Very hard arithmetic problem.

This question is a typical GMAT style divisibility question.

Since 125 = 5³, what the question is really asking is how many 5's there are in the prime factorization of N.

Among the factors of N are the following that include 5's among their prime factors.

30³� - Has 30 5's.
25²� - Has 50 5"s.
20²� - Has 20 5's.
15¹� - Has 15 5's.
10¹� - Has 10 5's.
5� - Has 5 5's.

So the total number of 5's in the prime factorization of N is 130.

To get an integer result for via dividing N by (125)ᴷ = (5³)ᴷ, divide 130/3.

130/3 = 43, remainder 1.

So if K can be a fraction, the highest value of K such that N/125ᴷ is an integer is [spoiler]130/3[/spoiler].

If K must be an integer, the highest is 43.

chrisdkoning wrote:(30^30) × (29^29) × (28^28) × . . . × (3^3) × (2^2) × (1^1) = N

What is the highest value of K, such that N/(125^K) is an integer?

How do I solve this, and is this question a GMAT level problem? Any help would be appreciated!

(30^30) × (29^29) × (28^28) × . . . × (3^3) × (2^2) × (1^1).

125 = 5³.
To determine how many times 5³ can divide into N, count how many 5's are contained within the product above.

30³� = 5³�6³� --> thirty 5's.
25²� = (5²)²� = 5�� --> fifty 5's.
20²� = (5²�)(4²�) --> twenty 5's.
15¹� = 5¹�3¹� --> fifteen 5's.
10¹� = 5¹�2¹� --> ten 5's.
5� = five 5's.
Total number of 5's = 30+50+20+15+10+5 = 130.

Implication:
N = (5¹³�)(other factors).
Thus:
N/(125^k) = [(5¹³�)(other factors)]/[(5³)^k].

The values in blue reveal that the greatest possible value for k is 43, as follows:
[(5¹³�)(other factors)]/[(5³)�³]

= [(5¹³�)(other factors)]/5¹²�

= (5)(other factors).

GMATGuruNY wrote:
(30^30) × (29^29) × (28^28) × . . . × (3^3) × (2^2) × (1^1) = 30!30!.

Hi Mitch,

Can you elaborate how you got 30!30!. it is not clear.

Thanks

I'm a little hung up on 30^30, for example. Seems that there would be 30 instances of 5.

For example, 30^30 = (2^30)(3^30)(5^30)

Hi Mitch,

i thought that from (30^30) X (29^29) X (28^28) X . . . X (3^3) X (2^2) X (1^1) = N

we have 6 multiple of 5 which are 30^30, 25^25, 20^20, 15^15, 10^10 and 5^5

=> 30^30= (5^30)*(6^30)
=> 25^25= 5^50
=> 20^20= 5^20*(4^20)
=> 15^15= 3^15*(5^15)
=> 10^10= 2^10*(5^10)

so in total N= (5^30)(5^50)(5^20)(5^15)(5^10)(5^5)(other factors)= 5^130(other factor)

N/125^k= 5^130/5^3k so K=43

i'm not sure where i did the mistake

joealam1 wrote:Hi Mitch,

i thought that from (30^30) X (29^29) X (28^28) X . . . X (3^3) X (2^2) X (1^1) = N

we have 6 multiple of 5 which are 30^30, 25^25, 20^20, 15^15, 10^10 and 5^5

=> 30^30= (5^30)*(6^30)
=> 25^25= 5^50
=> 20^20= 5^20*(4^20)
=> 15^15= 3^15*(5^15)
=> 10^10= 2^10*(5^10)

so in total N= (5^30)(5^50)(5^20)(5^15)(5^10)(5^5)(other factors)= 5^130(other factor)

N/125^k= 5^130/5^3k so K=43

i'm not sure where i did the mistake

That's what I came up with as well

regor60 wrote:I'm a little hung up on 30^30, for example. Seems that there would be 30 instances of 5.

For example, 30^30 = (2^30)(3^30)(5^30)

Good catch!
I've amended my post accordingly.

We can ignore all the bases that don't contain 5s, so we've got

30³� * 25²� * 20²� * 15¹� * 10¹� * 5�

Then, we can ignore all the factors in THESE bases that don't contain 5s, giving us

5³� * (5²)²� * 5²� * 5¹� * 5¹� * 5�

or

5¹³�

We want (5¹³�) / (5³)�, with the largest value of m. This reduces to 130/3m. The nearest multiple of 3 to 130 is 129, or 3*43. So m = 43, and we're set.