A telephone company needs to create a set of 3-digit area

Question

A telephone company needs to create a set of 3-digit area codes. The company is entitled to use only digits 2,
4 and 5, which can be repeated. If the product of the digits in the area code must be even, how many
different codes can be created?

A. 20
B. 22
C. 24
D. 26
E. 30

Show Spoiler

D

A. 20
B. 22
C. 24
D. 26
E. 30

walker · Accepted Answer

D

N = 3^3 - 1(exclude 555) = 26

fameatop · Answer

we have 3 places (---)to fill & we can use only digits 2, 4, 5
First place can be filled in 3 ways
2nd place can be filled in 3 ways because we can repeat the digits
3rd place can be filled in 3 ways because we can repeat the digits
Total no of codes = 3 x 3 x 3= 27

but one combination (555) will not be even so total no of even codes = 27 - 1 = 26
Answer D

Ralphcuisak · Answer

Can anyone please elaborate on how this was solved? Bunuel ?

Bunuel · Answer

A telephone company needs to create a set of 3-digit area codes. The company is entitled to use only digits 2,
4 and 5, which can be repeated. If the product of the digits in the area code must be even, how many
different codes can be created?

A. 20
B. 22
C. 24
D. 26
E. 30

Show Spoiler

D

A. 20
B. 22
C. 24
D. 26
E. 30

Total # of codes possible is 3*3*3 = 27. Out of those 27 codes only the product of 555 will be odd, the remaining 26 will have either 2 or 4 in them, which ensures that their product will be even. Therefore the number of codes where the product of the digits is even = (total) - (restriction) = 27 - 1 = 26.

Answer: D.

Hope it's clear.

wikdwik · Answer

Hi All,

This is the easiest way to do it : All (minus) the odd one = Total codes

But I did it the following way by using the basics of the counting the cases (was explaining my self while I was doing this problem - otherwise close to half of PnC and probability problems I do wrong by not understanding the type of cases or favorable events and the total outcomes - so this solution is for those who are having hard time understanding concept and application of PnC like me- Excuse me, if I wrote too much):

Conditions given in the questions : the product has to be even and repetition is allowed
Numbers we have : 2, 4 and 5 => two evens and one odd
Total places : 3
Cases : EEE + EEO + EOO
Counting: 
EEE = 2*2*2*(3!/3!) (repetition allowed=>2*2*2 and by anagram method even numbers become identical so EEE can be arranged in 3!/3! ways = 1 way)
EEO = 2*2*1*(3!/2!) (repetition allowed=>2*2*1 and by anagram method even numbers become identical so EEO can be arranged in 3!/2! ways = 3 ways) 
EOO = 2*1*1(3!/2!) ( same as the case as above)

Total Different codes : Total Sum = 26

Regards,
wikdwik

pranab4om · Answer

can you please help to clear my doubt.why only 555 is excluded what about 245 and 425? These are also not even number. or we need to find that there will be no even number available in 3-digit number, no matter 3-digit even or not.

Kurtosis · Answer

Hi  pranab4om ,

You have misunderstood the question. The question states that the product of the digits must be even. The question doesn't ask us to check for even numbered codes.

The product of the digits is odd only when the code is 555. In the rest of the cases we will have a even number in the code and hence the product will be even.

Take 245 --> 2*4*5 = 50 --> even
Take 555 --> 5*5*5 = 125 --> odd

Hope your doubt is resolved!!

megha_2709 · Answer

Hi, Thank you for posting such good question , but I have a doubt, instead of doing it this way why cant we follow this method : first position can be filled in 3 ways,2nd position can be filled in 3 ways and 3rd Pos can be filled in 2 ways (2 or 4 ) , since product need to be even. Therefore , 3*3*2 = 18. Please help me understand why this is incorrect. Thanks for your help in advance

Regards Megha

chetan2u · Answer

Hi,
 you are going wrong because you are working on the CODES to be EVEn rather than the PRODUCT of the numbers..
Example ..255 is a valid option BUT not as per your solution..

so safer bet is-- 
 total ways codes can be generated = 3*3*3=27..
Ways the product is not EVEN = 1*1*1=1 way when number is 555 
so answer = 27-1=26...

If you want to find the EVEn straight -- 
1) numbers when all are EVEN = 2*2*2 = 8
2) Only ONE is ODD-- 1*2*2= 4,  but this ODD can take any of three position , so 4*3=12
3) TWO odd--1*1*2=2,  but these 2 odd can occupy 3!/2! ways  , So 2*3=6 ways

total= 8+12+6=26

Archit3110 · Answer

total pairs possible ; 3^3 ; 27
and not even ; 5*5*5 ; only 1 such option
so 27-1 ; 26 all even'IMO D

davidbeckham · Answer

Hi  Bunuel  can you please explain what would be the scenario if the question is asking that the number should be even (instead of the product to be even)?

Bunuel · Answer

In this case each of the first two digits can take any of the three value (2, 4 or 5) but the third digit must be even (2 or 4), so 3*3*2 = 18.

CEdward · Answer

Good question.

There are 3 broad possibilities

even x even x even : 2 x 2 x 2 = 8 ways
even x odd x odd = (2 x 1 x 1) x 3 = 6 ways
even x even x odd = (2 x 2 x 1) x 3!/2! = 12 ways

8 + 6 + 12 = 26

D.

RastogiSarthak99 · Answer

Pretty straightforward once you realize its more about arriving at:

All Ways - Odd ways = Even

3 * 3 * 3 - 1 * 1 * 1 = 26

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A telephone company needs to create a set of 3-digit area

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