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30^30 yields 30 zeros

20^20 yields 20 zeros

10^20 yields 10 zeros

Total=60 zeros

The other numbers which when multiplied gives a 10 are 2 and 5. There are plenty of ‘2s’ there; wise would be to count ‘5s’.

5^5 yields five 5s

15^15 yields fifteen 5s

25^25 yields twenty five 5s

All together there are seventy 5s. There are definitely seventy 2s there in the product.

Total=70 zeros from this product

TOTAL = 70+30=130

Sorry a small mistake 10^10 yields 10 zeros

Hi Mohammad,

You are exactly correct here! Great job! Your work almost exactly aligns with our explanation. Keep it up, and have a great day! 🙂

hi chris,

105 is answer but it’s much time consuming question.can you give us short cut for this type of question?

A garment order consists of jackets costing $36 each and shirts costing $26 each. If the total

cost of the order is $1,200, and if the average (arithmetic mean) cost per garment is $30, how

many more shirts than jackets are in the order?

this kind of question also take much time

Regards,

pavan.

130 but to deal with problem take lot of time in exam.

Hi Pavan,

Once you see that you are just counting zeroes (30 zeros in 30, 20 zeroes in 20, etc. and realizing that you don’t need to count any ‘2’s just the fives, 50 fives in 25^25, 15 fives in 15, etc.), this problem can be done in less than two minutes.

Of course, that’s no easy feat!

I thought 90 would be

The answer is 130. The explanation is up on the blog 🙂

63 zeros:

1) 0 in power of 30 yields 30 zeros

2) 20 ^20 – 20 zeros

3) 10 ^10 – 10 zeros

now you have 60 zeros

4) 5^5 (a number that ends in 5), multiplied by an even number (2^2 or 4^4) – 1 zero

5) same with 15^15 multiplied by an even number (e.g.14^14) – 1 zero

6) and 25 ^ 25 multiplied by an even number – 1 zero

Total 63 zeros

Olha,

The part with the zeroes is correct. With the ‘5’s, though, each five (meaning the five 5’s in 5^5) can be multiplied by some even number–and there are plenty of those spread throughout this sequence of numbers. So you want to count all the fives. There are 80 of those, so the answer is 130.

Hope that helps!

3^30 2^20 10^10 are seemingly he only candidates that could give rise to trailing zeros.

(3×10)^30

(2×10)^20

(1×10)^10

powers of 3 are 3, 9, 12, 15, 18, 27

powers of 2 are 2, 4, 8, 16, 32, 64,128

any power of 1 is 1

Therefore we can ignore 3^30, 2^20 and 1^10.

So what we have left is 10^30, 10^20 and 10^10.

So I am guessing the answer is 30 + 20 + 10 = 60.

What about the ‘5’s 🙂

Any 5 multiplied by an even number will give you a zero.

See if you can get the answer now 🙂

60

You may have forgotten to count the ‘5’s 🙂

Also, the explanation is now up!

In this case we have to find the pair (2*5) in the above sequence and we can break each number separately for example:

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

20^20=(2*5*2)^20

10^10=(2*5)^10

At this time we have obviously 60 zero’s however we can create more pairs of (2*5) especially the number of five’s it will determine the number of zero’s due to the number of two’s is bigger* :

25^25=(5*5)25——->50 five’s

15^15=(3*5)^15——>15 five’s

5^5—->5 five’s

Eventually if we add how many number we have, then we will find that we have 130 which can much with 130 2’s in order to fix the pair (2*5)

*(Each even number is a multiple of two with something on this case you have 2^30,2^28,2^26,2^24,2^22,2^20,2^18,2^16,…….2^2)——>240 two’s however you will need only 130 from the total of 240.

So, we have 130 zero’s!!!!

Perfect explanation…but when working this problem out in real time, you don’t want to count out every ‘2’. Instead, you should have the realization that there will be a superfluity of ‘2’ spread throughout the sequence with which to match with the ‘5’s.

Again, good job 🙂

30

Don’t forget 20^20 and 10^10…the explanation is now up 🙂

105!

5+10+15+20+25+30=105

Close, but I think you forgot that extra batch of 25’s in 25^25. That would give you a total of 130 zeroes. BTW, the full explanation is now up on our blog 🙂

130

We have to look for the total number of factors of 2 and 5 (10 = 2*5).

Since the factors of 5 is the most restrictive, once we know the number of factors of 5 we can easily determine the total number of zeros at the end

(5)^5 (5 factors of 5)

(10)^10 = (2 * 5)^10 = (2^10)*(5^10) (10 factors of 5)

(15)^15 = (3*5)^15 = (3)^15 * (5)^15 (15 factors of 5)

(20)^20 = (2*2*5)^20 = (2^2)^20 * (5)^20 = (2^40) * (5)^20 (20 factors of 5)

(25)^25 = (5^2)^25 = (5)^50 (50 factors of 5)

(30)^30 = (2*3*5)^30 = (2)^30 * (3)^30 * (5)^30 (30 factors of 5)

Total no of factors of 5 = 5 + 10 + 15 + 20 + 50 + 30 = 130

Similarly we can calculate the number of factors of 2 (which suffice 130) could be greater

So total number of zero’s at the end = 130

You start from 5, but what about factors of 2 and 4? Also, the product of even numbers and 5, 15 etc, will yield 0 as the last digit.

I think what Jenil is doing is matching the 5’s with 2’s, since 5 x 2 = 10. Thus each 5 x 2 pairing will result in an extra ‘2’.

Hope that makes sense 🙂

I just want to confirm one thing — are those x’s in between the numbers or the multiplication sign? Assuming the latter.

Even I had that the same doubt..hover over the image the “alt” attribute shows multiplication

Hi Samy,

Yes, those are multiplication signs 🙂

They are multiplication signs 🙂