Stuck with factorials, some tips would be great!

[sunset_chill]

Dear all,

I am totally puzzled about how to approach the following problems, can anybody help??
1.If (13!)^16 - (13!)^8/(13!)^8 + (13!)^4 = a , what is the unit’s digit of a/(13!)^4?
a)0
b)1
c)3
d)5
e)9

2.What is the highest integer power of 12 that divides 27! evenly?

3.For every positive integer n, the function h(n) is defined to be the product of all the even integers from 2 to n inclusive. If p is the smallest prime factor of h(100) + 1, then p is? a)between 2 and 10
b)between 10 and 20
c)between 20 and 30
d)between 30 and 40
e)greater than 50

4.If 17! + 1≤ K ≤ 17! + 17, is K ever a prime number?

5.If n is the product of the integers from1 to 20 inclusive, which of the following is the greatest integer k for which 2k is a factor of n?
a) 408
b) 437
c) 486
d) 532
e) 1242

Thank you very much in advance!
sunset_chill

[larbgai]

Here's the answer to number 1. If you want to explore this type of problem in more depth (and understand the GMAT's approach to these), I suggest taking a look at this problem as well. Beat The GMAT Forum :: View topic - GMAT Prep Stumper
It explores some similar themes (although is harder than this one).

On to the problem:
1.If (13!)^16 - (13!)^8/(13!)^8 + (13!)^4 = a , what is the unit’s digit of a/(13!)^4?

Our first job is to figure out how to get a/(13!)^4
I always just start factoring out common numbers, and 100% of the time you will see what you are looking for quite quickly. So let's start with the left side of the equation. To save time, I am not going to use parentheses for everything, but you should be able to follow me....

13!^16 - (13!)^8/(13!)^8 + (13!)^4

Right away, it is clear that middle term = 1. So just make that 1.

13!^16 - 1 + 13!^4 = a

Now subtact the 1.

13!^16 + 13!^4 = a+1

Now, that we have simplified, this is a common GMAT trap. They want you to assume you are multiplying, and add the exponents. Don't do it. You need to figure out how to factor out only the common terms.

Here is how:

(13!)^16 = (13!)^4 * (13!)^12

So as far as our equation is concerned: (13!)^12 * (13!)^4 + (13!)^4 = a+1

NOW, we can factor out (13!)^4:
So..... (13!)^4 * (13!^12 + 1) = a+1

Now divide by (13!)^4, so: (13!)^12 + 1 = (a+1)/(13!)^4

Now we're pretty close to what we want. Let's slit up the fraction on the right....

(13!)^12 + 1 = a/(13!)^4 + (1/(13!)^4)

Now lets move 1/(13!^4) to the other side:

(13!)^12 - 1/((13!)^4) + 1 = a/((13!)^4)


Ok, so now we have what we want on the right side. The last part is to figure out the last digit of everything on the left side.

Remember, the factorials can't be calculated. The GMAT is making you painfully aware of this by using 13! That number is way too large for you to memorize, and to calculate it would probably take you 15 min if you're good at multiplication. Even if you did calculate it, then you would have to find the rule for taking that number to whatever exponent we are talking about. By that time, the test will be over. Anyway, the point is don't even think about calculating the number.

What do we know about 13! that will allow us to find its last digit?
Well, 13! = 1*2*3*4*5*6*7*8*9*10*11*12*13

Notice that the number has 10 as a multiple. ANY number * 10 has zero as its last digit. So if we multiply all the other numbers together, and save 10 for last, our last digit must be 10. Since the order of multiplication is irrelevant, our final digit must be 10 (I am going into detail behind the logic for this, because that is what the GMAT is asking you to do here).

So (13!)^12 has a last digit of 0. Now we have to subtract 1/(13!)^4 from this number.

Again, we need to use logic. 1/(13!)^4 is some ridiculously small fraction. If we were to convert it to a decimal, it would look something like .0000000000000008734 as an example (I made up the numbers after the zeros - the point is they don't matter).

