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 $def$.

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