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Get access to the detailed solutions to the previous years questions asked in IIM IPMAT exam
Looking at the additional information about the prime numbers should make one realise that they are the key to solving the question.
f(16000) can be written as f(28 × 54)
Now, we can try to find these individual values:
For any prime p: f(p) = 1
f(p2) = f(p) f(p) + f(p) + f(p) = 1 + 1 + 1 = 3
f(p3) = f(p2) f(p) + f(p2) + f(p) = 3 + 3 + 1 = 7
This way, we can find the function output for any prime number raised to a power.
We can see that each new exponent is twice the previous output +1, solving this way till prime raised to power 8
f(p4) = 7 + 7 + 1 = 15
f(p5) = 15 + 15 + 1 = 31
f(p6) = 31 + 31 + 1 = 63
f(p7) = 63 + 63 + 1 = 127
f(p8) = 127 + 127 + 1 = 255
Using these values in the original expression of f(28 × 54) = f(28) f(54) + f(28) + f(54) we get
f(28 × 54) = (255 × 15) + 255 + 15 = 4095