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# Pleas help! I tried many ways!

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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?

Thanks!

- TealSeal

May 12, 2021

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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?

I don't think that there are any....

Since there are numbers up to 7 in there, the base, p,  would have to be 11 or more.  (because it is prime)

Looking at just the last digit.

7+6+3+5+6=27                               2+1+0=3

27=3+24

This means that p must be 24 or less (not that that matters)

Anyway,

prime factors, 11 or bigger, of 24 don't exist.

So I think that there are no possible values of p that will satisfy this.

May 12, 2021
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CORERECT!!!!!!!!!!!!!!!!!!!!!

TYSM!

TealSeal  May 12, 2021
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That's so smart.

My first insinct was to set up p^3 + 7 + 3p^2 + 6.... etc.

=^._.^=

catmg  May 13, 2021
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Thanks catmg

Melody  May 13, 2021