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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## Tough Reminder Question ##### This topic has 3 expert replies and 1 member reply ## Tough Reminder Question In the first 1000 positive integers, how many integers exist such that they leave a remainder 4 when divided by 7, and a remainder 9 when divided by 11? A. 10 B. 11 C. 12 D. 13 E. 14 OA: D What is the best short cut to this problem? ### GMAT/MBA Expert GMAT Instructor Joined 25 May 2010 Posted: 15029 messages Followed by: 1859 members Upvotes: 13060 GMAT Score: 790 A quick lesson on remainders: Quote: When x is divided by 5, the remainder is 3. In other words, x is 3 more than a multiple of 5: x = 5a + 3. When x is divided by 7, the remainder is 4. In other words, x is 4 more than a multiple of 7: x = 7b + 4. Combined, the statements above imply that when x is divided by 35 -- the LOWEST COMMON MULTIPLE OF 5 AND 7 -- there will be a constant remainder R. Put another way, x is R more than a multiple of 35: x = 35c + R. To determine the value of R: Make a list of values that satisfy the first statement: When x is divided by 5, the remainder is 3. x = 5a + 3 = 3, 8, 13, 18... Make a list of values that satisfy the second statement: When x is divided by 7, the remainder is 4. x = 7b + 4 = 4, 11, 18... The value of R is the SMALLEST VALUE COMMON TO BOTH LISTS: R = 18. Putting it all together: x = 35c + 18. Another example: When x is divided by 3, the remainder is 1. x = 3a + 1 = 1, 4, 7, 10, 13... When x is divided by 11, the remainder is 2. x = 11b + 2 = 2, 13... Thus, when x is divided by 33 -- the LCM of 3 and 11 -- the remainder will be 13 (the smallest value common to both lists). x = 33c + 13 = 13, 46, 79... Onto the problem at hand: Mo2men wrote: In the first 1000 positive integers, how many integers exist such that they leave a remainder 4 when divided by 7, and a remainder 9 when divided by 11? A. 10 B. 11 C. 12 D. 13 E. 14 When x is divided by 7, the remainder is 4. x = 7a + 4 = 4, 11, 18, 25, 32, 39, 46, 53... When x is divided by 11, the remainder is 9. x = 11b + 9 = 9, 20, 31, 42, 53... Thus, when x is divided by 77 -- the LCM of 7 and 11 -- the remainder will be 53 (the smallest value common to both lists). x = 77c + 53 = 53, 130, 207... In the resulting list, the smallest value is yielded when c=0: 77*0 + 53 = 53. The greatest value less than 1001 is yielded when c=12: 77*12 + 53 = 977. Implication: c can be any integer between 0 and 12, inclusive, yielding a total of 13 options. The correct answer is D. _________________ Mitch Hunt Private Tutor for the GMAT and GRE GMATGuruNY@gmail.com If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon. Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. Student Review #1 Student Review #2 Student Review #3 Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now. ### GMAT/MBA Expert GMAT Instructor Joined 12 Sep 2012 Posted: 2636 messages Followed by: 114 members Upvotes: 625 Target GMAT Score: V51 GMAT Score: 780 In a pinch, another approach is to find two numbers that behave like this, then see what they have in common and generalize from there. If x leaves a remainder of 4 when divided by 7, we can say x = 7m + 4 where m is some integer whose value we don't care about. Similarly, if x leaves a remainder of 9 when divided by 11, we can say x = 11n + 9 where n is some integer whose value we don't care about. Since x = x, we also have 7m + 4 = 11n + 9, or 7m = 11n + 5. This suggests that we're looking for a multiple of 7 that is 5 greater than a multiple of 11. Checking the multiples of 11, plus 5, we find 16, 27, 38, 49, 60, 71, 82, 93, 104, 115, 126, ... Having found two of these, we can generalize that each solution is likely to be (126 - 49) units apart, or 77 units apart. (That should make sense, since 77 = 7*11). That means that our initial solutions will all be of the form 77*(some integer) + 49, meaning that we have 77*0 + 49 77*1 + 49 77*2 + 49 ... and so on, with our ceiling coming at 1000. 77*13 = 1001, so that's too big, meaning we must only go to 77*12 + 49. Since we started at 77*0 and made it to 77*12, we have 0 to 12, or 13 solutions, and we're set. To find the actual answers, add 4 to each, since we began by subtracting 4 to simplify. (In other words, our solution set becomes 49+4, 126+4, etc.) This seems like a little more remainder theory than the GMAT expects of you, however. Enroll in a Veritas Prep GMAT class completely for FREE. Wondering if a GMAT course is right for you? Attend the first class session of an actual GMAT course, either in-person or live online, and see for yourself why so many students choose to work with Veritas Prep. Find a class now! Master | Next Rank: 500 Posts Joined 13 Mar 2015 Posted: 410 messages Followed by: 7 members Upvotes: 120 GMAT Score: 770 Mo2men wrote: In the first 1000 positive integers, how many integers exist such that they leave a remainder 4 when divided by 7, and a remainder 9 when divided by 11? A. 10 B. 11 C. 12 D. 13 E. 14 OA: D What is the best short cut to this problem? remainder 4 when divided by 7 - Numbers can be 4, 11, 18, 25, 32, 39, 46, 53, ... remainder 9 when divided by 11 - Numbers can be 9,20, 31, 42, 53, ... Common numbers less than 1000 will be 53, 77*1 + 53, 77*2 + 53, ...., 77*12 + 53 Hence a total of 13 numbers. Correct Option: D ### GMAT/MBA Expert GMAT Instructor Joined 25 Apr 2015 Posted: 1808 messages Followed by: 14 members Upvotes: 43 Mo2men wrote: In the first 1000 positive integers, how many integers exist such that they leave a remainder 4 when divided by 7, and a remainder 9 when divided by 11? A. 10 B. 11 C. 12 D. 13 E. 14 We need to find the smallest integer that satisfies both conditions: Numbers that leave a remainder of 4 when divided by 7 are: 4, 11, 18, 25, 32, 39, 46, 53, ... Numbers that leave a remainder of 9 when divided by 11 are: 20, 31, 42, 53, â€¦ We see that 53 is the first number that satisfies both conditions. To find the subsequent numbers that also satisfy both conditions we keep adding the LCM of 7 and 11, i.e., 77, to (and beginning with) 53. So the numbers, including 53, are: 53, 130, 207, 284, 361, 438, 515, 592, 669, 746, 823, 900, and 977 So there are a total of 13 such numbers. Answer: D _________________ Scott Woodbury-Stewart Founder and CEO • Get 300+ Practice Questions 25 Video lessons and 6 Webinars for FREE Available with Beat the GMAT members only code • Award-winning private GMAT tutoring Register now and save up to$200

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