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Value of d

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by kanha81 » Sat Jun 06, 2009 2:22 pm
If d is positive integer, f is the product of the first 30 positive integers, what is the value of d?

1). 10^d is a factor of f
2). d>6

How to solve such problem efficiently? This is my attempt-

1) (10^d) * k = f, k is an integer
(10^d) * k = 30! = 30*29*28*27*...*1

so, we know that d can be at least 1.
Insuff

2) d>6
Insuff

1) & 2)
now what?
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Source: — Data Sufficiency |
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by Domnu » Sat Jun 06, 2009 2:29 pm
[spoiler]Okay, we need to find the total number of 0's at the end of 30!. This is how you do so:

We can make a 10 out of each factor of 10 less than or equal to 30. There are 3 of these.

We can make extra factors of 10 by putting 2's and 5's together. We have: (2, 5) (12, 15) (22, 25). HOWEVER, note that 25 has TWO fives in it... we can put this together with an 8 and make yet ANOTHER 10. This is all we can do.

So, we have that there are 7 factors of 10 in 30!, and 30! ends in 7 zeros. From this, we see that (1) isn't sufficient, but if d > 6, then d has to be 7. So both together are sufficient. So, C.[/spoiler]
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by yogami » Sun Jun 07, 2009 6:46 am
Tricky but I will give you a clue: 25 is 5*5
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by Pierreha » Mon Jun 08, 2009 12:54 am
Wow that's a brilliant solution. :shock:
What percentile do you think this question is (i.e., what level)?
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by lunarpower » Wed Jun 17, 2009 1:23 am
here's all you have to do:
forget entirely about 10, 20, and 30, and ONLY THINK ABOUT PRIME FACTORIZATIONS.
(TAKEAWAY: this is the way to go in general - when you break something down into primes, you should not think in hybrid terms like this. instead, just translate everything into the language of primes.)

each PAIR OF A '5' AND A '2' in the prime factorization translates into a '10'.

there are seven 5's: one each from 5, 10, 15, 20, and 30, and two from 25.

there are waaaaaaayyyyy more than seven 2's.

therefore, 30! can accommodate as many as seven 10's before you run out of fives.

--

statement 2 is clearly insufficient.

statement 1, by itself, means that d can be anything from 1 to 7 inclusive.

together, d must be 7.

ans (c)
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by shashank.mehra » Wed Jun 17, 2009 5:40 am
Ok here is an easier way. 10 is because of 2 and 5. The number of 5s are o course greater than number of 2s. therefore number of 0s will be determined by number of 5s. Simply calculate the number of 5 by the following:-

[x] : denotes the greatest integer less than x. i.e [3.2] = 3

Number of 5s = [30! / 5] + [30! / 5^2] + [30! / 5^3] .... = 6 + 1 = 7. This is only an easier way of calculating number of 0s in a given number.
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