{"id":4672,"date":"2014-05-12T09:00:47","date_gmt":"2014-05-12T16:00:47","guid":{"rendered":"https:\/\/magoosh.com\/gmat\/?p=4672"},"modified":"2020-01-15T10:48:40","modified_gmt":"2020-01-15T18:48:40","slug":"consecutive-integers-and-multiples-on-the-gmat","status":"publish","type":"post","link":"https:\/\/magoosh.com\/gmat\/consecutive-integers-and-multiples-on-the-gmat\/","title":{"rendered":"Consecutive Integers and Multiples on the GMAT"},"content":{"rendered":"

\"\"<\/p>\n

To begin, here are four reasonably challenging practice problems.<\/p>\n

1) S is a set of n consecutive positive integers.\u00a0 Is the mean of the set a positive integer?<\/p>\n

Statement #1<\/span>: the range of S is an even integer<\/p>\n

Statement #2<\/span>: the median of S is a positive integer<\/p>\n

2) If N = 255 is the lowest of a set of 23 consecutive multiples of 15, what is the range of this set?<\/p>\n

    \n
  1. \n
      (A) 315<\/ol>\n<\/li>\n<\/ol>\n

       <\/p>\n

        \n
      1. \n
          (B) 330<\/ol>\n<\/li>\n<\/ol>\n

           <\/p>\n

            \n
          1. \n
              (C) 345<\/ol>\n<\/li>\n<\/ol>\n

               <\/p>\n

                \n
              1. \n
                  (D) 360<\/ol>\n<\/li>\n<\/ol>\n

                   <\/p>\n

                    (E) 375<\/ol>\n

                    \"ciamotg_img1\"<\/p>\n

                    I. 4
                    \nII. 6
                    \nIII. 18<\/p>\n

                      \n
                    1. \n
                        (A) I only<\/ol>\n<\/li>\n<\/ol>\n

                         <\/p>\n

                          \n
                        1. \n
                            (B) II only<\/ol>\n<\/li>\n<\/ol>\n

                             <\/p>\n

                              \n
                            1. \n
                                (C) I and II only<\/ol>\n<\/li>\n<\/ol>\n

                                 <\/p>\n

                                  \n
                                1. \n
                                    (D) II and III only<\/ol>\n<\/li>\n<\/ol>\n

                                     <\/p>\n

                                      (E) I, II, and III<\/ol>\n

                                      \"ciamotg_img2\"<\/p>\n

                                        \n
                                      1. \n
                                          (A) 5<\/ol>\n<\/li>\n<\/ol>\n

                                           <\/p>\n

                                            \n
                                          1. \n
                                              (B) 10<\/ol>\n<\/li>\n<\/ol>\n

                                               <\/p>\n

                                                \n
                                              1. \n
                                                  (C) 25<\/ol>\n<\/li>\n<\/ol>\n

                                                   <\/p>\n

                                                    \n
                                                  1. \n
                                                      (D) 30<\/ol>\n<\/li>\n<\/ol>\n

                                                       <\/p>\n

                                                        (E) 35<\/ol>\n

                                                        Explanations for these will come at the end of the article.<\/p>\n

                                                         <\/p>\n

                                                        Consecutive Integers<\/h2>\n

                                                        The word “consecutive” means “in a row; one after the other.”\u00a0 A set of three consecutive integers might mean {3, 4, 5} or {137, 138, 139} or {\u201325, \u201324, \u201323}.\u00a0 As long as the integers are in a row, it doesn’t matter whether they are big or small, positive or negative.\u00a0\u00a0 In fact, the set {\u20131, 0, +1} contains one positive number and one negative number (of course, zero is neither positive nor negative).<\/p>\n

                                                         <\/p>\n

                                                        Properties of Consecutive Integers<\/h2>\n

                                                        a) Any set of n consecutive integers will contain exactly one number divisible by n.\u00a0 For example, any three integers in row must contain a multiple of 3; any 17 integers will contain one multiple of 17, etc.\u00a0\u00a0 Now, you may look at the set {\u20131, 0, +1}, a set of three consecutive integers, and wonder: where is the multiple of 3?\u00a0 This is tricky.\u00a0 As it turns out, zero is a multiple of every integer, because (any integer) times zero equals zero.<\/p>\n

                                                        b)\u00a0 In a set of three consecutive integers, we could have two evens and one odd, or two odds and one even, depending upon where we started.\u00a0 In a set of four consecutive integers, we would have to have two evens and two odds.\u00a0 More generally, if we have an odd number of consecutive integers, we could have more evens or more odds, depending on the starting value, but if we have an even number of consecutive integers, the evens and odds have to be evenly split.<\/p>\n

