functions problem?

Question

P(a,b) = P(a-b,b) if a>= b and
P(a,b) = a if a < b


it is given that P(a,7) = 4; P(a,11) = 6 and P(a,13) = 3.
Find P(a,17), if ‘a’ is a positive integer between 2000 and 3000.

(1) 6(2) 8 (3) 11 (4) 12

Answer 1

Supose a>0.
P(a,b) = P(a-b ,b) = P(a-2b, b) = P(a-3b, b) = ... = P(a-qb, b) = r
where q is the biggest integer such that a-qb > 0. By the definition, P(a,b) is the remainder of the division a/b (ie, a = qb+r).
Therefore
P(a,7) = 4 implies 7 divides a-4
then 7 divides a+192 = (a-4)+7(28)
P(a,11) = 6 implies 11 divides a-6
then 11 divides a+192 = (a-6)+11(18)
P(a,13) = 3 implies 13 divides a-3
then 13 divides a+192 = (a-3)+13(15)

therefore (7)(11)(13)=1001 divides a+192
ie, a+192 = 1001n (n is a integer)
taking n=3, a+192 = 3003 and a = 2811

but 2811 = (17)(165)+6, hence P(a,17) = P(2811,17) = 6.
The number (1) 6 is correct

Answer 2

2000 a mod 7=4=>a=7*x1+4=>(a-4)=7*x1
a mod 11=6=>a=11*x2+6=>(a-6)=11*x2
a mod 13=3=>a=13*x3+3=>(a-3)=13*x3


From this three equtions you find a
The P(a,17) is the remainder of a/17