So, it is almost irrelevant as far as affecting (13!)^12. However, since we know that 13!^12 has a zero as its last digit, when it subtracts .00000000008734, that zero will become a 9.

So now we are almost done. Finally, we just need to add the remaining 1. Since our last subtraction flipped us from a 0 to 9, adding 1 is going to flip the number back to 0 (it will be very very close to 1, since the fraction we subtracted is so small).

So the last digit is zero. Answer is A. Official Answer = ?

[larbgai]

Here is the answer to #2. The link I posted at the top of my response to #1 is the answer to #3.

#2)

27! is another incalculable number. They are asking for the greatest power of 12 that will divide this number. 12 = 2*3*3

So to reword the problem, how many prime factors of 2 and 3 can we eliminate, that are in the ratio 2:1 (two 2's to 3 ones).

At this point, I opted to just factor each number from 2 to 27, and count the primes. I'm sure there is a more proof like approach than this, but if I were taking the test, this process only takes about a minute, so I would probably choose it over trying to find an equation that gives me the answer. In addition, since many of the numbers are prime, and some don't have 2 or 3 as a factor, we only end up factoring a portion of them.

So let's do that:


2
3
4 = 2,2
5
6 = 3,2
7
8= 2,2,2
9 = 3,3
10 = 5,2
11
12 =2,2,3
13
14 = 7,2
15 = 5,3
16 = 2,2,2,2
17
18 = 3,3,2
19
20 = 5,2,2
21 = 7,3
22 = 11,2
23
24 = 3,2,2,2
25
26 = 13,2
27 = 3,3,3

If you count all the 2's, you get: 23
If you count all the 3's, you get: 13

The ratio between 2's and 3's is 2:1 for every multiple of 12, so if we had 13 3's, we would need 26 2's. Since all the primes need to evenly cancel, the max number of multiple's of 12 is 11, thereby cancelling out 22 2's. That would also cancel out 11 3's, which works since there are 13 3's available.

So the answer is 12^11 is the largest possible to get an integer.

To elaborate on this approach, all we are really saying is how do we cancel out everything in the denominator.

To take a simple example, if we want to know if 4 is divisible by 2, we know it is, because 4 = (2*2)

So (2*2)/2 = 2, because the top 2 and the bottom 2 cancel each other out. That is what we are doing here as well. Since prime numbers are only divisible by themselves and 1, if we break 27! into primes, and then cancel as many multiples of the primes of 12 as possible, what we have left is an integer. However, let's say we had 12^12 as our divisor. All 23 2's for 27! would cancel, and then we would have an extra 2 on bottom. Since there are no factors of 2 on top anymore, our final answer CANNOT be an integer. Anyway, hope that helps.

[larbgai]

4) The answer is that it is a prime number @ 17!+1, but never after that.

Again, take a look at the post I referenced for #3. This same theme comes up in there as well.

Here, we know that 17! is divisible by EVERY number from 1 to 17. By extension, 17!+1 is not divisible by any of these numbers. This is the "common sense" part of factorials that the GMAT wants you to understand (but of course, you won't be able to come up with this rule on test day when you are trying to solve the problem in 2 minutes or less).

Here is the logic behind this rule:
If a number is divisible by 2, is it also divisible by 2+1? No.
If a number is divisible by 3, is it also divisible by 3+1? No.
If you keep going, you will see that any number that is divisible by a number over 1, is NOT divisible by that number + 1.


By this same logic, if 17! is divisible by every number from 1 to 17, then 17! + 1 is divisible by none of the numbers from 1 to 17.

Ok, so now we know that 17!+1 is definitely prime, and is NOT divisible by any of the numbers from 1 to 17.

By this logic, if 17!+1 is not divisible by 2, then 17!+2 MUST BE divisible by 2. You can extend this logic to every number from 17!+1 to 17!+17

Since everyone of those numbers has a divisor that results in an integer, none of the numbers over 17!+1 are prime. However, it asks for greater than or equal to, so at 17!+1 the number is prime, and therefore there is a prime in that range.

[larbgai]

The Answer choices for question 5 don't make sense to me.

If n = 20!, and 2k is a factor of n, then what is the greatest value of k?

