BTGmoderatorDC wrote:If x is the product of the integers from 1 to 150, inclusive, and 5^y is a factor of x, what is the greatest possible value of y ?

A) 30
B) 34
C) 36
D) 37
E) 39

OA D

Source: GMAT Prep

To get the greatest possible value of y, we must count how many 5s are there in x = 1*2*3...150.

An easier way to do this is as follows.

Divide 150 by 5 and collect quotient. Again, divide the revised quotient by 5 and collect quotient. Do this process till the quotient is greater than 0.

"¢ 150/5 = 30;
"¢ 30/5 = 6;
"¢ 6/5 = 1
"¢ 1/5 = quotiet is 0. We should stop here.

Count of quotients = 30 = 6 + 1 = 37

Thus, there are thirty-seven 5s are there in x, thus, the maximum value of y = 37.

The correct answer: D

Hope this helps!

-Jay
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BTGmoderatorDC wrote:If x is the product of the integers from 1 to 150, inclusive, and 5^y is a factor of x, what is the greatest possible value of y ?

A) 30
B) 34
C) 36
D) 37
E) 39

-----ASIDE---------------------
A lot of integer property questions can be solved using prime factorization.
For questions involving divisibility, divisors, factors and multiples, we can say:

If N is a factor by k, then k is "hiding" within the prime factorization of N

Consider these examples:
3 is a factor of 24, because 24 = (2)(2)(2)(3), and we can clearly see the 3 hiding in the prime factorization.
Likewise, 5 is a factor of 70 because 70 = (2)(5)(7)
And 8 is a factor of 112 because 112 = (2)(2)(2)(2)(7)
And 15 is a factor of 630 because 630 = (2)(3)(3)(5)(7)
-----BACK TO THE QUESTION!---------------------

x is the product of the integers from 1 to 150, inclusive, and 5^y is a factor of x, what is the greatest possible value of y ?
If 5^y is a factor of x, and we want the greatest possible value of y, then we need to determine the MAXIMUM number of 5's "hiding" in the prime factorization of x.

x = (1)(2)(3)(4)(5)(6)(7).....(148)(149)(150)
We can see 5's hiding in several places
x = (1)(2)(3)(4(5)(6)(7)....(10)........
Since 10 = (2)(5), we can see that there's one 5 hiding in 10
Since 15 = (3)(5), we can see that there's one 5 hiding in 10
etc

So, it would SEEM that we need only find the multiples of 5 from 1 to 150 inclusive.
There are 30 such multiples of 5
HOWEVER, we've missed a few 5's since there are some values in which there's more than one 5 hiding.
For example, 25 = (5)(5), so we missed ONE 5
And 50 = (2)(5)(5), so we missed ONE 5 here
And 75 = (3)(5)(5), so we missed ONE 5 here
And 100 = (2)(2)(5)(5), so we missed ONE 5 here
And 125 = (5)(5)(5), so we missed TWO 5's here
And 150 = (2)(3)(5)(5), so we missed ONE 5 here
TOTAL number of 5's missed = 1 + 1 + 1 + 1 + 2 + 1 = 7

Total number of 5's = 30 + 7
= 37

In other words, 5^37 is a factor of x

Answer: D

Cheers,
Brent

BTGmoderatorDC wrote:If x is the product of the integers from 1 to 150, inclusive, and 5^y is a factor of x, what is the greatest possible value of y ?

A) 30
B) 34
C) 36
D) 37
E) 39

OA D

Source: GMAT Prep

The product of the integers from 1 to 150, inclusive, is 150!. To determine the number of factors of 5 within 150!, we can use the following shortcut in which we divide 150 by 5, and then divide the quotient of 150/5 by 5 and continue this process until we can no longer get a nonzero integer as the quotient.

150/5 = 30

30/5 = 6

6/5 = 1 (we can ignore the remainder)

Since 1/5 does not produce a nonzero quotient, we can stop.

The final step is to add our quotients; that sum represents the number of factors of 5 within 150!.

Thus, there are 30 + 6 + 1 = 37 factors of 5 within 150!.

Answer: D