Tough problem -please try

[guerrero]

Hello All.

Please help me understand the approach to solve the problem. I am finding it so tough to understand

g(x) is defined as the product of all even integers K such that 0<k<x.For e.g ,g(x)=2*4*6*8*10*12*14.
If g(y) is divisible by by 4^11,what is the smallest possible value of Y?

a)22
b)24
c)28
d)32
e)44

thanks in advance !

[mcressy]

Hi Guerrero,

The key here is realizing that you have to figure out how many times four divides. 4^11 is, by definition, divisible by 4 eleven times, and by extension, by 2 a total of 22 times (4^11 = (2^2)^11, or 2^22. Since we know that g(y) is going to be equal to 2x4x6x8x...y, in order to solve this we just need to figure out how many of these even numbers we need until reach a point where they cumulatively are divisible 22 times. Thus, we can track:

Number. How many times div by 2. Cumulative
2. 1. 1
4 2. 3
6. 1. 4
8. 3. 7
10. 1 8
12. 2. 10
14. 1. 11
16. 4. 15
18. 1. 16
20. 2. 18
22. 1. 19
24. 3. 22

As you can see, when you get to 24, you have a total of 22 '2's cumulatively, this that is your answer.

--Matt

[mcressy]

One other thing, Guererro, apologies for the formatting issues, the columns should be read across, like a spreadsheet. It's hard to get to look correct on an iPad.

[GMATGuruNY]

Hello All.

Please help me understand the approach to solve the problem. I am finding it so tough to understand

g(x) is defined as the product of all even integers k such that 0<k≤x. For example, g(14)=2*4*6*8*10*12*14. If g(y) is divisible by 4^11, what is the smallest possible value of Y?

a)22
b)24
c)28
d)32
e)44

thanks in advance !

The problem seems to have been transcribed incorrectly.
The changes in red (see above) reflect what I believe is intended.

4^11 = (2²)^11 = 2^22.
For g(y) to be divisible by 2^22, its prime-factorization must include twenty-two 2's.

We can plug in the answers, which represent the smallest possible value of y.
To determine the number of 2's included in the prime-factorization of g(y), count how many times EACH POWER OF 2 can be divided into g(y).

Answer choice A: 22
g(22) = 2*4.....20*22.

In the prime-factorization of g(22):
Every multiple of 2 provides at least one 2:.
22/2¹ = 11.
The calculation above indicates that g(22) includes 11 multiples of 2¹.

Every multiple of 2² adds a second 2:
22/2² = 22/4 = 5.
The calculation above indicates that g(22) will include five multiples of 2², increasing the total number of 2's by 5.

Every multiple of 2³ adds a third 2:
22/2³ = 22/8 = 2.
The calculation above indicates that g(22) will include two multiples of 2³, increasing the total number of 2's by 2.

Every multiple of 2^4 adds a fourth 2:
22/2^4 = 20/16 = 1.
The calculation above indicates that g(22) will include one multiple of 2^4, increasing the total number of 2's by 1.

Thus, the total number of 2's that can be divided into g(22) = 11+5+2+1 = 19.
Three more 2's are needed.
Thus, the prime-factorization must be expanded to include 24=2*2*2*3, implying that the smallest possible value of y is 24.

The correct answer is B.