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Find the absolute value of the difference of single-digit integers \(A\) and \(B\) such that 

 

Express your answer in base 6.

 Oct 8, 2018
 #1
avatar+575 
+1

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.

 Oct 8, 2018
 #2
avatar+101872 
+2

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

 

 

cool cool cool

 Oct 8, 2018

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