Question

Given:

where , , , , are unique positive single-digit numbers, what is the number of possible combinations for and to satisfy this equation?

Answer 1

Answer:

• 4 combinations ⇒ XYZ = 358, 457, 754, 853

Step-by-step explanation:

Sum of two digits can be less than 20, therefore A = 1 is the only option for A, since AB ≤ 19.

Substitute A:

• XYZ + ZYX = 1B11

Then Y = 0 or Y = 5.

Y = 0 doesn't work as it leads to B = 1, which is a repeat of A.

So Y = 5.

Substitute Y:

• X5Z + Z5X = 1B11

We see B = 2 is the only option for B, since X + Z = 11, add 1 to 1 (5 + 5 gives us 1 for 10).

Substitute B:

• X5Z + Z5X = 1211

Possible options for X and Z:

• 2 + 9 = 11, not considered as 2 is repeated
• 3 + 8 = 11, it gives us 358 + 853 = 1211
• 4 + 7 = 11, it gives us 457 + 754 = 1211
• 5 + 6 = 11, not considered as 5 is repeated
• 6 + 5 = 11, same as above
• 7 + 4 = 11, it gives us one more option 754 + 457 = 1211
• 8 + 3 = 11, it gives us 853 + 358 = 1211
• 9 + 2 = 11, not considered as 2 is repeated