If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest value of

Bunuel, I don't understand:

the number of trailing zeros for 10! is 2 according to the formula you explained there:
gmat-club-m12-100599.html

The number of trailing zeros for \((5!)^2\) is 2 (one for 5!, again, according to the same formula and common sense).

So the difference of two numbers should give 2 trailing zeros:
xxx00
-
yyy00
--------
zzz00

Could you please explain again? Thank you.

The red part is not correct. Consider this:

123,000
-23,000
-------
100,000

By the way: 10!-2*(5!)^2=3,628,800-28,800=3,600,000 - five trailing zeros.

10! = 2^8*3^4*5^2*7= 2 zeros (10*5*2)

after rearranging 10! = 100*2^6*3^4*7

2*5!^2 = 2^7*3^2*5^2=2 zeroes(5*2*5*2)

after rearranging 100*2^5*3^2

now 10!-2*5!^2 = 100(2^6*3^4*7 - 2^5*3^2)
= 100*2^5*3^2(2*3^2*7 - 1)
=100*2^5*3^2*125
=100 *2^5*3^2 *5^3
= 10^5 *2^2*3^2

OA:E

LOGIC: In fact this is a divisibility problem. In order 10! - 2*(5!)^2 to be divisible by 10^n , 10! - 2*(5!)^2 and 10^n must have the same number of 10's.
10 in terms of prime factors can be written as 10 = 2*5 --> 10^n = (2^n)* (5^n).
So the problem asks how many 2's equals to 5's the 10! - 2*(5!)^2 has.

Since we will loose valuable time by doing all these multiplications the easiest way is to find common terms.
10! - 2*(5!)^2 = 5!*6*7*8*9*10 - 2(5!)^2 = 5!*(2*3)*7*(2*4)*9*(2*5) - 2(5!)^2 = 5!* 5!* (7*9*2*2) - 2(5!)^2 = (5!)^2*252 -2(5!)^2 =(5!)^2 (252-2)=(5!)^2* 250

Since it is a product we can take its term separately.
# of 2's and 5's in (5!)
#of 2's in 5!: 5!/2 + 5!/4 = 2+1 = 3
#of 5's in 5!: 5!/5=1
BUT it is (5!)^2 --> There are 2 5's AND 6 2's in in (5!)^2
(5! has more factors of 2 than factors of 5 since every second number is multiple of 2. Thus, it would have been more effective to search only for factors of 5)

# of 2's and 5's in 250
With prime factorization 250 = 2*5^3. Thus there are 3 5's AND 1 2 in 250

In sum: (5!)^2* 250 = 2^6 * 5^2 * 2 * 5^3 * .... = 2^7 * 5^5. We want equal number of 5's and 2's so the greatest number of 5's and 2's in the (5!)^2* 250 is 5.

answer: E

Dear Bunuel (or whoever sees this)

in this step : 250 * (5!)^2 = 250 x 120 x 120 = 5^2 x 12 x 12 x 1000, I don't get why you multiply by 1000 and also why is 250 turning to 5^2 , and the 120's turning to 12's?

Thanks

 

\(250*(5!)^2=\)

\(=250*120^2=\)

\(=250*120*120=\)

\(=(25*10)*(12*10)*(12*10)=\)

\(=25*12*12*1000\).­

If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest value of n?

A. 1
B. 2
C. 3
D. 4
E. 5

The equation can be simplified as
=>(5!)(10*9*8*7*6*5 - 2*(5!))
=>120(30240-240)
=>(10^5) * K (No need to calculate the remaining part as we knew the power of 10)

Hence n=5 which is option E.

Could someone please tell me if this line of thinking is correct?

250*(5!)^2

Here 250 = 5*5*5*2 = 3 trailing zeros because 5 appears 3 times

and (5!)^2 has 2 trailing zeros, which brings the total = 3+2 = 5 trailing zeros.

The number 250 has just one "trailing zero", since there is just one zero at the end of the number. In general, if you can write a number as (2^a)(5^b)(other primes not equal to 2 or 5), then the number of trailing zeros will be the smaller of the two exponents a and b.

And I think when you're considering the number of zeros at the end of the product (250)(5!)^2, you're adding the number of zeros at the end of 250 to the number of zeros at the end of (5!)^2. That won't generally give you the correct answer (it will sometimes, but not always, and not here). For example, the numbers 4 and 25 don't have any trailing zeros, but 4*25 = 100 has two. You genuinely need to find how many 2s and how many 5s are in the product, and use the method I mention above.

All of that said, the question in this thread is either at the very upper end of GMAT difficulty, or is beyond the scope of the test. If any of the relevant concepts are fairly new to you, I would not recommend even attempting this question right now, and I'd generally only suggest spending time on questions like this one if you've mastered all of the commonly tested material, since a question like this one is extremely unlikely to appear on your actual test.

Ok, I really don't want to break my head doing all this stuff, so : here's what I would do:

10! = 3628800
2*5!*5! = 2*120*120 = 28800
3628800-28800=3,600,000
Now
the above is divisible by 10^5

... Had to google 10! I think remembering the value is quite useful on the test..

I'd recommend you not waste time memorizing things like this -- the chance it would be helpful to know the numerical value of 10! on a real GMAT question is zero, I suspect. I've seen several thousand official questions, and I can't recall even one where it would have helped to even know 7! numerically, let alone a number as big as 10!.

If someone can explain me how did Bunuel derive step 2 from from step 1
Step 1 - 252 * ((5!)^2) - 2((5!)^2)
Step 2 - 250 * (5!)^2

Bunuel, IanStewart or if any of you can guide, thanks

It's 252*x - 2*x = 250*x, where x is (5!)^2.

Hope it's clear.

I tried it as below:

Breaking down the numbers into prime factors:

10.9.8.7.6.5.4.3.2 - (2.5.4.3.2.5.4.3.2)
= 2.5.9.8.7.6.5.4.3.2 - (2.5.4.3.2.5.4.3.2)

Taking out common terms

= 2.5.9.8.7.6.5.4.3.2 - (2.5.4.3.2.5.4.3.2)

Leaves us with -

= (2.5.4.3.2.5.4.3.2)(2.9.7 - 1)
= 2.5.4.3.2.5.4.3.2.125
= 2.5.4.3.2.5.4.3.2.5.5.5

We can see we have 7 2's and 5 5's and hence 5 becomes limiting factor giving us 5 as the answer.

 

­
Hi Bunuel I saw your method and it is a great method. However, I was trying to use the power of prime method. I could not get the correct answer.

\(10! - 2*(5!)^2\)­
To find the geatest value of n such that the expression is divisible by \(10^n\)­, first I thought to find out the power of 5 in each component of the numerator. (Because: \(10^n = 2^n*5^n\))­.

For 10!, power of 5 is 10/5 = 2
For \(2*(5!)^2\)­, power of 5 is 5/5 = 1 for 5! and (1+1) = 2 for \(5!^2\). Therefore, n = 2.

Where did I go wrong? Thank you.­

All is correct until the red part. We need to find the number of trailing zeros of 10! - 2*(5!)^2, and the difference between numbers each having two trailing zeros does not necessarily imply the result will also have two. For example, 1,100 - 100 = 1,000, which has three trailing zeros. So, you should use the method described above.