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EMPOWERgmat Instructor
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Re: How many times will the digit 7 be written when listing the integers
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22 Jan 2018, 16:08
Hi All, There are a couple of different way to conceive of this question, and each has its own 'organization' to it, so you should try to think in whatever terms are easiest for you. To start, it shouldn't be hard to figure out how many 3DIGIT numbers will START with 7... 700 to 799 inclusive... so that's 100 appearances of a 7 right there. Next, you might find it easiest to think about the UNITs DIGIT. Consider the following pattern... 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 etc Notice how 1 out of every 10 numbers has a unit's digit of 7? That pattern repeats overandover. Since we have 1000 total consecutive integers in our list, 1/10 of them will have a unit's digit of 7... (1000)(1/10) = 100 appears of a 7... Now, think about what you've seen so far... a group of 100 and another group of 100. I wonder what will happen when we deal with the TENS DIGITS... 10 20 30 40 50 60 70 80 90 100 __ 71 72 73 74 75 76 7_ 78 79 80 In the first 100 integers, there are 10 additional 7s in the TENS 'spot' (I removed 70 and the units digit 7 from 77 since I already counted those). There are 10 sets of 100 to consider, so there are (10)(10) = 100 additional 7s... Total = 100 + 100 + 100 = 300 Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: How many times will the digit 7 be written when listing the integers
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10 Mar 2018, 06:24
seekmba wrote: How many times will the digit 7 be written when listing the integers from 1 to 1000?
(A) 110 (B) 111 (C) 271 (D) 300 (E) 304 _ _ _ 1. When all are 7, we have 1 combination i.e. 7 7 7 and since 7 appears 3 times we have 1*3= 32. When we have 2 7s, 7 7 _, we can fill the dash with 9 remaining digits and these can be further arranged in 3 ways. Total ways = 9*3=27; but 7 appears 2 times in each combination; thus we have 27*2 = 543. When we have 1 7; 7 _ _ , we can fill the remaining dashes with remainng 9 digits. Total ways = 9*9*3= 243Thus Final Total = 243+54+3 = 300
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Re: How many times will the digit 7 be written when listing the integers
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12 Mar 2018, 08:26
seekmba wrote: How many times will the digit 7 be written when listing the integers from 1 to 1000?
(A) 110 (B) 111 (C) 271 (D) 300 (E) 304 Here's one way to look at it. Write all of the numbers as 3digit 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 Answer: D Cheers, Bren
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Re: How many times will the digit 7 be written when listing the integers
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11 Sep 2018, 17:21
seekmba wrote: How many times will the digit 7 be written when listing the integers from 1 to 1000?
(A) 110 (B) 111 (C) 271 (D) 300 (E) 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, …, 970s). As the units digit, it appears 100 times also (7, 17, 27, …, 997). Therefore, the digit 7 has appeared a total of 300 times. Answer: D
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Re: How many times will the digit 7 be written when listing the integers
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30 Nov 2018, 03:58
Solution using permutation combination:
Digits needed to form numbers: 0,1,2,3,4,5,6,7,8,9 (10 digits)
Case 1: 7 comes in exactly one position:
7 _ _ , _ 7 _ and _ _ 7 For the above three sub cases, the number of possibilities are (1 x 9 x 9, 9 x 1 x 9 and 9 x 9 x 1) which is 81 for each sub case. Total number of 7s : 3 x 81 = 243 Note: (a) 9 possibilities (09, excluding 7) (b) This covers 2digit 7numbers as well since we have included 0.
Case 2: 7 comes in exactly 2 positions:
77_ , 7_7 and _77 For the above sub cases the number of possibilities are: (1x1x9 , 1x9x1 , 9x1x1) Each sub case has 9 possibilities. But the question asks about the number of 7s so each sub case will have 2x9 = 18 possibilities. Total number of 7s = 3x18 = 54
Case 3: 7 comes in all places:
777 : 1 possibility ; Total number of 7s=3
Adding 243 +54 +3 = 300
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Re: How many times will the digit 7 be written when listing the integers
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01 Nov 2019, 01:08
How many times will the digit 7 be written when listing the integers from 1 to 1000? (A) 110 (B) 111 (C) 271 (D) 300 (E) 304 Solution: 1000 will be ruled out so we'll just consider 3 digits, 2 digits and 1 digit. __ , __, __ > this is our slots to fill if we have one 7 then 7(any)(any) and any can be 0,1,2,3,4,5,6,8,9 ==> 9 possibilities. Therefore, total possible combinations for 1 occurence of 7 is 1x9x9x3 [7()(), ()7() , ()()7] if we have two 7 then 7(7)(any) and any can be 0,1,2,3,4,5,6,8,9 ==> 9 possibilities. Therefore, total possible combinations for 2 occurence of 7 is 1x9x3x2 [77() 7()7 ()77] 2 because two occurance of 7. So 7 will be written twice in this case if we have three 7 then 777 that's just 1 possibility and 7 is written thrice so 1x3 Answer = 1x9x9x3 + 1x9x3x2 + 1x3 = 300 (D) Bunuel is my approach correct? I haven't looked at any solution. Solved this under 90 seconds




Re: How many times will the digit 7 be written when listing the integers
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01 Nov 2019, 01:08



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