Find a base 7 three-digit number which has its digits reversed when expressed in base 9. (You do not need to indicate the base with a subscr

Let the number be XYZ       XY and Z are all 1 digit numbers between 0 and 6 inclusive

X and Z cannot by 0

Z must be smaller than X

So Y could possible be 0,12,3,4,5,6

So X could possible be 2,3,4,5,6

So Z could possible be 1,2,3,4,5

$$\\X*7^2+Y*7+Z=Z*9^2+Y*9+X\\\\
49X+7Y+Z=81Z+9Y+X\\\\$$

I might just use trial and error Mellie

Try Z=1

49X+7Y+1=81*1+9Y+X

48X-2Y=80

24X-Y= 40                   Multiples of 24 are 24,48,  Neither of these - an allowed y =40

so Z is not 1 

Try Z=2

49X+7Y+2=81*2+9Y+X

48X-2Y=160        

24X-Y=80                 Multiples of 24 are 24,48,72,96  Neither of these - an allowed y =80

so Z is not 2

Try Z=3

49X+7Y+3=81*3+9Y+X

48X-2Y=240          

24X-Y=120              The first multiple of 24 that is BIGGER than or equal to 120 is 120

SO Z can be 3, X=5 and Y=0

$$\\\mathbf{503_7=305_9}$$  

check

$${\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{49}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}} = {\mathtt{248}}$$

$${\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{81}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}} = {\mathtt{248}}$$                     (Edited: Thanks Chris )

There might be a much quicker way of doing this 

Nice answer, Melody.....

BTW....I think Melody meant to write.......  3(81) + 5    =  243 + 5  = 248