AB * CD = BA * DC

Number Theory Level 5

We know that the order of digits in a number matter, and we cannot randomly swap them around. Most of the time,

$\overline{AB} \times \overline{CD} \neq \overline{BA} \times \overline{DC} .$

How many ordered tuples of non-zero digits $(A, B, C, D)$ are there such that

$\overline{AB} \times \overline{CD} = \overline{BA} \times \overline{DC} ?$

The answer is 209.

2 solutions

Caleb Townsend
Feb 6, 2015

Java solution:

int numSolutions = 0;
for (int a = 1; a < 10; a++)
    for (int b = 1; b < 10; b++)
        for (int c = 1; c < 10; c++)
            for (int d = 1; d < 10; d++)
                if ((10*a + b)*(10*c + d) == (10*b + a) * (10*d + c))
                    numSolutions++;
System.out.println(numSolutions);

Note that the if statement simply checks if $\overline{AB} \times \overline{CD} = \overline{BA} \times \overline{DC}.$ Sure enough, the program says it's 209.

Calvin Lin Staff
Feb 2, 2015

This solution is incorrect as I missed out several cases. See the report forum for correct solutions.

(Solution deleted)

We also have cases where a=b and c=d

vinod kumar - 6 years, 4 months ago

Thanks. As it turns out, I missed out a lot of cases. Sorry about that. I have updated the answer to 209

Calvin Lin Staff - 6 years, 4 months ago

Here is how I solved it.

If we write the decimal expansion of the equation, we get

$(10A + B)(10C + D) = (10B + A)(10D + C)$

Simplifying this gives $A\times C = B\times D$

I then listed all the possible products $A \times C$ , where $A$ and $C$ are from 1 to 9.

These products can be divided into 4 cases

Case 1. $A \times C$ can only be written as a product of two same numbers ONLY.

We see that the following numbers satisfy this case

1.	$1$	$1 \times 1$
2.	$25$	$5 \times 5$
3.	$49$	$7 \times 7$
4.	$64$	$8 \times 8$
5.	$81$	$9 \times 9$

There is exactly one ordered tuple for each number (for example $(1,1,1,1)$ ) and since there are 5 numbers in this case, there are $5 \times 1 = 5$ ordered tuples.

Case 2 $A \times C$ can only be written as a product of two distinct digits

1.	$2$	$1 \times 2$
2.	$3$	$1 \times 3$
3.	$5$	$1 \times 5$
4.	$7$	$1 \times 7$
5.	$10$	$2 \times 5$
6.	$14$	$2 \times 7$
7.	$15$	$3 \times 5$
8.	$20$	$4 \times 5$
9.	$21$	$3 \times 7$
10.	$27$	$3 \times 9$
11.	$28$	$4 \times 7$
12.	$30$	$5 \times 6$
13.	$32$	$4 \times 8$
14.	$35$	$5 \times 7$
15.	$40$	$5 \times 8$
16.	$42$	$6 \times 7$
17.	$45$	$5 \times 9$
18.	$48$	$6 \times 8$
19.	$54$	$6 \times 9$
20.	$56$	$7 \times 8$
21.	$63$	$7 \times 9$
22.	$72$	$8 \times 9$

For each number, we have 4 possible ordered tuples, (since there are 2! permutations of $(A,C)$ and 2! permuations of $(B,D)$ ), and since there are 22 numbers, the total number of tuples = $22 \times 4 = 88$

Case 3 $A \times C$ can be written as a product of two same digits OR a product of two different digits.

1.	$4$	$1 \times 4$	$2 \times 2$
2.	$9$	$1 \times 9$	$3 \times 3$
3.	$16$	$2 \times 8$	$4 \times 4$
4.	$36$	$4 \times 9$	$6 \times 6$

For each number there are 9 possible tuples, ( $(A,C)$ and $(B,D)$ both have 3 choices each) therefore total tuples = $4 \times 9 = 36$

Case 4 $A \times C$ can be written as a product of two different digits in two ways.

1.	$6$	$1 \times 6$	$2 \times 3$
2.	$8$	$1 \times 8$	$2 \times 4$
3.	$12$	$2 \times 6$	$3 \times 4$
4.	$18$	$2 \times 9$	$3 \times 6$
5.	$24$	$3 \times 8$	$4 \times 6$

For each number there are 16 possible tuples, ( $(A,C)$ and $(B,D)$ both have 4 choices each) therefore total tuples = $5 \times 16 = 80$

Adding all of them gives, $5 + 88 + 36 + 80 = \boxed{209}$

Pranshu Gaba - 6 years, 4 months ago

The first 2 cases are all trivial. the next 2 cases provide 14 sets of tuples.

12 X 42 =21 X 24 // 13 X 92 =31 X 29 //24 X 84 =42 X 48 //46 X 96 =64 X 69

12 X 63 = 21 X 36// 13 X 62 =31 X 26// 12 X 84 =21 X 48 //14 X 82 =41 X 28

23 X 64 =32 X 46// 24 X 63 =42 X 36//23 X 96 =32 X 69 //29 X 36 = 92 X 63

34 X 86 =43 X 68// 36 X 84 = 63 X 48 //All these are distinct number sets.

Brahmam Meka - 6 years, 4 months ago

I think A=D ,B=C SOLutions are not correct

Brahmam Meka - 6 years, 4 months ago

AB X CD = BA X DC (10 A + B ) X (10 C +D) =(10 B + A)(10 D + C) 100 A C + 10 (AD +BC) + B D = 100 B D + 10(BC +AD) +A C 99 A C = 99 B D A C = B D 1 X6 = 2 X3 = 3X2 12 X63 =21 X 36 13 X62 = 31 X26 1X4 =2 X2 12X 42 =21 X24 1 X8 = 2 X4 = 4 X2 12 X 84 =21 X 48 14 X82 =41 X 28

like that we can write for

2 X6 = 3 X4 = 4 X3 2 X8 =4 X4 3 X 6 = 2X 9 = 9 X2 4 X6 =3 X8 =8 X3 etc. they give different pairs.

Brahmam Meka - 6 years, 4 months ago

Right. my solution is incorrect. Let me completely remove it. I was doing this problem in a daze, and didn't work through it completely.

Calvin Lin Staff - 6 years, 4 months ago

AB * CD = BA * DC

The answer is 209.

2 solutions

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