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Integer question

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by Baldini » Thu Mar 12, 2009 11:11 am
If d is a positive integer, and 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

OA is C

Can someone kindly explain whY?

Grazie mille
GMAX
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Source: — Data Sufficiency |
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by Alara533 » Thu Mar 12, 2009 2:31 pm
f is the product of first 30 numbers.
This product will have 7 zeros (Zero comes in the product when any number is multiplied by 10,20 and 30 and any even number is multiplied by 5, 15 and 25)

so we have

5 x 2 = 10
then the number 10
15 x 6 = 90
then the number 20
25 x 4 = 100
then the number 30

So altogether f has 7 zeros.

1) Says 10^d is a factor of f. 10 (d=1), 100 (d=2), 1000 (d=3), 10000 (d=4)...etc upto 10^7 can be a factor of f. Hence not sufficient.

2) Says d > 6. d can be any positive integer greater than 6.

Both combined, we have d = 7. Hence sufficient
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by tuscan21 » Wed May 27, 2009 1:25 pm
Alara533 wrote:f is the product of first 30 numbers.
This product will have 7 zeros
I still don't understand why this product will have 7 zeros. Could you please elaborate on how this happens?
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by ssmiles08 » Wed May 27, 2009 1:38 pm
tuscan21 wrote:
Alara533 wrote:f is the product of first 30 numbers.
This product will have 7 zeros
I still don't understand why this product will have 7 zeros. Could you please elaborate on how this happens?
f = 30!

the numbers in 30! that produce a zero at the end are
2*5*10*12*15*20*22*25*30 = 594x10^7
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by Svedankae » Wed Jun 03, 2009 12:43 pm
ssmiles08 wrote:
tuscan21 wrote:
Alara533 wrote:f is the product of first 30 numbers.
This product will have 7 zeros
I still don't understand why this product will have 7 zeros. Could you please elaborate on how this happens?
f = 30!

the numbers in 30! that produce a zero at the end are
2*5*10*12*15*20*22*25*30 = 594x10^7

Why "10" for instance and not "11"?
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by lunarpower » Fri Jun 05, 2009 2:36 am
here's what you 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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