Find the number of trailing zeros in the product of (1^1)*(5^5)*(10^10

We have a trailing zero when we multiplying 2 by 5. Now, in the product: (1^1)*(5^5)*(10^10)*(15^15) *(20^20)*(25^25)*...*(50^50) there will be obviously more 5-s than 2-s so the # of 2-s will be limiting factor for the # of trailing zeros.

So we should count # of 2-s for even bases, basically we should factor out 2-s: \(10^{10}*20^{20}*30^{30}*40^{40}*50^{50}=\)
\(=2^{10}*(2^2)^{20}*(2)^{30}*(2^3)^{40}*2^{50}*(something)=2^{10+40+30+120+50}*(something)=2^{250}*(something)\).

So there will be 250 trailing zeros in the above product as there are at least as many 5-s as 2-s.

Answer: C.

Theory on this topic: everything-about-factorials-on-the-gmat-85592.html
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Looking at the numbers it looks like

(1x5)^5
(2x5)^10
...
(10x5)^50

1. Determine the limiting factor. Is it 2 or is it 5? We know that all the numbers are multiple of 5 but not of 2. Thus, the limiting factor in this case is 2. Let's drop all the 5. Then, we count factors of 2 of even multiples.

2^10 = 10
4^20 = 20 + 20
6^30 = 30
8^40 = 40 + 40 + 40
10^50 = 50

250

Answer: C

More examples of Trailing Zeroes: Arithmetic: Trailing Zeroes

Thanks Bunuel.

My remark:
Because of conditions of the stem, I think, this question is an exclusion from the general approach where one must count 5s a number of which in factorials are usually less or equal to a number of 2s.

How come (2)^30 had no power inside with the 2?

\(30^{30}=(2*15)^{30}=2^{30}*15^{30}\)
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Hi

In this step:
So we should count # of 2-s for even bases, basically we should factor out 2-s: \(10^{10}*20^{20}*30^{30}*40^{40}*50^{50}=\)
\(=2^{10}*(2^2)^{20}*(2)^{30}*(2^3)^{40}*2^{50}*(something)=2^{10+40+30+120+50}*(something)=2^{250}*(something)\).

Why we counted only 2? what does it mean by limiting factor and whats the importance of it??
Please explain in detail

Thanks a lot

We have a trailing zero when we multiplying 2 by 5. So, each pair of 2 and 5 gives one more 0 at the end of the number. Our expression gives more 5's than 2's, so the number of 2 will determine the number of 0: for each 2 we have a 5, which when multiples will give 0.

Check here for theory: everything-about-factorials-on-the-gmat-85592.html

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Responding to a pm:

The method discussed in my post is useful while finding the maximum power of a number in a factorial. The given product is not in factorial form and hence the method needs to be suitably modified. That said, it should not be a big problem to modify the method if you understand the basics. Zeroes are produced by multiplying a 2 and a 5. So number of 0s in this product will depend on how many matching 2s and 5s we have here. In factorials, we have more 2s than 5s because we have consecutive numbers so we usually don't bother about finding the number of 2s. Here we have handpicked numbers so we need to ensure that we have both.

From where do we get 2s? From even multiples of 5.
10^10, 20^20, 30^30, 40^40, 50^50
\((2*5)^{10}, (2^2*5)^{20}, (2*15)^{30}, (2^3*5)^{40}, (2*25)^{50}\)
So number of 2s is 10 + 40 + 30 + 120 + 50 = 250

Now, let's see the number of 5s. Each term of the product has a 5. So the number of 5s is at least 5 + 10 + 15 + 20 + 25 + ... + 50
Then we also need to account for terms that have multiple 5s such as 25 and 50 but let's get to that later.

5 + 10 + 15 + 20 + 25 + ... + 50 = 5(1 + 2 + 3 + ..10) = 5*10*11/2 = 275

Note that number of 5s will be even more than 275 while the number of 2s is only 250. So there will be 250 trailing zeroes.

Is the answer 250 or 10²⁵⁰ ?

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The answer is 250.
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VeritasKarishma, shouldn't the number of 5s be 275 + 25 + 50 = 350 (additional 25 and 50 that get missed if you just sum the AP. Here is how : 25^25 = 5^50 and similarly 50^50 = {(5^2)*(2)} ^ 50 = 2^50*5^100)

A zero is formed by the multiplication of a 2 and a 5. Find the number of 2's and 5's and choose the lower of the 2 powers.


