Solution To Mensa Algebra Problem

 

 Which digits n (0 - 9) have a power x (greater than one) that is a sequence if the digit n?

n^x  =   n n n . . . n (x > 1)

Ans. Between 0 - 9, only 0 and 1 have a power x greater than 1 that is a sequence of the digit n.

E.g.  0 ^x  = 0 for any x

1 ^x  = 1 for any x

Try 2:

nnn. . . n = 222...2 = 222/2 = 111 (not a power of 2)

3) nnn. . . n = 333...3 = 333/3 = 111

If divisible by 3 is also divisible by 37 which is not a power of 3.

4) nnn. . . n = 444...4 = 444/4 = 111

which is not a power of 4.

5)nnn. . . n = 555... 5  = 555/5 = 111

which is not a power of 5.

6) nnn. . . n = 666... 6  = 666/6 = 111

which is not a power of 6.

7) nnn. . . n = 777... 7 

But the last 2 digits of  7 ^x  (x >1)are always:

49 (e.g  7 ² =  49),  43 (e.g  7 ³ =  343 ), or 01 (e.g.  7 ⁴ =  2,401) 

But never 77.

8)  nnn. . . n = 888... 8  = 888/8 = 111

which is not a power of 8.

9) nnn. . . n = 999... 9  = 999/9 = 111

But if 111...1 is divisible by  9 (when the number of 9s is a multiple of 9) then it is also divisible by 12, 345, 679 (which is not a power of 9)