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Joined: 17 Nov 2007
Posts: 2408
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
Can someone help me solve this problem
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Updated on: 15 Jul 2014, 05:35
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 6*6 = 36 numbers with the alternative sum of 11.
8) Taking into account #5, the answer is 2*36 = 72 numbers
It definitely isn't a GMAT question, but it's very good for practicing if you aim at 700+.
Originally posted by
walker on 15 Jul 2014, 05:29.
Last edited by
walker on 15 Jul 2014, 05:35, edited 1 time in total.
Edited typos