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If each of the digits can be used, as many times as [#permalink]
25 Feb 2008, 11:55

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 _________________

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

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

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

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

hmmm. whats the source?

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)

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)

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.

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

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