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Challenging Counting Problem

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Challenging Counting Problem

by knight247 » Mon Oct 10, 2011 1:05 pm
How many 4 digit number that are divisble by 4 can be formed using digits 0 to 7 if no digit is to occur more than once in each number.
A) 570
B) 370
C) 345
D) 440
E) 520

OA is B

Detailed explanations would be appreciated.
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by GmatMathPro » Mon Oct 10, 2011 1:24 pm
Here is one way:

To be divisible by 4 it must end in 04,12,16,20,24,32,36,40,52,56,60,64,72, or 76. 4 of these include zero and 10 don't.

Two groups:

1. number has 0 as one of last two digits:

4 ways to choose the ending, 6 ways to choose the first digit, 5 ways to choose the 2nd

4*6*5=120 ways

2. number does not have 0 has one of the last two digits:

10 ways to choose the ending, 5 ways to choose the first, 5 ways to choose the second:

5*5*10=250

250+120=370

B
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by shankar.ashwin » Mon Oct 10, 2011 1:42 pm
If I had exactly 20 seconds to do this sum, I would do (7*7*6*5)/4 which is approx equal to 370. No means an accurate method but an educative guess.
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by GMATGuruNY » Mon Oct 10, 2011 1:59 pm
knight247 wrote:How many 4 digit number that are divisble by 4 can be formed using digits 0 to 7 if no digit is to occur more than once in each number.
A) 570
B) 370
C) 345
D) 440
E) 520

OA is B

Detailed explanations would be appreciated.
Another approach:

Good = Total - Bad.

Total options if we disregard whether 0 appears in the thousands place:
The 2 rightmost digits must form a multiple of 4.
The total number of 2-digit multiples of 4 between 01 and 76 = 19.
Of these 19 multiples of 4, 5 options -- 08, 28, 44, 48, and 68 -- are not allowed.
Thus:
Number of options for the 2 rightmost digits = 19-5 = 14.
Number of options for the thousands digit = 6. (Any digit 0-7 other than the 2 digits already used.)
Number of options for the hundreds digit = 5. (Any digit 0-7 other than the 3 digits already used.)
Total options = 14*6*5 = 420.

Bad options:
A bad option puts 0 in the thousands place.
We need to count how many ways the remaining 3 digits can be chosen.
Since 0 is in the thousands place, the last 2 digits cannot include 0.
Thus, of the 14 multiples of 4 considered above for the 2 rightmost positions, 4 of these options -- 04, 20, 40, and 60 -- are not allowed here.
Thus:
Number of options for the 2 rightmost digits = 14-4 = 10.
Number of options for the hundreds digit = 5. (Any digit 1-7 other than the 2 digits used in the 2 rightmost positions).
Bad options = 10*5 = 50.

Good = 420-50 = 370.

The correct answer is B.
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by Proleefeek111 » Mon Oct 10, 2011 9:17 pm
knight247 wrote:How many 4 digit number that are divisble by 4 can be formed using digits 0 to 7 if no digit is to occur more than once in each number.
A) 570
B) 370
C) 345
D) 440
E) 520

OA is B

Detailed explanations would be appreciated.
Soln:
Divisibility test of 4 : the last two digits must be div by 4.

Hence we must focus on the last two digits first.
_X_X_X_X = expected number.

Last digit could be 0, 2, 4, 6 so 4 possible ways
Second last Digit can be filled in the following way:
20, 40, 60
12, 32, 52, 72
04, 24, 64
16, 36, 56, 76

So a total of 14 combinations are possible. For each of these 14 comboes the remaining two digits can be filled in 6*5 = 30 ways

Thus main sample size = 14*30 = 420,

But we cannot have a '0' in the 1000th position hence
the approach should be to find all the possibilities with 0 in the thousandth place and substract them from the sample size.

4 of 14 have 0 in units / tens place so neglecting those.

Thus we have 6 digits remaining to fill the 100th place. Thus total combos possible = 6*10 = 60
But 10 of these combos will have a '0' in the 100th place hence 50 possibilities are remaining which can have a '0' in the 1000th place.

Substract this number from the original sample i.e. 420 -50 = 370

Ans = B.
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