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Re: How many three digit numbers of distinct digits can be formed by [#permalink]

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04 Aug 2015, 07:48

Cadaver wrote:

462/264 , 143/34, 132/231,154/451 repitition??

There must be some easy way out....

When a question talks about distinct digits it means that the digits in 1 number should be distinct and not that they need to unique thoroughout.

Example, 110 is not an acceptable case as 1 is repeated but 132 and 231 will be 2 good values. 'Distinct' does not apply (usually) apply to different numbers.

CONCEPT: A number is divisible by 11 when Difference of Sums of even place digits and odd place digits is either zero or a multiple of 11.

i.e. a 3-Digit Number abc will be divisible by 11 if (a+c) - b = 0 or multiple of 11

@b=1, a+c should be 1 or 12 i.e. (a,c) can be (5, 7), (7, 5) --- 2 cases

@b=2, a+c should be 2 or 13 i.e. (a,c) can be (6, 7), (7, 6) ---- 2 cases

@b=3, a+c should be 3 or 14 i.e. (a,c) can be (1, 2), (2, 1) ---- 2 cases

@b=4, a+c should be 4 or 15 i.e. (a,c) can be (1, 3), (3, 1) ---- 2 cases

@b=5, a+c should be 5 or 16 i.e. (a,c) can be (1, 4), (2, 3), (3, 2), (4, 1) ---- 4 cases

@b=6, a+c should be 6 or 17 i.e. (a,c) can be (1, 5), (2, 4), (4, 2), (5, 1) ---- 4 cases

@b=7, a+c should be 7 or 18 i.e. (a,c) can be (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) ---- 6 cases

Total Cases = 2+2+2+2+4+4+6 = 22 cases

Answer: option E
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Did multiple mistakes due to mental calculations. Have updated options. There are 22 cases.

Thanks for notifying about my mistake.
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Re: How many three digit numbers of distinct digits can be formed by [#permalink]

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20 Sep 2017, 18:04

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Re: How many three digit numbers of distinct digits can be formed by [#permalink]

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22 Sep 2017, 17:54

the best approach to go via rule if a number is divisible by 11 then take every 2nd digit (in our case the 1 in the middle) and subtract it from the sum of the rest digits, if the result is 0 or a multiple of 11 then we are settled number x1x possible combination 7 and 5 (2 options as it could be 5 and 7) number x2x possible combination 7 and 6 (2 options) number x3x possible combination 1 and 2 (2 options) number x4x possible combination 1 and 3 (2 options) number x5x possible combination 3 and 2 also 4 and 1 (4 options) number x6x possible combination 5 and 1 also 4 and 2 (4 options) number x7x possible combination 5 and 2 also 4 and 3 also 6 and 1 (6 options) Anyways, I found this question a bit time-consuming, that's the best approach to solve it, however, even using that it took me 3 mins or more, to double-check the responses as all answers are pretty much close to each other

I'm curious what are the odd to get that kind of question on GMAT? (instead of 11 that can be 7 or 13, non-standard divisibility patterns)