IIFT 2019 Question 16

Question 16

The number $$37^{371}-26^{371}$$ is divisible by:

Solution

Option A: 

$$37^{371}-26^{371}$$ mod 10

=$$7^{371}-6^{371}$$

Since 7 and 10 are co-prime to each other

E(10) = 10*($$\left(1-\ \frac{\ 1}{2}\right)\left(1-\ \frac{\ 1}{5}\right)$$

=4

$$7^{4}$$ mod 10 = 1

$$7^{4*92}*7^{3}$$ mod 10 = $$1*7^{3}$$ mod 10 =3

$$6^{371}$$ mod 10 = 0

Hence $$37^{371}-26^{371}$$ is not divisible by 10

Option B: 

$$37^{371}-26^{371}$$ mod 11

$$4^{371}-4^{371}$$ mod 11

Hence $$37^{371}-26^{371}$$ is divisible by 11

Option C: 

$$37^{371}-26^{371}$$ mod 12

$$1^{371}-2^{371}$$ mod 12

Hence $$37^{371}-26^{371}$$ is not divisible by 12

Option D:

$$37^{371}-26^{371}$$ mod 15

$$7^{371}-11^{371}$$ mod 15

$$37^{371}-26^{371}$$ is not divisible by 15

B is the correct answer.


View Video Solution


Create a FREE account and get:

  • All Quant Formulas and shortcuts PDF
  • 40+ previous papers with solutions PDF
  • Top 500 MBA exam Solved Questions for Free

Comments

Register with

OR

Boost your Prep!

Download App