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 Author: questioner [ Sat Jun 19, 2010 12:44 am ] Post subject: GMAT Number Theory When 28 is divided by the positive integer x, the remainder is 1. What is the sum of all the possible values of x for which this is true?A. 2B. 3C. 9D. 30E. 39(E) First, we must find which numbers leave a remainder of 1. These numbers must be divisors of 27 (which is 1 less than 28) greater than 1. Such divisors of 27 are 3, 9, 27.The first number is 3: 28/3 = 9, remainder 1.The next number after that is 9: 28/9 = 3, remainder 1.27 also gives a remainder of 1, giving 3 possible valuesfor x.Adding these three values, we get 3 + 9 + 27 = 39.The correct answer is choice (E).-------------Why we need to add all the three possible divident for this question?

 Author: Gennadiy [ Sat Jun 19, 2010 12:45 am ] Post subject: Re: math (test 1, question 22): numbers theory We need to add all the three possible values because that is what questions asks us about. "What is the sum of all the possible values of x for which this is true?"

 Author: questioner [ Thu Aug 11, 2011 3:48 pm ] Post subject: Re: math (test 1, question 22): numbers theory 3 + 9 + 27 = 38 NOT 39.Answer E should state 38 NOT 39.

 Author: Gennadiy [ Thu Aug 11, 2011 3:54 pm ] Post subject: Re: math (test 1, question 22): numbers theory When you add several numbers it's better to group them to get some "good" terms, which are easy to calculate.3 + 9 + 27 = (3 + 27) + 9 = 30 + 9 = 39Another example:1 + 1 + 2 + 2 + 99 + 99 + 98 + 98 = (1 + 99) + (1 + 99) + (2 + 98) + (2 + 98) = 100 + 100 + 100 + 100 = 400This method is much faster than adding all the terms consecutively from left to right. It also reduces chances of making a mistake.

 Author: questioner [ Sun Apr 22, 2012 1:59 pm ] Post subject: Re: math (test 1, question 22): numbers theory There is the possibility that x can be equal to 1 so we have:1 + 3 + 9 + 27 = 40 which is not among the answers provided.

 Author: Gennadiy [ Sun Apr 22, 2012 2:03 pm ] Post subject: Re: math (test 1, question 22): numbers theory Quote:There is the possibility that x can be equal to 1…The divisor, x, cannot be 1, because the remainder, 1, must be LESS than the divisor.

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