\((1)^1 \space * \space (5)^5 \space * \space (10)^{10} \space * \space (15)^{15} \space * \space (20)^{20} \space * \space (25)^{25} \space * \space (30)^{30} \space * \space (35)^{35} \space * \space (40)^{40} \space * \space (45)^{45} \space * \space (50)^{50}\)


= \((5)^5 \space * \space (2 * 5)^{10} \space * \space (3 * 5)^{15} * (2^2 * 5)^{20} \space * \space (5^2)^{25} \space * \space (2 * 3 * 5)^{30} \space * \space (5 \space * \space 7)^{35} \space * \space (2^3 \space * \space 5)^{40} \space * \space (3^2 \space * \space 5)^{45} \space * \space (2 \space * \space 5^2)^{50}\)


Taking out the 2's and 5's, we have


\([2^{10} \space * \space (2^2)^{20} \space * \space 2^{30} \space * \space (2^3)^{40} \space * \space 2^{50}] \space * \space [5^5 \space * \space 5^{10} \space * \space 5^{15} \space * \space 5^{20} \space * \space (5^2)^{25} \space * \space 5^{30} \space * \space 5^{35} \space * \space 5^{40} \space * \space 5^{45} \space * \space (5^2)^{50}]\)


Separating the 2's and 5's



= \([2^{10} \space * \space 2^{40} \space * \space 2^{30} \space * \space 2^{120} \space * \space 2^{50}] \space * \space [5^5 \space * \space 5^{10} \space * \space 5^{15} \space * \space 5^{20} \space * \space 5^{50} \space * \space 5^{30} \space * \space 5^{35} \space * \space 5^{40} \space * \space 5^{45} \space * \space 5^{100}]\)


= \([2^{10 \space + \space 40 \space + \space 30 \space + \space 120 \space +\space 50} ] \space * \space [5^{5 \space + \space 10 \space + \space 15 \space \space + \space 20 \space + \space 50 \space + \space 30 \space + \space 35 \space + \space 40 \space + \space 45 \space + \space 100}]\)


= \(2^{250} \space * \space 5^{345}\)


Choosing the lower of the 2 powers, the number of trailing zeroes = 250


Option C

Arun Kumar
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Yes, there will be but we don't need to bother about the extras because we have already reached more than 250 through the AP. We just mark 250 and move on. See the highlighted part.

Usually, when looking for Trailing Zeroes (which is in effect looking for how many Factor Pairs of (2*5) we can Find in a Product Chain) there will be FEWER Prime Factor 5s < than Prime Factor 2s

However, since every Base being Multiplied is a Multiple of 5 (and ONLY EVERY OTHER Base being Multiplied is EVEN) there will definitely be fewer Prime Factor 2s in the Product Chain.

Thus, if we find the MAXIMUM Value of the Prime Factor 2s that can Evenly Divide into the Product Chain, there will be more than enough Prime Factor 5s to Pair off with and create a Trailing Zero.


10^10 ---- 2^10

20^20 ----- (2^2 * 5) ^ 20 ----- 2^40

30^30 ---- (2 * 3 * 5)^30 ---- 2^30

40^40 ---- (2^3 * 5)^40 ---- 2^120

50^50 ---- (25 * 2)^50 ----- 2^50


Tallying up the Prime Factor 2s:

2^10 * 2^40 * 2^30 * 2^120 * 2^50 = (2)^250


Answer is C - there will be 250 Trailing Zeroes in the Product

Thank you crackverbalGMAT for the detailed explanation. I had understood the part where 2 is the limiting factor and we need to take only even bases for 2(as odd multiples of 5 doesn't contain 2s) and calculated no. of 2s in each even base(I went by the 10/2 + 10/4 +... way) but never got the right answer- I got something which was not there in the option, couldn't figure out where I was going wrong. Your explanation helped me to understand the given answer and how we reached the solution. Thanks a lot

I think it will be 5^350, not 5^345.

Why do we the choose the lower of the 2 powers?

The number of trailing zeros in a number is the number of 2-5 pairs among the factors of that number. While we could determine both the number of 2's and the number of 5's in this product, it should be clear that there are more 5's in this product than there are 2's (every factor contains 5's, but only every other factor contains 2's). Therefore, the number of 2-5 pairs is determined by the number of 2's in the product.

Let's count the number of 2's in the product:

\(\Rightarrow\) 10^10 = (2^10)(5^10) (10 factors of 2)

\(\Rightarrow\) 20^20 = (4^20)(5^20) = (2^40)(5^10) (40 factors of 2)

\(\Rightarrow\) 30^30 = (2^30)(3^30)(5^30) (30 factors of 2)

\(\Rightarrow\) 40^40 = (8^40)(5^40) = (2^120)(5^40) (120 factors of 2)

\(\Rightarrow\) 50^50 = (2^50)(25^50) (50 factors of 2)

In total, there are 10 + 40 + 30 + 120 + 50 = 250 factors of 2 in the product. Therefore, there are 250 2-5 pairs in the product, which means the number of trailing zeros is 250.

Answer: C
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Concept of trailing zero has to do with a combination of 2 and 5. One factor to consider though is which of them is limiting,in this case it is 2 because the 5s are more.

Number of 2s will be
2^10 x 2^40 x 2^30 x 2^120 x 2^50

Which totals 250.
GMAT quant instructor, Lagos, Nigeria.

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