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# When $555_{10}$ is expressed in this base, it has 4 digits, in the form ABAB, where A and B are two different digits. What base is it?

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When $555_{10}$ is expressed in this base, it has 4 digits, in the form ABAB, where A and B are two different digits. What base is it?

Dabae  Jun 7, 2015

#1
+91462
+20

When $555_{10}$ is expressed in this base, it has 4 digits, in the form ABAB, where A and B are two different digits. What base is it?

Let the base be X

$$\\A*X^3+B*X^2+A*X+B=555\\\\ AX^3+BX^2+AX+B=555\\\\ X^2(AX+B)+(AX+B)=555\\\\ (X^2+1)(AX+B)=555\\\\ So\;\;X^2+1\;\;must be a factor of 555\\\\$$

$${factor}{\left({\mathtt{555}}\right)} = {\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{37}}$$

factors of 555 are   3,5,37,15,111,185,555

5  and 37 would both work so far

5 would  be base 2 And base 2 is to little,  37 would give base 6 (X=6) so that is probably correct.

Lets see

$$\\(X^2+1)(AX+B)=555\\\\ (6^2+1)(6A+B)=555\\\\ 37(6A+B)=555\\\\ 6A+B=15\\\\ If A=1 B=8 \qquad no good A and B cannot be more than 5\\\\ If A=2 B=3 \qquad Great\\\\ so our number is  2323_6$$

Check

$${\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{6}}}^{{\mathtt{3}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{6}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}} = {\mathtt{555}}$$

And that is excellent

So               $$555_{10}=2323_6$$

Melody  Jun 8, 2015
Sort:

#1
+91462
+20

When $555_{10}$ is expressed in this base, it has 4 digits, in the form ABAB, where A and B are two different digits. What base is it?

Let the base be X

$$\\A*X^3+B*X^2+A*X+B=555\\\\ AX^3+BX^2+AX+B=555\\\\ X^2(AX+B)+(AX+B)=555\\\\ (X^2+1)(AX+B)=555\\\\ So\;\;X^2+1\;\;must be a factor of 555\\\\$$

$${factor}{\left({\mathtt{555}}\right)} = {\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{37}}$$

factors of 555 are   3,5,37,15,111,185,555

5  and 37 would both work so far

5 would  be base 2 And base 2 is to little,  37 would give base 6 (X=6) so that is probably correct.

Lets see

$$\\(X^2+1)(AX+B)=555\\\\ (6^2+1)(6A+B)=555\\\\ 37(6A+B)=555\\\\ 6A+B=15\\\\ If A=1 B=8 \qquad no good A and B cannot be more than 5\\\\ If A=2 B=3 \qquad Great\\\\ so our number is  2323_6$$

Check

$${\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{6}}}^{{\mathtt{3}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{6}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}} = {\mathtt{555}}$$

And that is excellent

So               $$555_{10}=2323_6$$

Melody  Jun 8, 2015
#2
+81023
0

That's a nice approach and nice thinking, Melody....

CPhill  Jun 8, 2015
#3
+91462
0

Thanks Chris :)

Melody  Jun 8, 2015

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