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# If k is a positive integer and m is the product of the first 40 positi

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If k is a positive integer and m is the product of the first 40 positi  [#permalink]

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10 Jul 2018, 21:37
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If k is a positive integer and m is the product of the first 40 positive integers, what is the value of k ?

(1) 10^k is a factor of m.
(2) 10^k is a factor of n, where n is the product of the first 9 positive integers.

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If k is a positive integer and m is the product of the first 40 positi  [#permalink]

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10 Jul 2018, 21:48
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Bunuel wrote:
If k is a positive integer and m is the product of the first 40 positive integers, what is the value of k ?

(1) 10^k is a factor of m.
(2) 10^k is a factor of n, where n is the product of the first 9 positive integers.

Basically $$m=40!$$

Statement 1: $$10^k$$ is a factor of $$40!$$. but $$40!$$ has $$\frac{40}{5}+\frac{40}{25}=8+1=9$$ trailing $$0$$s. so $$k$$ could be $$1$$ or $$2$$ or any other number. insufficient

Statement 2: $$n=9!=1*2*3*4*5*6*7*8*9$$. this number has only one trailing $$0$$ which will be a multiple of $$2*5=10$$. hence $$k=1$$. Sufficient.

Option B
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Re: If k is a positive integer and m is the product of the first 40 positi  [#permalink]

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10 Jul 2018, 21:58
From statement 1: k could be 0,1,2.. Insufficient

From statement 2: k could be 0 or 1. Question stem says k is positive. Hence sufficient

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If k is a positive integer and m is the product of the first 40 positi  [#permalink]

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10 Jul 2018, 22:43
Bunuel wrote:
If k is a positive integer and m is the product of the first 40 positive integers, what is the value of k ?

(1) 10^k is a factor of m.
(2) 10^k is a factor of n, where n is the product of the first 9 positive integers.

Given m=40!
Question stem, k=?

We need only 1 value of k to prove sufficiency & a minimum of 2 values of k to prove insufficiency.

St1:- $$10^k$$is a factor of 40!
40!=1*2*3*4*5*6*7*8*9*10*...*40
So $$10^k$$ is a factor of 10 & 10^2 where k is 1 and 2 respectively.

Insufficient.
st2:-$$10^k$$ is a factor of 9!
9!=1*2*3*4*5*6*7*8*9
So $$10^k$$ is a factor of 10 when k is 1 only.
Hence, sufficient.

Ans. (B)
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Re: If k is a positive integer and m is the product of the first 40 positi  [#permalink]

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11 Jul 2018, 00:31
Bunuel wrote:
If k is a positive integer and m is the product of the first 40 positive integers, what is the value of k ?

(1) 10^k is a factor of m.
(2) 10^k is a factor of n, where n is the product of the first 9 positive integers.

Ans is B...(1) says k<= 8
(2) k=1
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Re: If k is a positive integer and m is the product of the first 40 positi  [#permalink]

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11 Jul 2018, 06:55
Can we look at it as

1) since there is 10 and 100 in 40!, we unable to determine k
2) there is only 2*5=10 in 9!, hence we can determine k=1
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If k is a positive integer and m is the product of the first 40 positi  [#permalink]

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11 Jul 2018, 07:06
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ETLim wrote:
Can we look at it as

1) since there is 10 and 100 in 40!, we unable to determine k
2) there is only 2*5=10 in 9!, hence we can determine k=1

Hi ETLim,
In your quotation (1), we are able to determine more than one value of 'k'. Hence given statement-1 is insufficient to answer the question stem. Please note: In DS questions those ask the value of variable or things, if we get one value from the statement then that statement is sufficient and if we get more than one value from the statement, then that statement is insufficient. For both st1 & st2 ,individually to be sufficient, we must obtain an unique value from both the statements.

In your quotation (2), since we obtained one value of k, hence st2 is sufficient.

Hope it helps.
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Re: If k is a positive integer and m is the product of the first 40 positi  [#permalink]

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11 Jul 2018, 08:02
If k is a positive integer and m is the product of the first 40 positive integers, what is the value of k ?

(1) 10^k is a factor of m.
m=1*2*3*.....*40=40!
Number of 10s will depend on the bigger prime factor that is 5.
# of 5s =$$\frac{40}{5}+\frac{40}{25}= 8+1=9$$
So k can take any values till 9 as 10^1, 10^2,....10^9 all are factors of 40!
Insuff

(2) 10^k is a factor of n, where n is the product of the first 9 positive integers.
Now n=9! And number of 5s are 9/5..=1
So k can take Only one value 1
Sufficient

B
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Re: If k is a positive integer and m is the product of the first 40 positi  [#permalink]

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11 Jul 2018, 18:54
chetan2u niks18 pushpitkc pikolo2510

How about below analysis of option A?

