SOLUTION: Show that "ab" (scale of ten) = "ba" (scale of twelve) is impossible, where a and b are digits common to scales of ten and twelve. Show, that "ab" (scale of ten) = "ba" (scale o

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: Show that "ab" (scale of ten) = "ba" (scale of twelve) is impossible, where a and b are digits common to scales of ten and twelve. Show, that "ab" (scale of ten) = "ba" (scale o      Log On


   

Question 394976: Show that "ab" (scale of ten) = "ba" (scale of twelve) is impossible, where a and b are digits common to scales of ten and twelve.
Show, that "ab" (scale of ten) = "ba" (scale of seven) is possible and give an example, where a and b are digits common to scales ten and seven.
Answer by Edwin McCravy(20059) About Me  (Show Source):
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Show that "ab" (scale of ten) = "ba" (scale of twelve) is impossible, where
a and b are digits common to scales of ten and twelve.



Assume a and b are digits in scale ten, such that

!0a + b = 12b + a 

9a = 11b

The smallest positive integer solution
to that is a=11 and b = 9, and 11 is not
a digit in base ten.
 
So there can be no solution where a and b are both digits of scale ten.

Show, that "ab" (scale of ten) = "ba" (scale of seven) is possible and give an example, where a and b are digits common to scales ten and seven.


Assume a and b are digits in scale seven.

!0a + b = 7b + a 

9a = 6b

3a = 2b

a has to be even. So,

the smallest solution in positive integers is a=2 and b=3,

so one solution is 23 (scale of ten) = 32 (scale of seven)

The next to smallest solution in positive integers is a=4 and b=6,

so another solution is 46 (scale of ten) = 64 (scale of seven)

The next solution in positive integers is a = 6 and b = 9.

But 9 is not a digit in scale 7.

So those are the only two solutions

Edwin