                                                        c) If n is an odd number, then the sum of n consecutive integers is divisible by n.\u00a0 For example: for any three integers in a row, the sum is divisible by 3; for any 7 integers in a row, the sum is divisible by 7, etc.<\/p>\n

                                                        Algebraic Representations<\/h2>\n

                                                        Anyone can recognize that {6, 7, 8, 9, 10, 11} is a set of consecutive integers.\u00a0 When you are given plain old numbers, it’s easy to see whether they are consecutive integers.\u00a0 That’s too easy.\u00a0 The GMAT will not ask about that.\u00a0 Instead, the GMAT will give you algebraic representations of consecutive integers.<\/p>\n

                                                        The following are examples of algebraic representations of consecutive integers:<\/p>\n

                                                        {n, n + 1, n + 2, n + 3, n + 4, n + 5}<\/p>\n

                                                        {n \u2013 2, n \u2013 1, n, n + 1}<\/p>\n

                                                        {n + 12, n + 13, n + 14}<\/p>\n

                                                        For all of those, if n equals any integer, then the set will be a set of consecutive integers.\u00a0\u00a0 For simplicity, let’s pick n = 10.\u00a0 The first set becomes the set of integers from 10 to 15; the second, from 8 to 11; and the third, from 22 to 24.\u00a0 Even without knowing the value of n, we can apply the properties of consecutive integers to the set: for example, the second is a set of four consecutive integers, so it must have two evens and two odds; the third is a set of three consecutive integers, so the sum of those three numbers must be divisible by three.<\/p>\n

                                                         <\/p>\n

                                                        Summary<\/h2>\n

                                                        If you had some insights reading this article, you may want to give the problems above a second look before reading the solutions below.\u00a0 Here’s another practice question from inside Magoosh:<\/p>\n

                                                        5) http:\/\/gmat.magoosh.com\/questions\/943<\/a><\/p>\n

                                                        If you have questions about anything unclear in this post, please let us know in the comments section.<\/p>\n

                                                        \"ciamotg_img3\"<\/p>\n

                                                         <\/p>\n

                                                        Practice problem explanations<\/h2>\n

                                                        1) First of all, for a set of consecutive integers, or for any set of evenly spaced numbers, the mean and the median are equal.\u00a0 If there’s an odd number of members of the list, then the median is the middle number.\u00a0 If there’s an even number of members of the list, then the median is the average of the two middle numbers.\u00a0 For example, the median of {1, 2, 3, 4, 5} is 3, a positive integer and member of the set.\u00a0 For consecutive integers, an even number of members would mean that the mean or median is the average of the two middle integers.\u00a0 For example, the median of {1,2, 3, 4} is the average of 2\u00a0and 3, that is, 2.5, not an integer.\u00a0 The only way the mean or median can be an integer is if the set of consecutive integers has an odd number of members.<\/p>\n

                                                        Statement #1<\/span>: If there are an even number of consecutive integers, then the evens and odds are balanced in the set, and the first and last number must be opposite: one must be even and the other must be odd.\u00a0 Thus, the range, the difference of (max) \u2013 (min) would be either (even) \u2013 (odd) or (odd) \u2013 (even), in either case, an odd number.\u00a0 If the range is odd, the number of consecutive integers is even.<\/p>\n

                                                        If there are an odd number of consecutive integers, then the first and last numbers are either both even or both odd.\u00a0 The range would be either (even) \u2013 (even) or (odd) \u2013 (odd), in either case, an even number.\u00a0 If the range is even, the number of consecutive integers is odd.\u00a0 That must be the case here.\u00a0 As we have seen above, this means the mean or median is a positive integer.\u00a0 This statement, alone and by itself, is sufficient<\/b>.<\/p>\n

                                                        Statement #2<\/span>: As we discussed above, the mean = the median.\u00a0 If the latter is a positive integer, so is the former.\u00a0\u00a0 This statement, alone and by itself, is sufficient<\/b>.<\/p>\n

                                                        Both statement are separately sufficient.\u00a0 Answer = (D)<\/b><\/p>\n

                                                        2) When we have a set of consecutive integers or consecutive multiples of the number, the range depends only on the size of the set, how many members, not where on the number line the set starts or ends.\u00a0 For example, any seven consecutive integers will have a range of 6, whether it’s 1 through 7 or 51 through 57.\u00a0 Thus, we can ignore the starting number, 255, which is just there to confuse us.\u00a0 We can pick any more convenient starting value.<\/p>\n

