Question

Find the value of n, where n is an integer and 2^n−5 × 6^2n−4 = 1/12^4×2 .
KINDLY EXPLAIN PROPERLY​

Answer 1

Answer:

f(n) = 2n is periodical for n = 4 in terms of last digit. i.e. last digit of 2n always repeat for next 4th value of n. (ex: 2, 4, 8, 16, 32, 64…)

f(n) = 3n is periodical for n = 4 in terms of last digit. i.e. last digit of 3n always repeat for next 4th value of n.(ex: 3, 9, 27, 81, 243, 729…)

f(n) = 4n is periodical for n = 2 in terms of last digit. i.e. last digit of 4n always repeat for next 2nd value of n.(ex: 4, 16, 64, 256..)

1n is going to be 1 always, independent of n.

So, If we will have a close look for periodicity of f(n) = (1n +2n + 3n + 4n ) we will get that its periodicity is also 4 and its last digits occurs as :

for n = 1, f(n) = 10

for n = 2, f(n) = 30

for n = 3, f(n) = 100

for n = 4, f(n) = 354

for n = 5, f(n) = 1300

Observing above periodicity we can see that if (n%4==0) result of f(n)%5 is going to be 4 other wise result = 0. So, rather than calculating actual value of f(n) and then obtaining its value with mod 5 we can easily get result only be examine value of n.