Well, the greatest value of k is going to be the value that makes it equal to n when multiplied times 2.

So where 2k=n, k is going to be at its greatest value, if 2k is a factor of n where n/2k =1.

Since 20! = 20*19*......*2*1, when we divide by 2k, the 2 cancels.

So k should be equal to 20*19*18.....*3 if we want its greatest value.

This obviously isn't an answer choice. So I don't get it.

Where did you get this question?

[sharad22]

Q5
I think this question is asking that from available choices in answer, find the greatest integer k for which 2k is a factor of n
If we pick each answer one by one
1. 2^3 X 3 X 17 - 2k is factor of n
2. 19 X 23 -> clearly not a factor of n
3. 2 X 3^5 -> so, 2k is factor of n
4. 2^2 X 131 -> clearly not a factor of n
5. 2 X 621 -> clearly not a factor of n

So, I got 2 choices. I am not sure which one is right.

[ihsuw]

Originally Posted by larbgai

So (2*2)/2 = 2, because the top 2 and the bottom 2 cancel each other out. That is what we are doing here as well. Since prime numbers are only divisible by themselves and 1, if we break 27! into primes, and then cancel as many multiples of the primes of 12 as possible, what we have left is an integer. However, let's say we had 12^12 as our divisor. All 23 2's for 27! would cancel, and then we would have an extra 2 on bottom. Since there are no factors of 2 on top anymore, our final answer CANNOT be an integer. Anyway, hope that helps.

>>larbgai
I think you are outstanding in math.
then, I propose another view of this tough question, please check it.

I also checked the number of 2's , 3's and 4's included in 27!.
as you did, each number is 23,13, and 7.

27! must be divided by 2^12 and 3^12, but not by 4^12
because 27! includes both 23 2's and 13 3's.
so, the highest interger of X^12 = 3.

but I doubt I can manage this in the real GMAT......

[ihsuw]

Originally Posted by sunset_chill

3.For every positive integer n, the function h(n) is defined to be the product of all the even integers from 2 to n inclusive. If p is the smallest prime factor of h(100) + 1, then p is?

a)between 2 and 10
b)between 10 and 20
c)between 20 and 30
d)between 30 and 40
e)greater than 50

h(100)=2x4x6x......x100
=2(1x2x3x.....x50)
here, you can divide 2(1x2x3x....x50) by each integer of 1 through 50.
this means that you cannot divide 2(1x2x3x....x50)+1 by each integer of 1 -50.

so, p, the smallest prime factor of h(100)+1 must be greater than 50.
the answer is e)

it's time to give us OA.

[ihsuw]

Originally Posted by sunset_chill

5.If n is the product of the integers from1 to 20 inclusive, which of the following is the greatest integer k for which 2k is a factor of n?
a) 408
b) 437
c) 486
d) 532
e) 1242

The anwser is D)532

n=20!

a)2x204=(2^3)x51
b)437 cannot be divided by 2. so eliminate this.
c)2x243=2x(3^5)
d)2x266=2x7x19
e)2x621=2x(3^3)x23

let's go through the answer choices.
a)this includes 51, so this cannot be the factor of n.
b)as mentioned above, this is not the factor of n.
c)this can be the answer.
d)this can be the answer.
e)this includes 23, so this cannot be the factor of n.

here, c) and d) are candidate.
d) is constituted by 2,7, and 19, therefore, this can be the factor of n.
in addition, k is greater than that of c).......... 266 > 243

thus 2x266 satisfy all the conditions.
the answer is d)

[larbgai]

ihsuw:
Are you saying that since 4^12 is over 27! (which we know from counting, since there are 2^23 in 27!, and 4^12 = (2^2)^12 = 2^24), and that 3^12 < 27! (also from counting), and since 4x3 = 12, that 3^11 x 4^11 is our maximum multiple of 12? It also follows that 3^11 * 4^11 = (12)^11.

That is what I got from your explanation.


sharad22: Good job on #5. I missed that they were asking for the largest amongst the answers. Your explanation makes sense to me.