If m is the product of all integers from 1 to 40, inclusive

The factors of 10 are 1, 2, 5, and 10.

To find what k is, find how many times you can make 10 with the numbers in 1-40.

First, find the 10's as they are going to be most restrictive. Only 10, 20, 30 and 40 have 10 as a factor. 4 total.

Next find 5's. 5, 15, 25 and 35 each have a 5, with 25 having 2. 5 total.

You dont need to find the twos as you know there are more than 5's.

Add the two you get 9.

I would suggest that you refrain from actually counting the 5s/10s. Not only can it be tedious with bigger numbers, but it is also prone to errors. With smaller numbers, I don't think it will matter very much what you do.
For some theory on these kind of questions, check out this link:
[url]
if-n-is-the-product-of-integers-from-1-to-20-inclusive-106289.html#p834162[/url]

Once you understand such problems, you wouldn't care how big or small the numbers are... You would be able to solve such questions easily.

This is just an amazing way to solve it. so simple its crazy.

i like this approach, but could you clarify where you wrote "you don't need to find the two's"

As for your question, forget 40!. Let's look at 10!

What is the maximum value of p such that 10^p is a factor of 10!
To make a 10, you need a 2 and a 5. In 10!, will you have more 2s or more 5s?

1*2*3*4*5*6*7*8*9*10 - Every alternate number has at least one 2 in it. Every fifth number has at least one 5 in it. You certainly have more 2s than 5s. If you have eight 2s and two 5s, how many 10s can you make? You certainly need a 5 to make a 10. Since you have only two 5s, you can make only two 10s. It doesn't matter how many 2s are extra. You cannot make a 10 with only 2s.

We can also understand it as follows:

The products that give rise to 0's.

(2 * 5) * (12 * 15) * (22 * 25) * (32 *35) * 10 * 20 *30 * 40
= (2 * 5) * (12 * 15) * (22 * 5 *5) * (32 *35) * 10 * 20 *30 * (8 * 5)
= (2 * 5) * (12 * 15) * (22 * 5 ) * (32 *35) * 10 * 20 *30 * (8 * 5* 5)

The first four terms give rise to four 0's, the next three to three 0's and the last one to two 0's and the total is 9.

Note: You may also multiply 5 with 4 and/or 6 and/or 8 as they contain a 2. Similarly for 15, 25 and 35. But the number of 0's is restricted by the number of 5's.

Why divided 5 though?

Check out this link: https://anaprep.com/number-properties-h ... actorials/

It explains this method in detail. Also, you divide by 5 because to make a 10, you need a 2 and a 5. The number of 5s will be fewer than the number of 2s hence all you need to do is find the number of 5s you have. You can make that many 10s.

For similiar questions check Trailing Zeros Questions and Power of a number in a factorial questions in our Special Questions Directory.



Hope this helps.

If m is the product of all integers from 1 to 40, inclusive, what is the greatest integer p for which
10^p is a factor of m?

10^p is a factor of 40! i.e., how many zeros(trailing zero's) are at the end of integer if 40! is expanded in full form.

then it becomes very simple all of a sudden as we use

The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer n, can be determined with this formula:

\(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where k must be chosen such that 5^(k+1)>n

\(\frac{40}{5}+\frac{40}{5^2}+\frac{40}{5^3}\)......................

8+1+0=9

Thus C is the correct answer.

I know this is an old post but
Bunuel I understand your answer and I did remember to use the formula too however I didnt know what to do with 40/25. To take that as a 2 or a 1. Is it that we have to take an INT value or do we round it up?
Just stuck there

Thanks
Parth





1. Trailing zeros:
Trailing zeros are a sequence of 0s in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.

125000 has 3 trailing zeros;

The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer n, can be determined with this formula:

\(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where k must be chosen such that 5^(k+1)>n

It's more simple if you look at an example:

How many zeros are in the end (after which no other digits follow) of 32!?
\(\frac{32}{5}+\frac{32}{5^2}=6+1=7\) (denominator must be less than 32, \(5^2=25\) is less)

So there are 7 zeros in the end of 32!

The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.

Back to the original question:
If m is the product of all integers from 1 to 40, inclusive, what is the greatest integer p for which 10^p is a factor of m?
(A) 7
(B) 8
(C) 9
(D) 10
(E) 11

Given: \(m=40!\). So we should find the # of trailing zeros in 40!, as it'll be the greatest value of p for which 40!/10^p will be an integer.

40! has \(\frac{40}{5}+\frac{40}{5^2}=8+1=9\) trailing zeros, which means that 40! ends with 9 zeros so p=9 is the greatest integer for which 10^p (10^9) is a factor of 40!.

Answer: C.

You should take only the quotient into the account, thus 40/25 = 1.



Check Trailing Zeros Questions and Power of a number in a factorial questions in our Special Questions Directory.

Hope it helps.

Hey Bunuel,
You mentioned the value of K should be picked by 5^(K+1)>n...although in verbal explanation for the question, you said 5^(K+1) should be LESS than n, in this case 40.
Im guessing the sign in the equation for K should be <n?

Hi All,

We're asked to find largest value of P so that 10^P divides into the product of the first 40 positive integers. This question comes down to "prime factorization" and finding all the "5s" and "10s" in the first 40 positive integers.

10 = (5)(2)…so we're looking for all of the 10s that can be "made" in this big product.

Let's look at all the multiples of 5….

5, 10, 15, 20, 25, 30, 35, 40

So we have some obvious 10s

10
20 = (2)(10)
30 = (3)(10)
40 = (4)(10)

We can "make" some more 10s with the numbers that end in 5….

