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For each set of three distinct nonzero digits, consider the sum of all [#permalink]
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Too tough for A real GRE question.

I, as Carcass, would not practice such questions.

Personal opinion though
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Re: For each set of three distinct nonzero digits, consider the sum of all [#permalink]
Carcass wrote:
The sum of the 6 different permutations of abc =

100a+10b+c

+100a+10c+b

+100b+10a+c

+100b+10c+a

+100c+10a+b

+100c+10b+c

=222(a+b+c)

Therefore, the sum of a+b+c determines the sum of the 6 different permutations of abc

How many different sums of a+b+c are there, and how many sums of 6 different permutations of abc are there?

The minimum case of the sum of a+b+c, 1+2+3=6, the sum of six different permutations=6*222

The maximum case of the sum of a+b+c, 7+8+9=24, the sum of six different permutations=24*222

1, 2, 3, 4, 5, 6, 7, 8, 9 numbers are used arbitrarily, then the sum of a+b+c can be obtained continuously from 6 to 24, and there are 24-6+1=19 possibilities

There are 19 different sums of a+b+c, so there are 19 sums of 6 different permutations of abc


Brilliant explanaiton, thanks Carcass :please: :thumbsup:
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Re: For each set of three distinct nonzero digits, consider the sum of all [#permalink]
Carcass wrote:
Too tough for A real GRE question.

I, as Carcass, would not practice such questions.

Personal opinion though


That's a great relief to know.
Thanks Carcass :please: :thumbsup: :D
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Re: For each set of three distinct nonzero digits, consider the sum of all [#permalink]
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