X and Y are the digits at the unit's place of the numbers (408X) and (789Y) where X ≠ Y. However, the digits a

Question

X and Y are the digits at the unit's place of the numbers (408X) and (789Y) where X ≠ Y. However, the digits at the unit's place of the numbers $(408X)^{63}$ and $(789Y)^{85}$ are the same. What will be the possible value(s) of (X + Y)?

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Accepted Answer

10. All numbers from 1 to 9 repeat their last digits over a cycle of 4. 63 can be written as 4k+3. 85 can be written as 4k+1. Some number's third power should yield the first digit as some number's first power. 2 3 2^3 2 3 will yield 8 8 8 as the last digit (2 and 8 is a possible solution).X+Y = 10 3 3

X and Y are the digits at the unit's place of the numbers (408X) and (789Y) where X ≠ Y. However, the digits at the unit's place of the numbers (408X)63(408X)^{63}(408X)63 and (789Y)85(789Y)^{85}(789Y)85 are the same. What will be the possible value(s) of (X + Y)?

X and Y are the digits at the unit's place of the numbers (408X) and (789Y) where X ≠ Y. However, the digits at the unit's place of the numbers (408X)63(408X)^{63}(408X)63 and (789Y)85(789Y)^{85}(789Y)85 are the same. What will be the possible value(s) of (X + Y)?

X and Y are the digits at the unit's place of the numbers (408X) and (789Y) where X ≠ Y. However, the digits at the unit's place of the numbers $(408X)^{63}$ and $(789Y)^{85}$ are the same. What will be the possible value(s) of (X + Y)?

X and Y are the digits at the unit's place of the numbers (408X) and (789Y) where X ≠ Y. However, the digits at the unit's place of the numbers $(408X)^{63}$ and $(789Y)^{85}$ are the same. What will be the possible value(s) of (X + Y)?