Question

Just how do I do this exactly?
Question attached:

Answer 1

@mathstudent55
@mathslover

Answer 2

@AravindG
@whpalmer4
@.Sam.

Answer 3

@experimentX may help you!

Answer 4

find the cubes in 29!/28

Answer 5

only two cubes of course.

Answer 6

Why? What about the multiples?

Answer 7

Like 4*16 ?

Answer 8

hmm ... okay, i see i ignored that.

Answer 9

7 and 5

Answer 10

(1) We are looking for two prime numbers as big as possible whose cubes each divide 30! + 29!/28.
(2) In order to do this we need to write down the prime factorization of 30! + 29!/28

(3) The first step is to recognize the GCF(s) (greatest common factor of the two terms namely 27!. So 30!+29!/28 = 27! (30*29*28 + 29).

(4) Okay, so we realize at this point our GCF wasn't quite the greatest one possible and that we can factor 29 again. So we have: 27!*29(30*28+1)

(5) The convenient part is that factorial numbers like 27! are essentially already factored for us! So then we need to check what the prime factors of (30*28+1) are. We can do this by hand, which is a bit tedious, or use some computer algebra system. You can take my word for it that 30*28+1 = 841
is prime.

Answer 11

841 = 29^2 ?

Answer 12

Does it? Let me check

Answer 13

Yep.

Answer 14

Damn, you are right

Answer 15

<scratches my head>

Answer 16

hehe...

@experimentX how did you get 7 and 5 ?

Answer 17

but 29 is definitely prime

Answer 18

fundamental theorem of arithmetic.

Answer 19

Mind explaining?

Answer 20

(6) So at this point, we can rewrite 27! as a prime factorization. Intermediate to this, I just list the factors that compose 27! and their prime factorizations.
So (6) for example, 1*2*3*4 is part of 27!. But just rewrite 4 as 2^2
The reason we rewrite it like this, at least I would is because it becomes must easier to see the exponent of the prime factors

(7) so now all we have to do is read the list and look for the biggest primes with exponent 3 or more starting from the right

Answer 21

look for multiples of primes ...
7, 14, 21 ...
11, 22, ??
so 7^3 will divide 29!/28 ... 11^3 will not

Answer 22

What about 29 @experimentX ?

Answer 23

... so the highest prime in the string is 29. Check if 29 appears three times, that means, has a power of 3 or more. It does. So it's the greatest prime that's cubed which divides our number. Since 29 was the highest prime appearing in our factorization, we know the next highest one must be a prime less than it.

27 is the next greatest prime, but it does not appear thrice. 23 is nice, but it does not appear thrice. 19 is nice, but again no. Continuing down until 7, prime, which appears thrice.

Answer 24

29 is prime ... but cube of 29 will not divide that thing.

Answer 25

So 7 is the second greatest prime whose cube divides 30! + 29!/28

Answer 26

It does right?

30! + 29!/28

29! ( 30 + 1/28)

\[\frac{29! * 841}{28}\]

\[\frac{29*29*29*27!}{28}\]

Answer 27

get your answer from here
http://www.wolframalpha.com/input/?i=prime+factorization+29%21%2F28+

Answer 28

p^3| 27! (28.29.30)+27! (29)
p^3|27! 29 (28.30 +1)

Answer 29

I think you missed the 30! @experimentX

Answer 30

yes
29^3 divides it

Answer 31

ignore that 30! ...something that divides 29!/28 will definitely divide 30!

Answer 32

huh?? what the hell ... how come??

Answer 33

oh lol wait I think we are misunderstanding each other so

Answer 34

probably ...
you need to do this,
\[ 29!\left( \frac 1 {28} + 30 \right) \]

Answer 35

so 29^3 divides this
because the result is integer
Know what I mean?