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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 ## How many 7s tagged by: Brent@GMATPrepNow ##### This topic has 4 expert replies and 9 member replies ## How many 7s How many times will the digit 7 be written when listing the integers from 1 to 1000? 1. 110 2. 111 3. 217 4. 300 5. 304 Legendary Member Joined 15 Jun 2010 Posted: 543 messages Followed by: 3 members Upvotes: 147 anuptvm wrote: How many times will the digit 7 be written when listing the integers from 1 to 1000? 1. 110 2. 111 3. 217 4. 300 5. 304 All the numbers (0,1,2,...9) will be appearing equal number of times. The numbers are : 000 001 002 ..... ..... 999 So, there are a total of 3*1000 digits. Now the number of times 7 will appear (or in fact any number 0,1,2,...9) will appear is = 3*1000/10 = 300. Answer is 4. _________________ Thanks Anshu (Every mistake is a lesson learned ) Senior | Next Rank: 100 Posts Joined 12 Aug 2009 Posted: 52 messages Followed by: 1 members Test Date: 12/30/2010 Target GMAT Score: 690 GMAT Score: 650 Thanks Anshumishra. Is there another method using Permutations? What is a good general strategy for such problems? Legendary Member Joined 15 Jun 2010 Posted: 543 messages Followed by: 3 members Upvotes: 147 anuptvm wrote: Thanks Anshumishra. Is there another method using Permutations? What is a good general strategy for such problems? Here is another method (using Combinations, which I read on one of the blogs, and modified for the current question) :- Since 7 does not occur in 1000, we need to count the number of times 7 occurs when we list the integers from 1 to 999. Now any integer between 1 and 999 is of the form abc, where 0 â‰¤ a, b, c â‰¤ 9. We first calculate the number of integers in which 7 occurs exactly once. Now 7 can occur at one place in 3C1 ways and the other 2 places can be occupied by any other 2 digits from 0 to 9 except 7. Hence number of such integers is 3C1 X 9 X 9 = 243. Since 7 occurs exactly once in each of them, there will be 243 listings of 7 in such integers. We next calculate the number of integers in which 7 occurs exactly twice. The two places in which 7 occurs can be selected in 3C2 ways. The remaining 1 place can be occupied by any other digit from 0 to 9 except 7. Hence number of such integers is 3C2 X 9 or 27. Since 7 occurs exactly twice in each of them, there will be 2 X 27 = 54 listings of 7 in such integers. Lastly we calculate the number of integers in which 7 occurs exactly once. Now there will be only 1 such integer which is 777 and there are 3 listings of 7 in this integer. Hence, the total number of times 7 occurs in all integers from 1 to 1000 is 243 + 54 + 3 = 300. My strategy is you must know the basic way to solve the problem (like the one mentioned here), apart from that if you can find some symmetries that would be awesome (But make sure that the symmetry is there, like it is in current case, that is why I could use the first method as shown in my previous post). Hope that helps ! _________________ Thanks Anshu (Every mistake is a lesson learned ) Master | Next Rank: 500 Posts Joined 26 Nov 2010 Posted: 139 messages Followed by: 1 members Upvotes: 20 Test Date: Feb 3, 2011 GMAT Score: 720 anuptvm wrote: Thanks Anshumishra. Is there another method using Permutations? What is a good general strategy for such problems? Anshu's solution is more elegant. You could also solve the problem using permutations as follows total count = count in single digit number (1-9) + count in two digit number(10-99) + count in 3 digit number(100-999) count in single digit = 1 count in two digit = 7x + y7 = 1x10(because x can be any number from 0 to 9) + 9(y could be any number from 1 to 9)x1 = 10+9 = 19 count in 3 digit = 7ab + c7d + ef7 = 1x10x10 ( both a and b could be any of the numbers from 0 to 9) + 9x1x10 (c could be any number from 1 to 9, d could be any number from 0 to 9) + 9x10x1 (e could be 1 to 9, f could be from 0 to 9) = 100+90+90=280 Total count = 1+19+280 = 300 Legendary Member Joined 15 Jun 2010 Posted: 543 messages Followed by: 3 members Upvotes: 147 stormier wrote: anuptvm wrote: Thanks Anshumishra. Is there another method using Permutations? What is a good general strategy for such problems? Anshu's solution is more elegant. You could also solve the problem using permutations as follows total count = count in single digit number (1-9) + count in two digit number(10-99) + count in 3 digit number(100-999) count in single digit = 1 count in two digit = 7x + y7 = 1x10(because x can be any number from 0 to 9) + 9(y could be any number from 1 to 9)x1 = 10+9 = 19 count in 3 digit = 7ab + c7d + ef7 = 1x10x10 ( both a and b could be any of the numbers from 0 to 9) + 9x1x10 (c could be any number from 1 to 9, d could be any number from 0 to 9) + 9x10x1 (e could be 1 to 9, f could be from 0 to 9) = 100+90+90=280 Total count = 1+19+280 = 300 Nice work stormier ! I liked the part where you used 3 different colors to make it easier to understand ! _________________ Thanks Anshu (Every mistake is a lesson learned ) Junior | Next Rank: 30 Posts Joined 03 Jun 2011 Posted: 23 messages Followed by: 1 members Upvotes: 8 Test Date: June 2009 GMAT Score: 750 There's a slightly easier way to look at it if we consider the digit positions. The number of times 7 occurs in the hundreds' position is 100. (From 700 to 799). The number of times 7 occurs in the tens' position is 100. (70-79, 170-179,...,970-979). The number of times 7 occurs in the units' position is 100. (7,17,27,...107,....,997). 