What is the greatest value of integer n such that 4^n is a factor of 2

LHC8717

what if question is to find 4^n in 200! ??

It took me around 1 min 30 to 2 min, but with a number like 200! Your technique will be definitely faster, but I found it easier for me to have a picture of that.

First I solved using below method. From 26!
24 = 4x6
20 = 4x5
16 = 4x4
12 = 4x3
8 = 4x2
4 = 4x1
So n = 7
Why this is wrong? Thanks

You forgot all the multiple of 2, remember that a number divided by 4 it's the same than divided by 2 twice

For example you have 2 and 6 in 26! 2*6 = 12 it's divided by 4 (equal 3)

I also highlight in red the remaining multiple of 2 (8 could be divided by 4 once and after by 2 = 1)

Hi. Great method.
Do we just take the closest integer to be multiplied by? For example: 23/2= 11.5 which is approximately 12 but you took 11. We do not have to round off?

Shiv2016

yes 11 will be cosidered and rremainder is ignored

Since 4 = 2^2, we are actually trying to determine the largest value of n such that 2^(2n) is a factor of 26!.

Let’s first determine the number of factors of 2 within 26!. To do that, we can use the following shortcut in which we divide 26 by 2, and then divide the quotient of 26/2 by 2 and continue this process until we can no longer get a nonzero integer as the quotient.

26/2 = 13

13/2 = 6 (we can ignore the remainder)

6/2 = 3

3/2 = 1 (we can ignore the remainder)

Since 1/2 does not produce a nonzero quotient, we can stop.

The final step is to add up our quotients; that sum represents the number of factors of 2 within 26!.

Thus, there are 13 + 6 + 3 + 1 = 23 factors of 2 within 26!

However, we are not asked for the number of factors of 2; instead we are asked for the number of factors of 4. Since there are 23 factors of 2 within 26!, and since 23/2 = 11 remainder 1, there are 11 factors of 4 with a factor of 2 left over.

In other words, 2^23 = 4^11 x 2 (which equals 2^22 x 2^1 = 2^23).

Answer: D

Can somebody explain why this shortcut works?

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