JEE ADVANCED 2016 (7)

The method I followed is to find individual remainders for all terms and add them together and find the resultant remainder.

So let's first add the terms less than 100 together as if we divide them individually we will just get the same number.

f(1) + f(2) + .........f(7) = 99 (mod 100)

This can be simply evaluated by multiplying the last 2 digits of the value of the previous function and the next odd number to be multiplied. Then all these terms can be added together.

Now using the same method / trick of multiplying only the last 2 digits by the next odd number we can evaluate f(8),f(9) and so on....

While doing this we can notice a pattern that is :

f(8) = f(9) = 25,

f(10 )= f(11) = 75,

f(12) = f(13) = 25,

f(14) = f(15) = 75 and so on......

So f(8) + f(9) is 50 (mod 100) and f(10) + f(11) is also 150 (mod 100) = 50 (mod 100) which follows for all the upcoming terms.

So we can add 2 terms of them to get 100 = 0 (mod 100)

Now we see that 8th term is 25 , 10th term is 75 , 12th term is 25 ,14th term is 75 and so on.....

From this we can find what f(2016) mod 100 is by using AP (Arithmetic Progression) with a = 8 , d = 2 , tn = 2016 [As we can see that all odd positioned terms are 25 (mod 100) and all even positioned terms are 75 (mod 100).]

We get n = 1005 which is an odd position so the remainder will be 25. (as 8, 12 ..... are all in odd positions and their remainders from 100 are equal to 25.)

But we know that all the terms before it from f(8) to f(2015) will get cut [..or more appropriately will become 0 (mod 100)]

So now we add the remainder 25 [remainder of f(8) to f(2016)] to the previous remainder 99 [remainder of f(1) to f(7)]

25 + 99 = 124 = 24 (mod 100)

Therefore the remainder is $\boxed{24}$ .

Overall it's just a matter of thinking and fast 2 digit multiplications !!!

From f (3) every term is divisible by 5.f (1)+f (2)=4 So the remainder must be 4 (mod 5) and must be even since even no. Of odd no. Are added so 4 (mod 10)....24 only matches so the answer.