2010/032) If p and q are distinct primes, show that p^q + q^p ≅ (p + q) mod pq

If p and q are distinct primes, show that p^q + q^p ≅ (p + q) mod pq.
as

q is prime

so p^(q-1) mod q = 1 (as per fermats little theorem)


so p^(q-1) = mq + 1

multiply by p on both sides p^q = mpq + p = p mod pq

similarly as q is prime q^p = p mod pq

adding we get (p^q + q^p) mod pq = (p+q)

proved

2010/031) 5-digit number whose half is a perfect cube and one-third is a perfect square?

5-digit number whose half is a perfect cube and one-third is a perfect square?

solution
we have
2n^3 = 3p^2
so n has to be multiple of 3 say 3x and p multiple of 2 say 2a
2(3x)^3 = 3(2a)^2
or 54x^3 = 12a^2
or 9x^3 = 2a^2
so a has to be divisible by 3 (say a = 3y) (sqrt of 9) and x by 2 say x = 2b
so n= 6y and p = 6b

and 9(2y)^3 = 2(3b)^2

or 4y^3 = b^2

so b has to be divisible by 2 say b = 2c that is p = 12c

then 4y^3 = 4c^2 or y^3 = c^2 so y has to be a prfect square and c a perfect cube that it

y= m^2 and c= m^3

so n = 6m^2 and p = 12m^3

now it shall not be difficult to find 2n^3 or 2(6m^2)^3 which is a 5 digit number

by checkin we find that it is m = 2 so n = 24 and ans = 2 * 24^3 or 27648