#Numbers-6

P(a,b) = { P(a-b , b)  if a ≥ b
               { a              if  a< b
Given that P(a,7) = 4, P(a,11) =6 and P(a,13) = 3, if "a" is a positive integer between 2000-3000. Find the value of P(a,17).

Solution-

The function works on the concept of the division
P(a,7) = 4 means when "a" is divided by "7" remainder is "4"
=>P(a,7) = 7x+4---------(1)
Similarly,P(a,11) = 6 =>11y+6--------(2)
 P(a,13) = 3c=> 13z+3----------(3)
Now,for finding the value of "a" we have to find the common term from these three equation
Equating equation (1) and (2),
7x+4 = 11y+6
x=(11y+2)/7
for y = 3,x=5 first solution
next solution,y=10,x=16
We can see y is increasing by 7(co-efficient  and x) and x by 11(co-efficient of y)
General term = 39+77r -----------(4)
Now,comparing equation (3) and (4), we get,
39+77r = 13z +3
(36+77r)/13 = z
(2+5r) + (10+12r)/13 = z
r = 10, z=62
Now z will increase by 77 and r by 13,
General term = 809+1001q
So for q=2
a=2002+809 = 2811
P(a,17) = 2811 mod 17 = 165*17+6
Hence, p(a,17) = 6 ANS

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