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If d is a positive integer and f is the product of the first [#permalink]
31 Jan 2007, 05:20

00:00

A

B

C

D

E

Difficulty:

(N/A)

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0% (00:00) correct
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If d is a positive integer and f is the product of the first 30 integers, what is the value if d?

(1) 10 (exp) d is a factor of f
(2) d > 6

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE
is sufficient.
EACH statement ALONE is sufficient
Statements (1) and (2) TOGETHER are NOT sufficient _________________

The GMAT, too tough to be denied.
Beat the tough questions...

Re: DS - Product of the first 30 integers [#permalink]
31 Jan 2007, 11:29

TOUGH GUY wrote:

If d is a positive integer and f is the product of the first 30 integers, what is the value if d?

(1) 10 (exp) d is a factor of f (2) d > 6

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. EACH statement ALONE is sufficient Statements (1) and (2) TOGETHER are NOT sufficient

1) alone is not enough:
Since f=1x2x3x4x5x...x28x29x30
Multipying only the numbers that end in 2,5 and 0 you would get:
2x5x10x12x15x20x22x25x30 with several 10 factors there, like this it will be easier:
2
5
10=2x5
12=2x2x3
15=3x5
20=2x2x5
22=2x11
25=5x5
30=2x3x5

We have 8 numbers 2 and 7 numbers 5, so d can be from 1 do 7 by just taking these numbers...

2)d>6 alone is not enough either.

Putting them together: you have that d>6 from the 2) and that you have 7 possible numbers 10

C is the answer.

PS) - the important thing is to watch for the 5s and not the 2s, because we have plenty of 2s...

well, lets see 10 has prime factors 5 and 2, so in the first 30 integers there are how many 2s? 30/2=15, 4 gives one more 2, 8 gives 2 more, 16 gives 3 more etc...

lets see how many 5s are there in first 30 integers...30/5=6, 25 gives us one more 5...so we have 7 fives...

now..if 10^d is a factor of F, then power of D cannot be greater than 7..
..
rewrite 10^d as 5^a 2^b, where the power of a determines d.

so..going back to statment 1) d can be anynumber from 0-7...Insuff

together..if d>6 then the only number that fits this is 7...