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Suppose  b and c are positive integers.

When b^2 is written in base c, the result is 121_c. When c^2 is written in base b, the result is 71_b.

What is b+c?

thisismyname  Sep 15, 2016

Best Answer 

 #1
avatar+93866 
+15

Hi thisismyname  :)

Your question is interesting :))

 

Suppose  b and c are positive integers.
When b^2 is written in base c, the result is 121_c. When c^2 is written in base b, the result is 71_b.
What is b+c?

 

\(b^2=121_c\\ b^2=1*c^2+2*c+1\\ b^2=c^2+2c+1\\ b^2=(c+1)^2\\ b=c+1\\~\\ c^2=71_b\\ c^2=7b+1\\ c^2=7(c+1)+1\\ c^2=7c+8\\ (c-8)(c+1)=0\\ c=8 \;\;or \;\;c=-1 \qquad \text{but c is positive so}\\ c=8\\~\\ b=8+1=9\\ so\\ b+c=17 \)

 

there might be a short cut so that you do not have to find both b and c seperately.....

I didn't notice one though.

Melody  Sep 15, 2016
 #1
avatar+93866 
+15
Best Answer

Hi thisismyname  :)

Your question is interesting :))

 

Suppose  b and c are positive integers.
When b^2 is written in base c, the result is 121_c. When c^2 is written in base b, the result is 71_b.
What is b+c?

 

\(b^2=121_c\\ b^2=1*c^2+2*c+1\\ b^2=c^2+2c+1\\ b^2=(c+1)^2\\ b=c+1\\~\\ c^2=71_b\\ c^2=7b+1\\ c^2=7(c+1)+1\\ c^2=7c+8\\ (c-8)(c+1)=0\\ c=8 \;\;or \;\;c=-1 \qquad \text{but c is positive so}\\ c=8\\~\\ b=8+1=9\\ so\\ b+c=17 \)

 

there might be a short cut so that you do not have to find both b and c seperately.....

I didn't notice one though.

Melody  Sep 15, 2016

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