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

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

I am getting A as answer....

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

Yes clear...

I did some silly mistake

Thanks a lot

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

Hi Bunnel

Isn't zero considered a positive integer as well?

Thanks

No. Zero is neither positive nor negative even integer.

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?

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?

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) AB<A+B
Option 1: If B=1 then A<A+1 or 0<1 -- CHECK
Option 2: If B=2 then 2A<A+2 or A<2 therefore A can be 0 -- CHECK {or if B=5 then 5A<A+5 or A<5/4... leads to A=0}

Therefore INSUFFICIENT

(2) AB>0 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

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

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.

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