Quote:
If k is a positive integer and m is the product of the first 40 positive integers, what is the value of k ?

$$k\geq{0}$$
m = 40!

Quote:
(1) 10^k is a factor of m.

So m also contains 2*5 ie 10 divisible by 10^1
and m also contains 25*4 which is divisible by 10^2
So k can be 1 or 2.

I could not understand above approach used for examinating statement 1
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If k is a positive integer and m is the product of the first 40 positi  [#permalink]

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11 Jul 2018, 22:22
chetan2u niks18 pushpitkc pikolo2510

How about below analysis of option A?

Quote:
If k is a positive integer and m is the product of the first 40 positive integers, what is the value of k ?

$$k\geq{0}$$
m = 40!

Quote:
(1) 10^k is a factor of m.

So m also contains 2*5 ie 10 divisible by 10^1
and m also contains 25*4 which is divisible by 10^2
So k can be 1 or 2.

I could not understand above approach used for examinating statement 1

The analysis of statement 1 as already explained by chetan2u is as follows

There is a formula for finding the highest power of a prime number in a factorial
If p is a prime number, then the highest power of p in a factorial n is {$$\frac{n}{p}$$} + {$$\frac{n}{p^2}$$} + {$$\frac{n}{p^3}$$}.....
where {$$\frac{a}{b}$$} is the quotient when integer b divides another integer a

Now coming to the problem at hand. Since p is 10, which can be prime-factorized as 2*5, it is
enough if we are able to find the highest power of 5(since it is the biggest prime number). n = 40

Therefore, the highest power of p(10) is nothing but the highest power of 5.
Substituting values in the formula $${\frac{40}{5^1}} + {\frac{40}{5^2}} = {\frac{40}{5}} + {\frac{40}{25}} = 8 + 1 = 9$$
This means that the 10^9 will divide 40!. So even the smaller powers of 10 will divide 40!

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Re: If k is a positive integer and m is the product of the first 40 positi  [#permalink]

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15 Jul 2018, 03:55
PKN wrote:
Bunuel wrote:
If k is a positive integer and m is the product of the first 40 positive integers, what is the value of k ?

(1) 10^k is a factor of m.
(2) 10^k is a factor of n, where n is the product of the first 9 positive integers.

Given m=40!
Question stem, k=?

We need only 1 value of k to prove sufficiency & a minimum of 2 values of k to prove insufficiency.

St1:- $$10^k$$is a factor of 40!
40!=1*2*3*4*5*6*7*8*9*10*...*40
So $$10^k$$ is a factor of 10 & 10^2 where k is 1 and 2 respectively.

Insufficient.
st2:-$$10^k$$ is a factor of 9!
9!=1*2*3*4*5*6*7*8*9
So $$10^k$$ is a factor of 10 when k is 1 only.
Hence, sufficient.

Ans. (B)

If the question would have asked for highest value of k, then A would be the answer. Right?
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Re: If k is a positive integer and m is the product of the first 40 positi  [#permalink]

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15 Jul 2018, 05:10
AkshdeepS wrote:
PKN wrote:
Bunuel wrote:
If k is a positive integer and m is the product of the first 40 positive integers, what is the value of k ?

(1) 10^k is a factor of m.
(2) 10^k is a factor of n, where n is the product of the first 9 positive integers.

Given m=40!
Question stem, k=?

We need only 1 value of k to prove sufficiency & a minimum of 2 values of k to prove insufficiency.

St1:- $$10^k$$is a factor of 40!
40!=1*2*3*4*5*6*7*8*9*10*...*40
So $$10^k$$ is a factor of 10 & 10^2 where k is 1 and 2 respectively.

Insufficient.
st2:-$$10^k$$ is a factor of 9!
9!=1*2*3*4*5*6*7*8*9
So $$10^k$$ is a factor of 10 when k is 1 only.
Hence, sufficient.

Ans. (B)

If the question would have asked for highest value of k, then A would be the answer. Right?

I don't think so. Each statement has its own greatest value of k. $$(st1(k_{max}=9, st2(k_{max}=1)$$

In DS value questions, each statement must produce same value. Even answer can't be D.

So, test maker would set question accordingly.
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Re: If k is a positive integer and m is the product of the first 40 positi &nbs [#permalink] 15 Jul 2018, 05:10
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