Each of p and q is a 6-digit base ten positive integer with no leading zero. The 12-digit number that is obtained by writing p and q side-by-side is divisible by the product p*q.
Determine all possible pair(s) (p, q) for which this is possible.
I am assuming that the order of the digit combination is p then q,
thus if p=123456 and q=654321 then we would get 123456654321
this can be expressed as 10^6*p + q thus we want
(10^6*p+q)/(pq) = k for some integer k
10^6p+q=kpq
10^6p=(kp-1)q
q=10^6p/(kp-1)
now kp-1 and p are coprime thus we need
kp-1 to divide 10^6
so let d be a divisor of 10^6
then we have
kp-1=d
kp=d+1
thus p is any divisor of d+1
thus we want to find all divisors d of 10^6 such that d+1 has a 6 digit divisor, for this to be true then d must have at least 6 digits.
the divisors of 10^6 with at least 6 digits are
100000, 125000, 200000, 250000, 500000, and 1000000
for each of these d+1 is prime thus we have k=1 and thus
q=10^6p/(p-1)
for each of the divisors d above if we set p=d+1 then we end up
with q being a 7 digit number. Thus unless there is an error in my math, then there are no solutions to this problem.
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Posted by Daniel
on 2010-06-29 13:55:19 |
analytical solution