3*100 = 300. _________________ Deirdre at testprepdublin.com Newbie | Next Rank: 10 Posts Joined 18 Jul 2016 Posted: 1 messages anshumishra wrote: anuptvm wrote: How many times will the digit 7 be written when listing the integers from 1 to 1000? 1. 110 2. 111 3. 217 4. 300 5. 304 All the numbers (0,1,2,...9) will be appearing equal number of times. The numbers are : 000 001 002 ..... ..... 999 So, there are a total of 3*1000 digits. Now the number of times 7 will appear (or in fact any number 0,1,2,...9) will appear is = 3*1000/10 = 300. Answer is 4. There are not 3000 digits from 1 - 1000. It should be 9 + 180 + 2700 + 4 ### GMAT/MBA Expert GMAT Instructor Joined 08 Dec 2008 Posted: 12831 messages Followed by: 1247 members Upvotes: 5254 GMAT Score: 770 anuptvm wrote: How many times will the digit 7 be written when listing the integers from 1 to 1000? A. 110 B. 111 C. 217 D. 300 E. 304 Here's another way to look at it. Write all of the numbers as 3-digit numbers. That is, 000, 001, 002, 003, .... 998, 999 NOTE: Yes, I have started at 000 and ended at 999, even though though the question asks us to look at the numbers from 1 to 1000. HOWEVER, notice that 000 and 1000 do not have any 7's so the outcome will be the same. First, there are 1000 integers from 000 to 999 There are 3 digits in each integer. So, there is a TOTAL of 3000 individual digit. (since 1000 x 3 = 3000) Each of the 10 digits is equally represented, so the 7 will account for 1/10 of all digits. 1/10 of 3000 = 300 So, there are 300 0's, 300 1's, 300 2's, 300 3's, . . ., and 300 9's in the integers from 000 to 999 Cheers, Brent _________________ Brent Hanneson â€“ Creator of GMATPrepNow.com Use our video course along with Sign up for our free Question of the Day emails And check out all of our free resources GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMATâ€™s FREE 60-Day Study Guide and reach your target score in 2 months! ### GMAT/MBA Expert Elite Legendary Member Joined 23 Jun 2013 Posted: 10129 messages Followed by: 494 members Upvotes: 2867 GMAT Score: 800 Hi All, If you find these types of questions to be challenging, then you might find it helpful to 'break down' the prompt into logical pieces... To start, you know that all of the 700s (700 through 799) will have a 7 in the 'hundreds spot', so that's 100 sevens right there. Next, you should be able to figure out that one out of every 10 numbers will end in a 7. List them out and you'll see... 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Etc. Since we're dealing with the numbers from 1 to 1000, we know that we'll have 100 'groups of 10.' One out of every 10 of those numbers will END in a 7, so that's another (100)(1) = 100 sevens. Now we just have to consider the "tens digits.' In the first 100 integers, there will be 10 numbers that have a 7 in that 'spot': 70 71 72 73 74 75 76 77 78 79 That pattern will occur in each 100 digits... 170 171 172 173 174 175 176 177 178 179 Etc. Since there are 10 'groups of 100' and each group has 10 sevens in the "tens digits", that's another (10)(10) = 100 sevens. Total = 100 + 100 + 100 = 300 Final Answer: D GMAT assassins aren't born, they're made, Rich _________________ Contact Rich at Rich.C@empowergmat.com ### GMAT/MBA Expert GMAT Instructor Joined 12 Sep 2012 Posted: 2635 messages Followed by: 116 members Upvotes: 625 Target GMAT Score: V51 GMAT Score: 780 jeffreyut wrote: There are not 3000 digits from 1 - 1000. It should be 9 + 180 + 2700 + 4 Anshumishra is treating each number from 0 to 999 as having leading zeros - a clever approach - so there are indeed 3000 digits to work with: 0 is 000, 1 is 001, etc. 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! Newbie | Next Rank: 10 Posts Joined 11 Sep 2017 Posted: 1 messages anuptvm wrote: How many times will the digit 7 be written when listing the integers from 1 to 1000? 1. 110 2. 111 3. 217 4. 300 5. 304 Newbie | Next Rank: 10 Posts Joined 18 Dec 2013 Posted: 1 messages I took the following approach but got the answer wrong: first I divided the numbers into ten equal groups of 100 digits each 0-99 100-199 and so on within each of those groups, the number 7 will appear exactly 20x times: 10x in the first position (7, 17, 27, ...) and 10x in the second position (70, 71, 72, ...) For numbers 700-799, the digital 7 will appear exactly 100 times. I then just added everything up: 0-99: 20x 100-199: 20x 200-299: 20x 300-399: 20x 400-499: 20x 500-599: 20x 600-699: 20x 700-799: 100x 800-899: 20x 900-999: 20x The result ist 280x. Where's my mistake? ### GMAT/MBA Expert GMAT Instructor Joined 25 Apr 2015 Posted: 2424 messages Followed by: 18 members Upvotes: 43 anuptvm wrote: How many times will the digit 7 be written when listing the integers from 1 to 1000? 1. 110 2. 111 3. 217 4. 300 5. 304 The digit 7 as the hundreds digit appears 100 times (700 to 799). As the tens digit, it appears 100 times also (ten times each in the 70s, 170s, 270s, â€¦, 970s). As the units digit, it appears 100 times also (7, 17, 27, â€¦, 997). Therefore, the digit 7 has appeared a total of 300 times. Answer: D/4 _________________ Scott Woodbury-Stewart Founder and CEO scott@targettestprep.com See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. 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