If N is a two-digit even integer, is N

Question

If N is a two-digit even integer, is N < 20?
 
(1) The product of the digits of N is less than the sum of the digits of N.
 
(2) The product of the digits of N is positive.

Bunuel · Accepted Answer

If N is a two-digit even integer, is N < 20?

(1) The product of the digits of N is less than the sum of the digits of N. This case is possible only if either digit is 0 or 1. For example, N can be 10, 12, 14, 16, 18, 20, 30, 40, ... So, N can be less as well as greater or equal to 20. Not sufficient.

(2) The product of the digits of N is positive. This implies that neither of the digits of N is zero. N still can be be less as well as greater than 20, for example, consider 12 or 22. Not sufficient.

(1)+(2) Since neither of the digits of N is zero, then values of N like 20, 30, 40, ... are not possible, thus N is definitely less than 20 (12, 14, 16, 18). Sufficient.

Answer: C.

Hope it's clear.

sanjoo · Answer

What if N is 21??

i chose E

Bunuel · Answer

You have to read the question carefully: "if n is a two-digit   even   integer..." 21 is not even.

HarveyS · Answer

I am getting A as answer....

Not sure how C....is OA..

HarveyS · Answer

Yes clear... I did some silly mistake

Thanks a lot

ravishankar1788 · Answer

Target question: Is N < 20?

Statement 1: The product of the digits of N is less than the sum of the digits of N. 
Under what circumstances is the product of the digits of N less than the sum of the digits? 
This occurs when one of the digits is either a 0 or a 1. 
So, for example, N could equal 10, 11, 12, ....,20, 21, ...30, 31, 40, 41, 50, 51etc. 
BUT the question says that N is EVEN. 
So, N can be 10, 12, 14, 16, 18, 20, 30, 40, 50, 60, 70, 80, or 90 
As you can see, N can be less than 20, or N can be greater than 20 
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: The product of the digits of N is positive. 
There are several possible values of N. Here are two: 
Case a: N = 12 (product is less than sum). Here, N is less than 20 
Case b: N = 21 (product is less than sum). Here, N is greater than 20 
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined 
Statement 1 tells us that N = 10, 12, 14, 16, 18, 20, 30, 40, 50, 60, 70, 80, or 90 
Statement 2 lets us exclude values of N such that one of the digits is zero (since the product of the digits is zero and zero is not positive) 
So, if we exclude values of N that have a zero digit, we're left with N = 12, 14, 16 or 18 
This means that N is definitely less than 20 
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer = C

pretzel · Answer

Hi Bunnel

Isn't zero considered a positive integer as well?

Thanks

Bunuel · Answer

No. Zero is neither positive nor negative even integer.

Hesham_87 · Answer

when i started to solve the problem, i assumed one of the integers of N is negative as per statement 1. for example N has X and Y if X or Y is negative so statement 1 would be fulfilled. for example X is -2, Y is 4 ==> -2*4 = -8 while summing both of them -2+4 = 2. so statement 1 is not sufficient. statement 2 is not sufficient as well. while combining both statement it would not be sufficient. why did we neglect positive and negative number assumption?

rjivani · Answer

Hi guys,

Can someone help explain why we are not considering negative numbers here? (Product will be + which will be > the sum) so it satisfies the initial condition. if we assume negative, then 1 and 2 both dont tell us anything either?

standyonda · Answer

Easier solution -- The correct Q indicates that N is positive! N=10B+A with A=0, 2, 4, 6, 8 Yes -- B=1 and A=0, 2, 4, 6, 8 No -- B>=2 and A=0, 2, 4, 6, 8 (1) AB0 or A≠0 and B≠0 A=2, 4, 6, 8 and B=1, 2, 3... Therefore INSUFFICIENT (1+2) The (2) cancels the Option 2 from (1) therefore SUFFICIENT! Bunuel check

Manu03 · Answer

Hi Everyone!
Can this be done with algebra?

Like... is 10t+u<20?

1) 10du<10d+u
2) 10du not 0

Thank you very much in advance!
Manu

marine2008 · Answer

That is pretty straightforward:

Let N = 10a + 2b

Where
(1) 0 < a < 9,
(2) 0 < b < 5,
(3) ab > 0,
(4) 2ab < a + 2b.

Zeros excluded by a*b > 0 condition.
From the bottom inequality (4) we get a < 2b/(2b-1).

Suppose the opposite to the initial statement: N greater or equal 20.

Then 20 <= N = 10a + 2b < 10*2b/(2b-1) + 2b
This leads to the quadratic equation 2b^2 - 11b + 10 >0.

One can see that it is satisfied only if b>=5 or b <=1.
But from (4) in case b=1 we get a < 2, which leads to N < 20.

Hence, the assumption was incorrect and N < 20 required.

bumpbot · Answer

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If N is a two-digit even integer, is N < 20?

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If N is a two-digit even integer, is N < 20?

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