NOTE: I have reworded the question so that the correct is, indeed,
B
RBBmba@2014 wrote:How many POSITIVE INTEGERS less than 1000 are divisible by 5 AND have no repeated digits?
(A) 144
(B) 154
(C) 155
(D) 182
(E) 214
OA: B
This question requires us to consider 1-digit, 2-digit and 3-digit numbers that are divisible by 5 AND have no repeated digits.
To be safe, I'll handle each case separately.
1-digit numbers
Only the integer 5 is divisible by 5.
So, the number of 1-digit integers that meet the condition is
1
2-digit numbers
We need to consider integers of the form _0 and _5
- for numbers of the form _0, we can place
9 different digits in the tens position (1,2,3,4,5,6,7,8 or 9).
- for numbers of the form _5, we can place
8 different digits in the tens position (1,2,3,4,6,7,8 or 9).
So, the total number of 2-digit integers that meet the condition =
9 +
8 =
17
3-digit numbers
We need to consider integers of the form _ _0 and _ _5
- integers of the form _ _0,
We can place 9 different digits in the hundreds position (1,2,3,4,5,6,7,8 or 9).
We can place 8 different digits in the tens position (to avoid repetition)
So, the number of integers of the form _ _ 0 = (9)(8) =
72
- integers of the form _ _5,
We can place 8 different digits in the hundreds position (1,2,3,4,6,7,8 or 9).
We can place 8 different digits in the tens position (to avoid repetition)
So, the number of integers of the form _ _ 5 = (8)(8) =
64
So, the total number of 3-digit integers that meet the condition =
72 +
64 =
136
---------------------------------
So, the TOTAL number of integers =
1 +
17 +
136
=
154
=
B
Cheers,
Brent
