Find the absolute value of the difference of single-digit integers \(A\) and \(B\) such that 
Express your answer in base 6.
+583 B=2 A=1
Look at hundred place
B+4+A <=14 for 0<=A,B<=5
So, A =0 or 1
Look at the ones place
Even B=1, the tens place is increment by one
So let's assume B>=1 ,then we ahve 1+B+1+1=5 \(\Rightarrow\)B=2
Then look at the left parts, 2+4+A=6*A+1\(\Rightarrow\)A=1
Verify it ,it works.
+128656 We have the following equation :
[ 36B + 6B + A } + [ 4*36 + 6 + B ] + [ 36A + 6 + 5 ] = [ 216A + 36 + 30 + 2 ]
[ 42B + A ] + [ 150 + B ] + [ 36A + 11 ] = 216A + 68
43B + 37A + 161 = 216A + 68
43B + 93 = 179A
This equation is true when B = 2 and A = 1
And the absolute value of their difference is l A - B l = l 1 - 2 l = l -1 l = 1
