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For each positive integer n, p(n)

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For each positive integer n, p(n)

by NandishSS » Sun Dec 18, 2016 6:04 am
For each positive integer n, p(n) is defined to be the product of the digits of n. For example, p(724)=56 since 7∗2∗4=56

Which of the following statements must be true?

I. p(10n)=p(n)

II. p(n+1)>p(n)

III. p(2n)=2p(n)

--

A. None
B. I and II only
C. I and III only
D. II and III only
E. I, II, and III

OA:A

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by GMATGuruNY » Sun Dec 18, 2016 6:18 am
NandishSS wrote:For each positive integer n, p(n) is defined to be the product of the digits of n. For example, p(724)=56 since 7∗2∗4=56

Which of the following statements must be true?

I. p(10n)=p(n)

II. p(n+1)>p(n)

III. p(2n)=2p(n)

A. None
B. I and II only
C. I and III only
D. II and III only
E. I, II, and III
Try to prove that I, II and III do NOT have to be true.

Case 1: n=1

I: p(10n) = p(n)
p(10n) = p(10*1) = p(10) = 1*0 = 0.
p(n) = p(1) = 1.
Since p(10n) ≠ p(n), Statement I does not have to be true.
Eliminate B, C and E.

II: p(n+1) > p(n)

In Case 1, p(10n) < p(n).
The reason is that 10n includes a digit of 0.
To show that Statement II does not have to be true, test a case in which n+1 includes a digit of 0.

Case 2: n=9, with the result that n+1 = 10

p(n+1) = p(9+1) = p(10) = 1*0 = 0.
p(n) = p(9) = 9.
Since p(n+1) < p(n), Statement II does not have to be true.
Eliminate D.

The correct answer is A.
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by Jay@ManhattanReview » Mon Dec 19, 2016 10:42 pm
NandishSS wrote:For each positive integer n, p(n) is defined to be the product of the digits of n. For example, p(724)=56 since 7∗2∗4=56

Which of the following statements must be true?

I. p(10n)=p(n)

II. p(n+1)>p(n)

III. p(2n)=2p(n)

--

A. None
B. I and II only
C. I and III only
D. II and III only
E. I, II, and III

OA:A

Source: GMATPrep EP2
This is a must-be-true type of question.

For these types of questions, our attempt must be to find at least one extreme case that would break the given claims (options).

The moment I saw p(724) = 7*2*4 = 56, I thought what would happen if there were a 0 digit in the function; for example, p(7240) = 7*2*4*0 = 0.

So p(724) > p(7240) => p(724) > p(724*10).

So, there could be a role of a multiplicand 10.

Let us try to break each of the given statements.

I. p(10n) = p(n): We have already seen above that p(724) > p(7240) OR p(10n) < p(n) for this case. Not true!

II. p(n+1) > p(n): It would fail if n = 9 => n+1 = 10.

III. p(2n) = 2p(n): It would fail if n = 5 => 2n = 10.

Thus, none is correct.

OA: A

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by Scott@TargetTestPrep » Wed Dec 21, 2016 8:59 am
NandishSS wrote:For each positive integer n, p(n) is defined to be the product of the digits of n. For example, p(724)=56 since 7 x 2 x 4=56

Which of the following statements must be true?

I. p(10n)=p(n)

II. p(n+1)>p(n)

III. p(2n)=2p(n)

--

A. None
B. I and II only
C. I and III only
D. II and III only
E. I, II, and III
We are given that for each positive integer n, p(n) is defined to be the product of the digits of n. Let's analyze each Roman numeral to determine which MUST be true.

I. p(10n)=p(n)

We can determine that 10n will always end in a zero (since any integer multiplied by 10 will always end in a zero). Thus, the p(10n) will always equal zero. Since p(n) does not always have to equal zero, p(10n) does not necessarily have to equal p(n). For example, if n = 24, then p(n) = 2 x 4 = 8; but p(10n) = p(240) = 2 x 4 x 0 = 0. Roman numeral I is not always true.

II. p(n+1) > p(n)

Roman numeral II does not always have to be true. For instance, if n = 9, then:

P(9 + 1) = p(10) = 0

and

p(9) = 9

In this case we see that p(n+1) IS NOT greater than p(n). Roman numeral II is not always true.

III. p(2n)=2p(n)

Roman numeral III does not always have to be true. For instance, if n = 9, then:

P(2 x 9) = p(18) = 1 x 8 = 8

and

2p(9) = 2 x 9 = 18

In this case we see that p(2n) DOES NOT equal 2p(n). Roman numeral III is not always true.

Answer: A

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