# 3d x 1d = 2d x 2d

Find the largest integer that is a product of three-digit number and a one-digit number and also a product of two two-digit numbers.

For example, 200 x 1 = 10 x 20. Of course, 200 is not the largest.

• 3d x 1d = 2d x 2d; d = 0 – Avigrail Jan 29 '15 at 14:34
• @Avigrail You're right :-) – P.-S. Park Jan 29 '15 at 14:35

$8928 = 992 \times 9 = 96 \times 93$.

Explanation:

The only way to get to a number larger than $8000$ is if the number is of the form $(100-n)(100-m)=9(1000-k)$. The last two digits of the result equal minus a multiple of $9$ modulo $100$: $19$, $28$, $37$, $46$, $55$, $64$, $73$, $82$, or $91$. The smallest number in this list that occurs in the standard multiplication tables is $28$. So, the solution sought is $n=4$, $m=7$ and $100-9k=28$ or $k=8$.

• So the second smallest is $n=1, m=19$, which results in $81 * 99 = 8019 = 9 * 891$? – durron597 Jan 29 '15 at 18:18
• @durron597 - for the above argument to hold, $m$ and $n$ need to be single-digit numbers. – Johannes Jan 29 '15 at 19:13
• So the second smallest is $8464=92*92=9*940$ cos $37,46,55$ can't be expressed as the product of two one digit numbers but $64$ can. Beyond that, this argument doesn't hold and it can be that the third is indeed $8019$ but that would require a different proof. – chx Jan 29 '15 at 20:17
• @chx except that $9 * 940 = 8460$, not $8464$. However, $64 = 4 * 16 \rightarrow 96 * 84 = 8064 = 9 * 896 > 8019$ – durron597 Jan 30 '15 at 13:10

8928 = 992 * 9 = 96 * 93
9 - is obviosly largest 1-digit number, so we need to maximize 3-digit number so that their product can be a product of two 2-digit numbers.
Candidates are 993, 994, 995, 996, 997, 998, 999.
Let's find divisors of their products with 9 to show that they can't be products of 2-digit numbers. Write them in a row and in second row in reversed order. Example for 9*993:
1, 3, 9, 27, 331, 993, 2979, 8937
8937, 2979, 993, 331, 27, 9, 3, 1
Now it's easy to see that it cannot be product of 2-digit number (there will be two 2-digit numbers one above second if it is possible). In same way you check other candidates (not showing here, because long lists of divisors).

It took just a couple of minutes to come up with a program that quickly calculates this. It takes about 0.2 seconds to run. The variables I use are simply A*B=C*D and the for loops take care of the number of digits in each variable. Here's the output:

Largest Product = 8928
Largest A = 992
Largest B = 9
Largest C = 93
Largest D = 96

Here's the C# code:

static void Main(string[] args)
{
int largestProduct = 200;
int[] largestABCD = { 200, 1, 10, 20 };

for(int a = 100; a <= 999; a++)
{
for(int b = 1; b <= 9; b++)
{
int product1 = a * b;

for(int c = 10; c <= 99; c++)
{
for(int d = 10; d <= 99; d++)
{
int product2 = c * d;
if(product1 == product2)
{
if (product1 > largestProduct)
{
largestProduct = product1;
largestABCD[0] = a;
largestABCD[1] = b;
largestABCD[2] = c;
largestABCD[3] = d;
}
}
}
}
}
}

Console.WriteLine("Largest Product = " + largestProduct.ToString());
Console.WriteLine("Largest A = " + largestABCD[0].ToString());
Console.WriteLine("Largest B = " + largestABCD[1].ToString());
Console.WriteLine("Largest C = " + largestABCD[2].ToString());
Console.WriteLine("Largest D = " + largestABCD[3].ToString());

}


I could probably come up with some optimizations for this code. I could constrain the D loop counter. Or I could break out of a loop once product2 is greater than product1. But the code runs so fast that it doesn't matter.

• The problem with writing software to solve these sorts of problems is that they does not scale to larger problems well, whereas more mathematical approaches do. – BlueRaja - Danny Pflughoeft Jan 29 '15 at 22:17

I have a simpler way to get to the answer. Clearly, if the 1 digit number is $8$, the highest possible product will be less than $8000$ because $8 * 999 < 8000$. Therefore, assume the 1 digit number is $9$.

This means the product of the two 2-digit numbers must, between them, one must be divisible by $9$, or both must be divisible by 3.

