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 Modular mathematics (Posted on 2012-12-28)
Determine all pairs of positive integers(a,p) where p is a prime and

(a modp)+(a mod2p)+(a mod3p)+(a mod4p)=a+p

 No Solution Yet Submitted by Danish Ahmed Khan Rating: 4.0000 (1 votes)

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 Definitely all the cases (spoiler) | Comment 2 of 4 |
Oh, Charlie.  You were so close to a complete analytical solution.  Well, thanks for the excellent start.  This will finish it, analytically

Note that
a mod p   < p
a mod 2p < 2p
a mod 3p < 3p
a mod 4p < 4p

Therefore a+p < 10p
a     < 9p

So there are only nine more cases to consider
a = 5p
a = 6p
a = 7p
a = 8p
4p < a < 5p
5p < a < 6p
6p < a < 7p
7p < a < 8p
8p < a < 9p

******************

a = 5p
0 + p + 2p + p = a+p
a = 3p, contradiction

******************

a = 6p
0 + 0 + 0 + 2p = a+p
a = p, contradiction

******************

a = 7p
p + p + p + 3p = a+p
a = 5p, contradiction

******************

a = 8p
0 + 0 + 2p + 0 = a+p
a = p, contradiction

******************

4p < a < 5p
(a-4p) + (a-4p) + (a-3p) + (a-4p) = a+p
a = p*(16/3), contradiction

******************

5p < a < 6p
(a-5p) + (a-4p) + (a-3p) + (a-4p) = a+p
a = p*(17/3)
gives solution of (17,3)

******************

6p < a < 7p
(a-6p) + (a-6p) + (a-6p) + (a-4p) = a+p
a = p*(23/3), contradiction

******************

7p < a < 8p
(a-7p) + (a-6p) + (a-6p) + (a-4p) = a+p
a = 8p, contradiction

******************

8p < a < 9p
(a-8p) + (a-8p) + (a-6p) + (a-8p) = a+p
a = p*(31/3), contradiction

SO, I HAVE PROVED THAT CHARLIE'S ANSWER IS COMPLETE.  THERE ARE NO OTHER SATISFACTORY VALUES of a and p

Edited on December 29, 2012, 8:34 pm
 Posted by Steve Herman on 2012-12-29 10:45:24

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