Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: divisors of 2^50 [#permalink]
13 Aug 2007, 09:35

vineetgupta wrote:

If n is the sum of all the divisors of 2^50 that are greater than 10, what is the units digit of n?

(A) 2 (B) 4 (C) 6 (D) 8 (E) 0 Can someone explain this one...

I get A for this one.

The divisors of 2^5 that are greater than 10 are 2^4, 2^5 and so on up to 2^50. Their sum will be
n = 2^4+2^5+2^6+...+2^50
or, n = 2^4*(1+2+2^2+2^3+...+2^46)
or, n = 2^4*(2^47-1)/(2-1) = 2^4*(2^47-1)

Units digit of (2^47-1) will be (8-1) = 7.
So, units digit of n = units digit of (2^4 * 7) = (16*7) = 2.

Re: divisors of 2^50 [#permalink]
13 Aug 2007, 09:48

brilliant!!!

can you explain the bolded piece please..thanks

sumande wrote:

vineetgupta wrote:

If n is the sum of all the divisors of 2^50 that are greater than 10, what is the units digit of n?

(A) 2 (B) 4 (C) 6 (D) 8 (E) 0 Can someone explain this one...

I get A for this one.

The divisors of 2^5 that are greater than 10 are 2^4, 2^5 and so on up to 2^50. Their sum will be n = 2^4+2^5+2^6+...+2^50 or, n = 2^4*(1+2+2^2+2^3+...+2^46) or, n = 2^4*(2^47-1)/(2-1) = 2^4*(2^47-1)

Units digit of (2^47-1) will be (8-1) = 7. So, units digit of n = units digit of (2^4 * 7) = (16*7) = 2.

Re: divisors of 2^50 [#permalink]
13 Aug 2007, 10:27

vineetgupta wrote:

If n is the sum of all the divisors of 2^50 that are greater than 10, what is the units digit of n?

(A) 2 (B) 4 (C) 6 (D) 8 (E) 0 Can someone explain this one...

A. I got 2 too but i would do differently.

2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16

2^5 = 32
2^6 = 64
2^7 = 128
2^8 = 256

2^9 = 512

we know the pattren is repeating after every 4th power.

so in 2^50, the pattren repeats 12 times and then still remains 2 more powers.

the sum of the unit digits of every 4 powers = 0. so 2^50 has a sum of 6 (2+4) but we need to take out 2, 4 and 8 from 6. so we substract 14 (2+4+8) from x6, then the unit digit is 2.

Not sure I understand the question correctly, but if I take a shot at it...

the units digit seem to be repeating every four times {2,4,8,6,2,4,8,6 etc.} so with that said, I think the answer is B! The units digit should be 4.

That reasoning makes sense to me (the others are over my head). But I get a different answer. Since the question asks for the units digit for the sum of all those numbers, (48 numbers in all, 12 sets of 4), shouldn't the digit be 0, since the sum of 16, 32, 64, and 128 has a units digit of 0? Actually the sum of any 4 of the numbers in the series has a units digit of 0.

I will go with C. The answer according to me is 6.

I agree with the approach of Fistail. However I differ from him in the last stmt. The question asks us to find the units digit of all divisors greater than 10 hence there is no reason why we need to subtract 2+4+8 from x6.

I will go with C. The answer according to me is 6.

I agree with the approach of Fistail. However I differ from him in the last stmt. The question asks us to find the units digit of all divisors greater than 10 hence there is no reason why we need to subtract 2+4+8 from x6.

the question says factors grater than 10. so 2^1, 2^2 and 2^3 should not be included.
therefore 2, 4 and 8 needs to be subtracted from x.....x6.

Re: divisors of 2^50 [#permalink]
13 Aug 2007, 19:38

Fistail wrote:

vineetgupta wrote:

If n is the sum of all the divisors of 2^50 that are greater than 10, what is the units digit of n?

(A) 2 (B) 4 (C) 6 (D) 8 (E) 0 Can someone explain this one...

A. I got 2 too but i would do differently.

2^1 = 2 2^2 = 4 2^3 = 8 2^4 = 16

2^5 = 32 2^6 = 64 2^7 = 128 2^8 = 256

2^9 = 512

we know the pattren is repeating after every 4th power.

so in 2^50, the pattren repeats 12 times and then still remains 2 more powers.

the sum of the unit digits of every 4 powers = 0. so 2^50 has a sum of 6 (2+4) but we need to take out 2, 4 and 8 from 6. so we substract 14 (2+4+8) from x6, then the unit digit is 2.

I followed your reasoning till the statement ,"the sum of the unit digits of every 4 powers = 0".I dont quite follow this part,"so 2^50 has a sum of 6 (2+4) but we need to take out 2, 4 and 8 from 6. so we substract 14 (2+4+8) from x6, then the unit digit is 2".

Since it is greater than 10.. the pattern is x6, x2, x4, x8... You can see that there are 47 terms between 2 ^ 4 and 2^ 50. Each 4 term would lead to 0.

Re: divisors of 2^50 [#permalink]
13 Aug 2007, 21:05

sumande wrote:

vineetgupta wrote:

If n is the sum of all the divisors of 2^50 that are greater than 10, what is the units digit of n?

(A) 2 (B) 4 (C) 6 (D) 8 (E) 0 Can someone explain this one...

I get A for this one.

The divisors of 2^5 that are greater than 10 are 2^4, 2^5 and so on up to 2^50. Their sum will be n = 2^4+2^5+2^6+...+2^50 or, n = 2^4*(1+2+2^2+2^3+...+2^46) or, n = 2^4*(2^47-1)/(2-1) = 2^4*(2^47-1)

Units digit of (2^47-1) will be (8-1) = 7. So, units digit of n = units digit of (2^4 * 7) = (16*7) = 2.

Re: divisors of 2^50 [#permalink]
13 Aug 2007, 23:13

fresinha12 wrote:

brilliant!!!

can you explain the bolded piece please..thanks

sumande wrote:

vineetgupta wrote:

If n is the sum of all the divisors of 2^50 that are greater than 10, what is the units digit of n?

(A) 2 (B) 4 (C) 6 (D) 8 (E) 0 Can someone explain this one...

I get A for this one.

The divisors of 2^5 that are greater than 10 are 2^4, 2^5 and so on up to 2^50. Their sum will be n = 2^4+2^5+2^6+...+2^50 or, n = 2^4*(1+2+2^2+2^3+...+2^46) or, n = 2^4*(2^47-1)/(2-1) = 2^4*(2^47-1)

Units digit of (2^47-1) will be (8-1) = 7. So, units digit of n = units digit of (2^4 * 7) = (16*7) = 2.

That is the Geometric Progression (a, a*r, a*r^2, a*r^3, and so on) sum. If a is the first term, r the common ratio and n the number of terms, then
sum = a*(r^n - 1)/(r-1); if r>1
or, sum = a*(1 - r^n)/(1-r); if r<1