If each of the digits can be used, as many times as

Question

If each of the digits can be used, as many times as necessary, what is the probability of creating a four-digit number that is divisible by four and that begins and ends with a prime number?
(A) 1/18
(B)1/25
(C)2/25
(D) 1/9
(E) 4/25

GMAT TIGER · Answer

= (4 x 10 x 5 x 1)/3000 
= 1/15

hmmm. whats the source?

bmwhype2 · Answer

can you explain your answer.

got it off the net.

maratikus · Answer

why are you dividing by 3,000?

GMAT TIGER · Answer

1000's place = 2, 3, 5 and 7. so 4 times
100's place = all 10 digits so 10 times
10's place = only 1, 3, 5, 7 and 9. so 5 times
unit place = only 2 so 1 time

so multiply them = 4x10x5x1 = 200.
total = 9x10x10x10 = 9000 (not 3000)

= 200/9000
= 1/45

GMAT TIGER · Answer

i am getting recovered from the lost world.

i do calculation in my mind. for a while i was out of practice so i am practicing it again.

bmwhype2 · Answer

how are u accounting for divisibility by 4?

maratikus · Answer

that's exactly how I solved it too.

maratikus · Answer

the last digit has to be equal to 2, the third digit has to be an odd number (22 is not divisible by 4, 12 is).

The second digit doesn't matter, the first digit is either 2,3,5 or 7. The second and third digits don't impact divisibility by 4 because 100 is divisible by 4.

bmwhype2 · Answer

i dont understand why the units digit must be the digit 2.

a number is divisible by 4 when the last two digits are divisible by 4
xx04
xx08
xx12
xx16
xx20
etc...

maratikus · Answer

the number is divisible by 4 -> the last digit has to be even
the last digit is a prime number -> the last digit is equal to 2 -> the only even prime number

elmagnifico · Answer

good question. we need more of these here. where exactly online did you get it from?

GMAT TIGER · Answer

remember: the question says the last digit can only be a prime. so nothing other than 2 can be in the unit digit.

bluegiant · Answer

Answer is A) 1/18

In this question, we should make 4-digits Number ; hereinafter OOOO

Based on the condition, the last 2-digits of number must be as follows,
if 4-digits numbers can be diviable by 4

OO12
OO32
OO52
OO72
OO92

The begining number must be the prime number, accodring to the condition.

2OOO
3OOO
5OOO
7OOO

the second number can be any number from zero to nine (o~9)

Here is the reasoning for this question.

Probability = (4X10X5) / (9X10X10X4) = 1/18

Is that clear?

bmwhype2 · Answer

thanks. this one was tougher than i expected.

If each of the digits can be used, as many times as

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