GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 07 Dec 2019, 20:33

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# A pass code is created by combining one-digit prime numbers. What is t

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 59588
A pass code is created by combining one-digit prime numbers. What is t  [#permalink]

### Show Tags

13 Jun 2019, 00:27
00:00

Difficulty:

95% (hard)

Question Stats:

30% (02:22) correct 70% (02:25) wrong based on 40 sessions

### HideShow timer Statistics

A pass code is created by combining one-digit prime numbers. What is the probability that the resulting four-digit number is a multiple of 12 ?

(A) 1/16
(B) 21/256
(C) 3/16
(D) 1/12
(E) 15/16

_________________
Intern
Joined: 03 Jun 2019
Posts: 5
Re: A pass code is created by combining one-digit prime numbers. What is t  [#permalink]

### Show Tags

18 Jul 2019, 19:34
pushpitkc wrote:
There are 4 one digit prime numbers - 2,3,5,7.

The total number of 4 digit numbers possible is 4*4*4*4(since digits can be repeated)
In order to determine how many of these 4 digit numbers are divisible by 12,
we employ the following method :
For a number to be divisible by 12, it has to be divisible by both 3 and 4.

Only numbers ending with 32,52 and 72 can be divisible by 4
as the rule for divisibility of 4 is that the last 2 digits must be divisible by 4.
For the number to be divisible by 3, the sum of the digits must add to give a multiple of 3 :

We have the possible total combinations for the numbers ending :

1. _ _ 32 are 2532,5232,5532,3732,7332, and2232(6 combinations)
2. _ _ 52 are 7752,3552,5352,2352, and 3252(5 combinations)
3. _ _ 72 are 5772,7572,2772,7272, and 3372(5 combinations)

P(Numbers divisible by 12) = P(A) = 6+5+5 = 16
Also P(T) = P(Total possible combinations) = 256

The probability is $$\frac{P(A)}{P(T)}$$ = $$\frac{16}{256} = \frac{1}{16}$$ (Option A)

Hi there,
I notice that this is an old thread, but anyway I more or less solved the question the same way as you did, but I feel it's too time consuming and error-prone, especially when it gets to the part of divisible by 3. Basically I had to try all the combinations of numbers to see which sum works and it took me probably a good 10 minutes to make sure I didn't miss or over-count any combinations.

So is there a faster way to solve this within 2 min? How did you count the possible combinations of sums (divisible by 3)?
Director
Joined: 28 Jul 2016
Posts: 670
Location: India
Concentration: Finance, Human Resources
GPA: 3.97
WE: Project Management (Investment Banking)
Re: A pass code is created by combining one-digit prime numbers. What is t  [#permalink]

### Show Tags

19 Jul 2019, 00:48
Bunuel
Can you help explain
why its only 16
--32 where first two can actually tally take 4*4 values =16 values
and thus for rest 2 also 16
thus total should be 16*3 = 48
VP
Joined: 19 Oct 2018
Posts: 1157
Location: India
A pass code is created by combining one-digit prime numbers. What is t  [#permalink]

### Show Tags

Updated on: 19 Jul 2019, 20:25
As the number must be divisible 4, last 2 digits must be divisible by 4
1. ab32
2. ab52
3. ab72

32=1 mod 3, 52=2 mod 3, and 72=0mod3. As the last 2 digits, when divided by 3, in 3 cases gives different and all possible remainders, starting 2-digits, hence, cover all possible combinations of 2-digits that can be formed by 4 prime numbers.

Total 2 digits number than can be formed by the use of 2, 3, 5 and 7, when repetition is allowed= 4*4=16

Probability= 16/256=1/16

Bunuel wrote:
A pass code is created by combining one-digit prime numbers. What is the probability that the resulting four-digit number is a multiple of 12 ?

(A) 1/16
(B) 21/256
(C) 3/16
(D) 1/12
(E) 15/16

Originally posted by nick1816 on 19 Jul 2019, 18:48.
Last edited by nick1816 on 19 Jul 2019, 20:25, edited 1 time in total.
VP
Joined: 19 Oct 2018
Posts: 1157
Location: India
Re: A pass code is created by combining one-digit prime numbers. What is t  [#permalink]

### Show Tags

19 Jul 2019, 19:00
xx32... xx xan't take all 16 values, because sum of 4 digits must be divisible by 3.
32 leaves 2 remainder when divided by 3, hence xx must leave 1 remainder when divided by 3.

Hence in this case xx can be combination of 22, 25, 55 or 73.
Total arrangements possible= 1+2+1+2=6.

Similarly you have to find for the rest of the 2 cases.

globaldesi wrote:
Bunuel
Can you help explain
why its only 16
--32 where first two can actually tally take 4*4 values =16 values
and thus for rest 2 also 16
thus total should be 16*3 = 48
Re: A pass code is created by combining one-digit prime numbers. What is t   [#permalink] 19 Jul 2019, 19:00
Display posts from previous: Sort by