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Joined: 02 Sep 2009
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43% (01:38) correct 57% (01:41) wrong based on 157 sessions
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Math Expert
Joined: 02 Sep 2009
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Re M0432
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16 Sep 2014, 00:23
Official Solution:How many fivedigit numbers can be formed using the digits 0, 1, 2, 3, 4 and 5 which are divisible by 3, without repeating the digits? A. 15 B. 96 C. 120 D. 181 E. 216 First step: We should determine which 5 digits from given 6, would form the 5 digit number divisible by 3. We have six digits: 0, 1, 2, 3, 4, 5. Their sum is 15. For a number to be divisible by 3 the sum of the digits must be divisible by 3. As the sum of the six given numbers is 15 (divisible by 3) only 5 digits good to form our 5 digit number would be 150={1, 2, 3, 4, 5} and 153={0, 1, 2, 4, 5}. Meaning that no other 5 from given six will total the number divisible by 3. Second step: We have two set of numbers: {1, 2, 3, 4, 5} and {0, 1, 2, 4, 5}. How many 5 digit numbers can be formed using these two sets: {1, 2, 3, 4, 5} This set gives \(5!\) numbers, as any combination of these digits would give us 5 digit number divisible by 3. \(5!=120\). {0, 1, 2, 4, 5}. Now, here we cannot use 0 as the first digit, otherwise number won't be any more 5 digit and become 4 digit. So, desired # would be total combinations \(5!\), minus combinations with 0 as the first digit (combination of 4) \(4!\): \(5!4!=4!(51)=4!*4=96\). Total \(120+96=216\) Answer: E
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Re: M0432
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05 Aug 2015, 06:31
Hello, Could you please explain me: Second step{0, 1, 2, 4, 5}. Now, here we cannot use 0 as the first digit, otherwise number won't be any more 5 digit and become 4 digit. So, desired # would be total combinations 5!, minus combinations with 0 as the first digit (combination of 4) 4!: 5!−4!=4!(5−1)=4!∗4=96. Total 120+96=216 Why not : 5*4*2*1 = 40 (3 is not included in this combination) Thanks a lot, Daniela
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Re: M0432
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05 Aug 2015, 07:21
trick is to know which numbers are divisible by 3 , whose sum of the digits are divisible by 3. with 1 , 2 ,3,4, and 5 we can make following numbers : 5*4*3*2*1 = 120 numbers ; here the sum is 15 for the digits of the numbers formed formed ; next are the numbers whose sum of the digits is 12 . taking 0 , 1 , 2 ,4 ,5 : we can make 96 numbers: 4*4*3*2*1 = 96 ( 4 IN STARTING AS A NUMBER CANT START WITH ZERO ) TOTAL NUMBERS FORMED = 120+96 = 216 OPTION E THEREBY .
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Joined: 16 Aug 2013
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Re: M0432
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09 Aug 2015, 22:40
hi BunuelI agree with you about the solution. but what if we calculate all 5digit numbers and then calculate 4digit number. subtract them, then divide the result by 3(to find multiples of 3). I think this approach is correct but the result is quite different. let's see : 1 5digit numbers: 5*5*4*3*2=600 2 4digit numbers: 5*5*4*3=300 3 (600300)/3=100 ?



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Re M0432
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17 Aug 2015, 08:11
I think this is a highquality question and I agree with explanation. I think that this question is great.



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Re: M0432
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20 Nov 2015, 11:27
Hello Bunuel,
I think question is flawed.
First Digit possibility 5 (except zero) Second digit possibility 5 (Zero + remaining four digits) Third digit possibility 4 Forth digit possibility 3 Fifth digit possibility 2
Total 5 X 5 X 4 X 3 X 2 = 600
Please assist.



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Re: M0432
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21 Nov 2015, 02:49



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Re M0432
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24 Jul 2016, 03:26
I think this is a highquality question and I agree with explanation. It is a super quality question and probably one of the best questions to understand number arrangements probability. It may not be possible for even smartest of minds to solve it correctly if it comes in 3037 range of question number.
But even if we are hitting q50 or 51 and compelling GMAT to throw harder questions will GMAT throw 19 nos. of 700 level questions?
If somebody could throw some light.



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Re: M0432
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18 Sep 2018, 19:30
Hi Bunuel, Thanks for the elegant solution. Is this solvable via the slot method? Since only 5 numbers can be used and because 0 cannot be the first digit: 4 x 4 x 3 x 2 x 1 gives me 96.










