Problem Solving

I do not know how to solve questions like these (quickly)! Can someone please explain!

1. What is the remainder when 2^91 is divided by 7? OA: 2

2. How many positive integers less than 1000 have no factors (other than 1) in common with 1000? OA: A
a) 400
b) 410
c) 411
d) 412
e) None of the above

3. How many divisors does 39690 have? OA: 60

garuhape wrote:
1. What is the remainder when 2^91 is divided by 7? OA: 2

2^3 = 8 which is when divided by 7 leaves us a reminder 1.

so write 2^91 = (2^90) * 2 = (8^30) * 2

When 8^30 is divided by 7, we have remainders as 1*1*.... 30 times, thus 8^30 divided by 7 leaves us 1 as remiader.

Now we are left with 2 which is the answer.

garuhape wrote: 2. How many positive integers less than 1000 have no factors (other than 1) in common with 1000? OA: A
a) 400
b) 410
c) 411
d) 412
e) None of the above

I hardly doubt this one if it can be tested in GMAT.
Essentially, we have to find out number of co-primes under 1000 above 1.
We have to use Euler's totient function here.

1. Factorize 1000.

1000 = (2^3)*(5^3)

2. Use the bases only not exponents for the calculation of number of co-primes.

N = 1000(1-1/2)(1-1/5) = 1000*1/2*4/5 = 400

garuhape wrote:
3. How many divisors does 39690 have? OA: 60

This is pretty well-discussed question in the forum.

1. Factorize the number.

39690 = (2^1)*(3^4)*(5^1)*(7^2)

2. Total number of divisors

N = (1+1) * (4+1) * (1+1) * (2+1) = 60

The formula is:

First break the number to all possible prime numbers (along with the powers) Then add one to each power and multiple.

N = a^x*b^y*c^z where a b c are distinct prime numbers and x y and z are corresponding powers then
Total number of factors = (x+1)*(y+1)*(z+1)

shovan85 wrote:

garuhape wrote:
1. What is the remainder when 2^91 is divided by 7? OA: 2

2^3 = 8 which is when divided by 7 leaves us a reminder 1.

so write 2^91 = (2^90) * 2 = (8^30) * 2

When 8^30 is divided by 7, we have remainders as 1*1*.... 30 times, thus 8^30 divided by 7 leaves us 1 as remiader.

Now we are left with 2 which is the answer.

I do not understand it. Any other suggestions?

Sie muss lernen "remainder theorm" hier...

https://www.purplemath.com/modules/remaindr.htm

nach dass...bitte schreiben (8^30)/7 = x^30/(x-1)...

solved it in a diff way. let me knw if im missing anythgn here.

1000 has 2 and 5 as prime factors, so all the numbers less than 1000 which are not divisible by 2 or 5 will give us the answer.
no. of 1 digit numbers not divisible by 2 or 5= 4
no. of 2 digits= last digit shud not be 0,2,4,5,6,8. total combinations = 9*4=36
similarly no. of 3 digit combinations = 9*10*4=360

ading up all the combos we get 400.

shovan85 wrote:

garuhape wrote: 2. How many positive integers less than 1000 have no factors (other than 1) in common with 1000? OA: A
a) 400
b) 410
c) 411
d) 412
e) None of the above

I hardly doubt this one if it can be tested in GMAT.
Essentially, we have to find out number of co-primes under 1000 above 1.
We have to use Euler's totient function here.

1. Factorize 1000.

1000 = (2^3)*(5^3)

2. Use the bases only not exponents for the calculation of number of co-primes.

N = 1000(1-1/2)(1-1/5) = 1000*1/2*4/5 = 400

1. What is the remainder when 2^91 is divided by 7? OA: 2


I do not understand it. Any other suggestions?

As i figured out 2^n follows a pattern when divided by 7. Lets see it more clearly:

2^1 = 2 /7 ---> Remainder 2
2^2 = 4/7 ----> Rem 4
2^3= 8/7 -----> Rem 1
2^4 = 16/7 ----> Rem 2
2^5 = 32/7 ----> rem 4

So it is repeating in a set of 4. So, 22*4=88. Then 3 more . So Remainder is 1.

Hope it helps.