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15 Jul 2014, 03:15
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How many 6-digit number formed by 1, 3, 5, 6, 7, 9 that is divisible by 11 and no repetitions of digits?
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15 Jul 2014, 05:29
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1) Divisibility by 11 rule: an alternative sum has to be divisible by 11. For example, if abcd is an integer and letters represent digits, then a-b+c-d is an alternative sum and it has to be divisible by 11.
2) Let's build the number with the largest alternative sum:
3 biggest digits: 6+7+9 = 22
3 smallest digits: 1+3+5 = 9
the alternative sum is 22 - 9 = 13
one of the numbers with the largest alternative sum is 617395
3) From #2 we can say that there are three possible alternative sums that are divisible by 11: -11, 0, and 11.
4) Rule out "0" because all alternative sums are odd
5) "-11" represents the same number of digit combinations but 1st, 3rd, 5th digits are switched with 2nd, 4th, and 6th digits
6) Let's try to build a number with the alternative sum of 11
The largest alternative sum is 13 (from #2). it means we need to find a way to decrease it by 2. The only way to decrease the alternative sum by 2 is two switch places of digits that have the difference of 1. So, it's 5 and 6 only. One of the numbers with the alternative sum of 11 is 517396
7) We can change places of 1,3,6 (3! = 6 combinations) and 5, 7, 9 (3!=6) combinations. Therefore, there are 12 numbers with the alternative sum of 2.
8) Taking into account #5, the answer is 2*12 = 24 numbers
It definitely isn't a GMAT question, but it's very good for practicing if you aim at 700+.
2) Let's build the number with the largest alternative sum:
3 biggest digits: 6+7+9 = 22
3 smallest digits: 1+3+5 = 9
the alternative sum is 22 - 9 = 13
one of the numbers with the largest alternative sum is 617395
3) From #2 we can say that there are three possible alternative sums that are divisible by 11: -11, 0, and 11.
4) Rule out "0" because all alternative sums are odd
5) "-11" represents the same number of digit combinations but 1st, 3rd, 5th digits are switched with 2nd, 4th, and 6th digits
6) Let's try to build a number with the alternative sum of 11
The largest alternative sum is 13 (from #2). it means we need to find a way to decrease it by 2. The only way to decrease the alternative sum by 2 is two switch places of digits that have the difference of 1. So, it's 5 and 6 only. One of the numbers with the alternative sum of 11 is 517396
7) We can change places of 1,3,6 (3! = 6 combinations) and 5, 7, 9 (3!=6) combinations. Therefore, there are 12 numbers with the alternative sum of 2.
8) Taking into account #5, the answer is 2*12 = 24 numbers
It definitely isn't a GMAT question, but it's very good for practicing if you aim at 700+.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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