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angel

 

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?

 

angel

 

Thanks!

- TealSealangel

 May 12, 2021
 #1
avatar+113787 
+3

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
 #2
avatar+106 
+2

CORERECT!!!!!!!!!!!!!!!!!!!!! 

TYSM! angel

TealSeal  May 12, 2021
 #3
avatar+1973 
0

That's so smart. 

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

 

=^._.^=

catmg  May 13, 2021
 #4
avatar+113787 
+1

Thanks catmg  laugh

Melody  May 13, 2021

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