2 GMAT Number system question

Question

Question 1. It is given that 2^32+1 is exactly divisible by a certain no. which one of following is also divisible by same no.?
a. 2^96+1
b. 2^16-1
c. 2^16+1
d. 2^64+1
e. 7*2^32

question 2. The HCF and LCM of 2 nos. are 13 & 455 respectively. If one of the number lies between 75 & 125, then that number is
a.91
b.78
c.117
d.104
e. 65

MrMicrostrip · Answer

which is the simplest way to solve such questions.???

ilhom1986 · Answer

To answer question 1.
Usage of last digits needed to start solving:
2^1=2
2^2=4
2^3=8
2^4=.6

2^5=.2
2^6=.4
2^7=..8
2^8=..6

So, 2^32=...6+1= ...7
It is prime number and it can be divided by 1, and itself.
7*2^32 is divisible by 2^32+1
Answer e

Question 2.
To me it is more like a logical question:
Number is between 75 & 125 so 65 out

91 is only number
455/91=5 (which has an integer result)
91/13=7 (also suit LCM)
Answer a

scbguy · Answer

For the LCM and HCF let me be a little more detailed. 
Step 1 
Assume numbers to be x and y
Now they must have a 13 in their prime factor boxes to have a 13 in their GCF.

Step 2
If you look at the prime factors of 455 it is composed of 13*7*5
hence x and y should be 13*7 and 13*5

Step 3
Look at the answer choices and look at the constraints in the question. Clearly the answer choices should have 455 as a multiple. The answer choices you are left with are 91 and 65. 65 is automatically disqualified because it is part of the constraints. 91 is the answer

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scbguy · Answer

Need a little more explanation on the first question.

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MrMicrostrip · Answer

Ilhom1986....I need more explanation in First question.
Answer is not option (e), it's (a)

gurpreetsingh · Answer

2^{96}+1  =  2^{(32)*3} + 1^3  =  (2^{32}+1)  * something..

apply  a^3+b^3  =  (a+b)(a^2+b^2-ab)

hence A

MrMicrostrip · Answer

Thanx

for explaining...Now it's seems to be a very simple one...

puneetj · Answer

Thanks for the explanation for 1....i was able to answer question 2 but was stuck between b and e for question 1.

sukhbarinder · Answer

what i would have done is this: i would have changed this number question to a variable based question.. 2^32 + 1 = 2^n + 1 assuming n=1 i would get 3.. and 3 is divisilbe by 3

.. now i would change the answer options to variables assume n=1 and see if the ersult is divisible by 3... 32 is n, 16 is n/2, 64 is 2n and so on ...: 1st option: 2^96 + 1 = 2^3n +1 and for n=1 it would be 9 which is divisible by 3.. if u dont have time nd clocks ticking mark it and move ahead!!! 2nd option : 2^16 + 1 = 2^n/2 +1 for n=1 it would be root2 + 1 which is not divisble by 3. same for option 3:2^64 + 1 = 2^2n +1 for n=1 would be 5 not divisible by 3.. eliminated same goes for option 5: 14 is nt divisible by 3.. answer is A

this approach can backfire: so assume 2 values and solve it!!

sukhbarinder · Answer

for quetion 2:

product 0f two numbers = lcm * hcf of these numbers

i just need to find product of 13 and 455 and see which number in the options divides this product perfectly..

one thing imp : the product will end in 5.. so u can eliminate option b and d straight up as any integer product with them wont end in 5..
option e is a filler as per the question constraints..

now we have option a and option c.. little calc and the answer is a..

2 GMAT Number system question

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