**All the questions solved here are relevant to Quantitative Aptitude questions such as those asked in examinations conducted by UPSC.**

**Read on to find out the complete solution to this problem.**

__Today's Problem(05/09/2015)__

**Questions Credit : http://careerbless.com**

**1.**

**{(481 + 426)2 - 4 x 481 x 426} = ?**

**A. 3025**

**B. 4200**

**C. 3060**

**D. 3210**

**2.**

**(64 - 12)2 + 4 x 64 x 12 = ?**

**A. 5246**

**B. 4406**

**C. 5126**

**D. 5776**

**3.**

**121 x 5^4= ?**

**A. 68225**

**B. 75625**

**C. 72325**

**D. 71225**

**4.**

**If (2^32 + 1) is completely divisible by a whole number, which of the following numbers is completely divisible by this number?**

**A. (2^96 + 1)**

**B. (7 x 2^23 )**

**C. (2^16 - 1)**

**D. (2^16 + 1)**

**5.**

**What is the unit digit in (6324)^1797× (615)^316× (341)^476?**

**A. 1**

**B. 2**

**C. 4**

**D. 0**

__Solutions__

**1.The numbers are in the format (a + b)^2 - 4ab = a^2 + b^2 -2ab = (a - b)^2**

**Therefore,**

**{(481 + 426)2 - 4 x 481 x 426} = (481 - 426)^2 = (55)^2 = 3025**

**2.The numbers are in the format**

**(a -b)^2 + 4ab =**

**a^2 + b^2 + 2ab = (a + b)^2**

**Therefore,**

**(64 - 12)2 + 4 x 64 x 12 = (64 + 12)^2 = 5776**

**D.5776**

**3. 121 x 5^4 = (11)^2 x (25)^2 = (275)^2 = 75625 (Without this note,it will seem like a huge,time-consuming task but it is nothing like that,all you need to multiply is 27x28 and 5x5.Click here for learning.)**

**B.75625**

**4. (2^96 + 1) = (2)^32*3 + (1)^3 = (2^32 + 1)(2^64 - 2^32 + 1),which is divisible by (2^32 + 1).**

**A.(2^96 + 1)**

**5. Just keep an eye on the last digits of each number. If you multiply odd numbers of 4,you will get 4 as the one's digit whereas multiplying even numbers of 4 will give you 6 as one's digit.**

**In the first place,we have**

**(6324)^1797,now 1797 is an odd number,so the one's digit you will get is 4.**

**No matter how many times you multiply 5,you will get only 5 as the one's digit.**

**Again,no matter how many times you multiply 1,you will get 1 as the one's digit.**

**So,now you multiply 4x5x1 = 20,where the unit digit (or one's digit) is 0.**

**Therefore,the correct answer is 0.**

**D.0**

For the 4th, there is an error in the answers. It gave me trouble.

ReplyDeleteIn the Fifth, it's much simpler.

615 is a multiple of 5. So 615^316 is a multiple of 5^316. The same deduction gives that (6324)^1797 is a multiple of 2^1797 with is a multiple of 2^316.

Hence, the big product is a multiple of 10^316, and there is 316 zeros!

There was error in the question itself.I missed the exponents.I have corrected it now.

DeleteAnd yes,this solution is definitely cooler! :)