).
• Calculating \(999 \times 868 = 868000 - 868 = 867132\).
• Comparing \(867132\) with \(def132\) determined \(d=8, e=6, f=7\).
• The final sequence of digits is \(8, 6, 8, 8, 6, 7\).

Additional Information on Number Properties

Understanding properties of numbers, especially numbers close to powers of 10 like 999 (which is \(10^3 - 1\)), can simplify multiplication problems. Multiplying a number \(N\) by \(10^k - 1\) can be done by calculating \(N \times 10^k - N\). This is often easier than direct multiplication.

In this problem, multiplying by \(999 = 1000 - 1\): $\(N \times 999 = N \times (1000 - 1) = N \times 1000 - N$\) If \(N\) is a 3-digit number \(abc\), then \(N \times 1000 = abc000\). Subtracting \(N=abc\) from \(abc000\) gives: $\(\text{abc}000 - \text{abc}$\) The last three digits of the result come from \(1000 - \text{abc}\) considering borrows across the places. Specifically, the hundreds, tens, and units digits of \(N\) are determined by the thousands, hundreds, and tens digits of \(1000\) and the corresponding digits of the result's last three digits.

• Units digit of result = \(10 - c\) (with borrow from tens of thousands)
• Tens digit of result = \((10-1) - b\) (after borrowing from hundreds of thousands)
• Hundreds digit of result = \((10-1) - a\) (after borrowing from thousands of thousands, i.e., the \(c\) place)

In our case, the last three digits are 132. \(10 - c \equiv 2 \pmod{10} \implies c = 8\). \((10-1) - b = 3 \implies 9 - b = 3 \implies b = 6\). \((10-1) - a = 1 \implies 9 - a = 1 \implies a = 8\).

This confirms \(abc=868\). The remaining digits of the result \(def\) come from the subtraction \(868000 - 868\). The thousands place of \(868000\) becomes \(867\) after the borrows necessary for the last three digits (\(868000\) becomes \(867\) thousand and \(1000\), where \(1000-868=132\)). So \(867\) forms the leading digits \(def\).

$\(868000 - 868 = 867132$\) \(d=8, e=6, f=7\).