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 subscript for this answer.)
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∗72+Y∗7+Z=Z∗92+Y∗9+X49X+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
5037=3059
check
5×49+3=248
3×81+5=248 (Edited: Thanks Chris )
There might be a much quicker way of doing this
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∗72+Y∗7+Z=Z∗92+Y∗9+X49X+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
5037=3059
check
5×49+3=248
3×81+5=248 (Edited: Thanks Chris )
There might be a much quicker way of doing this