If n is the sum of all the divisors of 2^50 that are greater

Question

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...

sumande · Answer

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.

FN · Answer

brilliant!!!

can you explain the bolded piece please..thanks

VAGMAT · Answer

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.

Fistail · Answer

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.

mdot · Answer

Using another approach, I got A as well.

2^50 = 2^5 * 2^10 = 2^5 * 2^5 * 2^2 = 32*32 + 32*4

Unitis Digits = (2*2) + (2*4) = 12; Units Digit = 2

empty_spaces · Answer

2^50 is not equal to 2^5 *2^10

Robin in NC · Answer

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.

subhen · Answer

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.

Fistail · Answer

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.

r0m3416 · Answer

will go with C - The last two number's unit digit ought to be 2 and 4..so the units digit is 6

Kiski · Answer

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 ".

Could you please explain?

ioiio · Answer

2^50 = 1024 * 1024 * 1024 * 1024 * 1024 = x6 *x4 * x4 * x4 = x4 * x4 * x4 = x6 * x4 = x4

Ans : 4

edit : sorry dint even read the question fully.

2, 4, 8, 6 is the pattern. 

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.

47 mod 4 = 3 => x6 + x2 + x4 = x2

Ans : 2

vineetgupta · Answer

hi...I followed the same approach.

But made a silly mistake.
The OA is A....thanks.

sumande · Answer

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

excelgmat · Answer

I go with A too.

I have a nonquant background so here's my attempt.

n = 2^4+2^5+2^6+...+2^50

if you add 2^4+2^4+2^5+2^6+2^7. the unit digits add up to 0.

multiple by 11 and you have added upto 2^47 with the digit 0 as total.
After that it is easy. Add 2^48+2^49+2^50. which is 6+2+4= 2.

But I like the formulae approach of sumande.

somethingbetter · Answer

Yupp, it gotta be A. 
OG has similar questions with solutions exactly like Fistail has given.

If n is the sum of all the divisors of 2^50 that are greater

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