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

#1**+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**+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