(5)(any even) = multiple of 10
(15)(any even) = multiple of 10
(35)(any even) = multiple of 10

I've separated the "25" for a reason; it creates an EXTRA 0 when multiplied by a multiple of 4…

(25)(4) = 100 = 10^2

So we have….4 + 3 + 2 = nine 10s…..P = 9

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Let's talk strategy here. Many explanations on this forum focus blindly on the math. While that isn't necessarily bad, remember: the GMAT is a critical-thinking test. For those of you studying for the GMAT, you will want to internalize strategies that actually minimize the amount of math that needs to be done, making it easier to manage your time. The tactics I will show you here will be useful for numerous questions, not just this one. My solution is going to walk through not just what the answer is, but how to strategically think about it. Ready? Here is the full "GMAT Jujitsu" for this question:

First of all, it is worth highlighting one of the commonly-found traps embedded in this question. I call it "Mathugliness" in my classes. (Get it? It's math. It's ugly. It acts like a thug. But, like most thugs, it's main game is to intimidate.) This question deliberately uses obnoxiously-large values, in the hopes of scaring novice test takers. After all, if you were to actually work out the "product of all integers from 1 to 40, inclusive", this would result in a massive number called \(40!\) (or "40 factorial.") It would turn out to be:

\(40*39*38*37*36*35*34*33*32*31*30*29*28*27*26*25*24*23*22*21*20*19*19*17*16*15*14*13*12*11*10*9*8*7*6*5*4*3*2*1=\)

\(815,915,283,247,897,734,345,611,269,596,115,894,272,000,000,000\)

Obviously, there is no way you are going to do this math. When you see "Mathugliness" on the GMAT, don't panic. There will always be a quicker, more conceptual way to look at the question, allowing you to avoid most of the messy math. The key is to focus on exactly what the question is asking, looking for logical leverage to strategically attack the question. In this case, the problem asks for the "greatest integer \(p\) for which \(10^p\) is a factor of \(m\)." In other words, "how many times could we factor a \(10\) out of \(40\) factorial?"

Of course, a possible trap would be to just think of the multiples of \(10\)s in the factorial:

\((40)\) \(*39*38*37*36*35*34*33*32*31*\) \((30)\) \(*29*28*27*26*25*24*23*22*21*\) \((20)\) \(*19*18*17*16*15*14*13*12*11*\) \((10)\) \(*9*8*7*6*5*4*3*2*1\)

There are four multiples of \(10\) in \(40!\), but this isn't even one of the answer choices. (To be honest, it should be. This is a missed opportunity on the part of the Test-maker!) But if we think about a "\(10\)" as a product of its parts (or \(2*5\)), it is obvious to see that there are a lot more \(10\)s embedded in the product. So the real question we should ask is: "how many \(2\)s and \(5\)s can we find in \(40!\)?"

There are A LOT of \(2\)s. After all, every 2nd integer will be divisible by \(2\). Since we are looking for the "greatest integer \(p\) for which \(10^p\) is a factor of \(m\)," the number of \(2\)s will not be our limiting factor. (In fact, we could actually factor out a "\(2\)" thirty-six times from \(40!\), but there is no way we could match up all of those \(2\)s with a corresponding \(5\) to create \(10\)s.) There are a lot fewer \(5\)s embedded in the expression.

Of course, we could count the multiples of \(5\), but doing this would also be a trap, based on a poor assumption:

\((40)\) \(*39*38*37*36*\) \((35)\) \(*34*33*32*31*\) \((30)\) \(*29*28*27*26*\) \((25)\) \(*24*23*22*21*\) \((20)\) \(*19*18*17*16*\) \((15)\) \(*14*13*12*11*\) \((10)\) \(*9*8*7*6*\) \((5)\) \(*4*3*2*1\)

If we were to merely do this, our answer would be \(8\). This is the most commonly chosen wrong answer. The way around this bad logic is to realize that the number of "multiples of \(5\)" embedded in a factorial isn't exactly the same thing as the number of \(5\)s. \(25 = 5*5\). Thus, it can contribute two \(5\)s to the total amount. The number of \(5\)s we could factor out of \(40!\) is therefore \(9\), and the answer is C.

Now, let’s look back at this problem through the lens of strategy. Your job as you study for the GMAT isn't to memorize the solutions to specific questions; it is to internalize strategic patterns that allow you to solve large numbers of questions. This problem can teach us patterns seen throughout the GMAT. The primary pattern this problem demonstrates is "Mathugliness" -- whereby the GMAT tries to "flex on you," using obnoxious or repetitive math to make the problem seem harder than it actually is. When you encounter Mathugliness, get excited. Look for leverage in the problem that will allow you to solve the problem conceptually, instead of working out all the math. In this particular problem, solid leverage can be found in the phrase "greatest integer." Questions that ask for maximum values are prime candidates for critical-thinking. You just need to look at how the problem is structured and determine what would limit or define the value you are looking for. Determine what those limits are, and you have your answer. And that is thinking like the GMAT.

What if the question read: "What is the greatest integer p for which 5^p is a factor of 40! ? Is there an efficient way to solve this?
Thanks a lot in advance

Hi SUV0508,

In your hypothetical question, we'd just have to 'find' all of the 5s within the product of 40! That's actually fairly straight-forward (since we'd just have to consider the multiples of 5). However, someone who was moving too quickly might not recognize that 25 = (5)(5) - meaning that THAT multiple of 5 actually has two 5's in it (while all of the others have just one 5).

5 = 1 x 5
10 = 2 x 5
15 = 3 x 5
20 = 4 x 5
25 = 5 x 5
30 = 6 x 5
35 = 7 x 5
40 = 8 x 5

There are nine 5s in 40!, so the largest possible value of P in your question would be 9.

GMAT assassins aren't born, they're made,
Rich

Contact Rich at: Rich.C@empowergmat.com