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# Base Number Arithmetic

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Suppose that p is prime and $$1007_p+306_p+113_p+125_p+6_p=142_p+271_p+360_p$$. How many possible values of  p are there?

Mar 28, 2018

### 1+0 Answers

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Suppose that p is prime and

$$1007_p+306_p+113_p+125_p+6_p=142_p+271_p+360_p.$$

1007_p+306_p+113_p+125_p+6_p=142_p+271_p+360_p.

How many possible values of  p are there?

$$\small{ \begin{array}{|rcll|} \hline 1007_p+306_p+113_p+125_p+6_p &=& 142_p+271_p+360_p \\\\ \underbrace{p^3+7}_{=1007_p}+\underbrace{3p^2+6}_{=306_p}+ \underbrace{p^2+p+3}_{=113_p} + \underbrace{p^2+2p+5}_{=125_p} +\underbrace{6}_{=6_p} &=& \underbrace{p^2+4p+2}_{=142_p}+ \underbrace{2p^2+7p+1}_{=271_p} + \underbrace{3p^2+6p}_{=360_p} \\ p^3+ 3p^2+ p^2+ p^2 +p +2p+7 +6 +3 +5 + 6 &=& p^2+ 2p^2+ 3p^2 +4p+ 7p+6p +2 + 1 \\ p^3+ 5p^2+3p +27 &=& 6p^2 +17p +3 \\ p^3+ 5p^2-6p^2+3p-17p +27-3 &=& 0 \\ p^3-p^2-14p +24 &=& 0 \\ (p-2)(p^2+p-12) &=& 0 \\ (p-2)(p-3)(p+4) &=& 0 \\ \hline \end{array} }$$

The only primes that will work are $$p=2$$ or $$p=3$$

Mar 28, 2018
edited by heureka  Mar 28, 2018