### One number is divisible by 9:

$$\frac{99 * (99, 98, 97, 96, 95, 94, 93, 92, 91, 90)}{9} \rightarrow \\ (1089, 1078, 1067, 1056, 1045, 1034, 1023, 1012, 1001, 990)$$

So the first solution candidate is $99 * 90 = 8910 = 9 * 990$.

### Both numbers are divisible by 3:

$$99 * 96 \rightarrow 9504 \rightarrow 1056 * 9\\ 96 * 96 \rightarrow 9216 \rightarrow 1024 * 9\\ 96 * 93 \rightarrow 8928 \rightarrow 992 * 9$$

Since $8928 > 8910$, we have our answer, and the second place answer.

These two processes can be continued to generate further answers.

1. $99$ times any two digit below $90$ will also result in a valid answer.
2. Any two multiples of 3 equal to or below $96$ and $93$ will also result in a valid answer.
3. Of course, once we get under $8000$, the product no longer needs to be a multiple of $9$ as the 1 digit number doesn't need to be $9$ anymore.

Exhaustive list of answers above $8000$:

$$8928 = 9 * 992 = 96 * 93 \\ 8910 = 9 * 990 = 99 * 90 \\ 8820 = 9 * 980 = 98 * 90 \\ 8811 = 9 * 979 = 99 * 89 \\ 8730 = 9 * 970 = 97 * 90 \\ 8712 = 9 * 968 = 99 * 88 \\ 8649 = 9 * 961 = 93 * 93 \\ 8640 = 9 * 960 = 96 * 90 \\ 8613 = 9 * 957 = 99 * 87 \\ 8550 = 9 * 950 = 95 * 90 \\ 8514 = 9 * 946 = 99 * 86 \\ 8460 = 9 * 940 = 94 * 90 \\ 8415 = 9 * 935 = 99 * 85 \\ 8370 = 9 * 930 = 93 * 90 \\ 8352 = 9 * 928 = 96 * 87 \\ 8316 = 9 * 924 = 99 * 84 \\ 8280 = 9 * 920 = 92 * 90 \\ 8217 = 9 * 913 = 99 * 83 \\ 8190 = 9 * 910 = 91 * 90 \\ 8118 = 9 * 902 = 99 * 82 \\ 8100 = 9 * 900 = 90 * 90 \\ 8091 = 9 * 899 = 93 * 87 \\ 8064 = 9 * 896 = 96 * 84 \\ 8019 = 9 * 891 = 99 * 81 \\ 8010 = 9 * 890 = 89 * 90 \\$$

A longer, more thorough attempt at finding the solution without doing actual multiplications (except for $32\times32 = 2^5\times2^5 = 2^10 = 1024$). Hopefully it's easier to understand.

Notations:

\begin{align} ab &=\text{number with first digit}\ a\ \text{and second digit}\ b\\ a \times b &= a\ \text{multiplied by}\ b \end{align}

First phase is to observe that

$1d$ must be $9$.

The highest solution that can be made with $1d = 8$ is:

$8\times999 = 8 \times 111 \times 9 = 9 \times 888 < 9 \times 900 = 90 \times 90$
Ergo there are solutions with $1d = 9$ higher than can be made with $1d = 8$.

The important thing to note is that the number is $9*3d$, which is divisible by $9$.

The second phase is to determine/guess the number starting with the highest possible factors. Note that

$2d\times2d$ is divisible by $9$, so either one of the factors is divisible by $9$ ($99,90,$etc) or both factors are divisible by $3$ ($96,93,$etc)

This leads to the following cases (we stop when it's obvious we can't find better solutions):

\begin{align}99\times9x &= 9 \times 11 \times 9x \\&= 9 \times 9(9+x)x \\&\implies x = 0 \\&\implies\text{solution}: 99 \times 90 = 9 \times 990\\ 96\times96 &= 9 \times 32 \times 32 \\&= 9 \times 1024 \\&\implies\text{no solution}\\ 96\times93 &= 9 \times 32 \times 31 \\&= 9 \times (1024-32) \\&= 9 \times 992 \\&\implies\text{solution}: 96 \times 93 = 9 \times 992\\ 93\times93 &\ \text{irrelevant as it's smaller than}\ 96\times93\\ 90\times9x &\ \text{the best solution is obviously}\ 99\times90=9\ \text{discovered above}\end{align}

Obviously the best solution is

$$\boxed{9\times992=96\times93}$$