                                                        Let’s start at a1 = 15 = 15*1.\u00a0 Then a2 = 15*2 = 30, and a3 = 15*3 = 45.\u00a0 Continuing in this pattern, the last number would be a23 = 15*23.\u00a0 Don’t multiply that yet.\u00a0 The range would be highest minus the lowest:<\/p>\n

                                                        range = (a23) \u2013 (a1) = 15*23 \u2013 15*1 = 15*(23 \u2013 1) = 15*22<\/p>\n

                                                        Now, use the doubling & halving trick<\/a>.\u00a0 Half of 22 is 11, and twice 15 is 30, so<\/p>\n

                                                        15*22 = 11*30 = 330<\/p>\n

                                                        Answer = (B)<\/b><\/p>\n

                                                        3) This is a tricky one.\u00a0 If you start plugging in values for n, you are sunk.\u00a0 The numbers are gigantic and unwieldy.\u00a0 This one begins with some clever factoring.\u00a0 Clearly, in the first factor, we can factor out a factor of n:<\/p>\n

                                                        \"ciamotg_img4\"<\/a><\/p>\n

                                                        In the second one, we can us the Difference of Two Squares<\/a>:<\/p>\n

                                                        \"ciamotg_img5\"<\/a><\/p>\n

                                                        Now, put all that together, and rearrange the order:<\/p>\n

                                                        \"ciamotg_img6\"<\/a><\/p>\n

                                                        Written in that order, we see this is a product of four consecutive integers.\u00a0 We absolutely know that two of the numbers are even and two are odd.\u00a0 If there are two even numbers, each one has a factor of two, so the product would have a factor of 2*2 = 4.\u00a0 The product must be divisible by 4.\u00a0 We know I<\/b> must be true.<\/p>\n

                                                        Any three consecutive integers contains one multiple of 3, so four consecutive integers would contain at least one.\u00a0 We have a multiple of 3, which has 3 as a factor, and at least one even number, which has 2 as a factor.\u00a0 They may be the same number or may not be: that doesn’t matter.\u00a0 As long as the entire product contains at least one factor of 3 and at least one factor of 2, then the product must be divisible by 6.\u00a0 We know II<\/b> must be true.<\/p>\n

                                                        To be divisible by 18=2*3*3, we need one factor of 2 and two factors of 3.\u00a0 We definitely have more than one factor of 2.\u00a0 The problem is: how many factors of three do we have? In a set of four consecutive integers, we could<\/i> have two factors of three, as in {12, 13, 14, 15} or {6, 7, 8, 9}.\u00a0 BUT, we could also have a set of four consecutive integers with only one factor of three, as in {4, 5, 6, 7}.\u00a0 Thus, we could<\/i> have two factors of three, and the product could<\/i> be divisible by 18, but we are not sure.\u00a0 We cannot say that this must<\/span> be true.\u00a0 Therefore, III<\/b> cannot be among the answer.<\/p>\n

                                                        Only I<\/b> and II<\/b> work.\u00a0 Answer = (C)<\/b><\/p>\n

                                                        4) We know that odd integers are spaced two apart: if we have one odd integer, we can add 2 to get the next one.\u00a0\u00a0 Starting from the given expression for the first member of the set, we can say:<\/p>\n

                                                        \"ciamotg_img7\"<\/a><\/p>\n

                                                        This highest member, we can set equal to 7n<\/p>\n

                                                        \"ciamotg_img8\"<\/p>\n

                                                        n = 2 or n = 5.\u00a0 Well, if n equals 2, then the first member would be 4, not an odd integer.\u00a0 That value doesn’t work with the problem.\u00a0 Therefore, we know n = 5.\u00a0 This means that first member is 25, and the whole set is {25, 27, 29, 31, 33, 35}.\u00a0 The median of this set is the average of the two middle numbers, 29 and 31; median = 30.<\/p>\n

                                                        Answer = (D)<\/b>.<\/p>\n

                                                         <\/p>\n","protected":false},"excerpt":{"rendered":"

                                                        To begin, here are four reasonably challenging practice problems. 1) S is a set of n consecutive positive integers.\u00a0 Is the mean of the set a positive integer? Statement #1: the range of S is an even integer Statement #2: the median of S is a positive integer 2) If N = 255 is the […]<\/p>\n","protected":false},"author":26,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[160],"tags":[],"ppma_author":[13209],"class_list":["post-4672","post","type-post","status-publish","format-standard","hentry","category-arithmetic"],"acf":[],"yoast_head":"\nConsecutive Integers and Multiples on the GMAT - Magoosh Blog \u2014 GMAT\u00ae Exam<\/title>\n<meta name=\"description\" content=\"Consecutive integer problems on GMAT Quant can be tricky. 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