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Of the three digits greater than 700, how many have two digi
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20 Nov 2009, 11:07
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56% (02:39) correct 44% (02:36) wrong based on 762 sessions
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Of the three digits greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two A) 90 B) 82 C) 80 D) 45 E) 36 Is there a faster method to find out the answer, because i listed down all the numbers and it took me hardly 1.5 mins. but what if there are more numbers ? OPEN DISCUSSION OF THIS QUESTION IS HERE: http://gmatclub.com/forum/ofthethree ... 35188.html
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Re: Of the three digits greater than 700, how many have two digi
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20 Nov 2009, 11:53
This can be solved in another way and maybe it's faster: Numbers between 700 and 999 =299; Numbers with all distinct digits = 3*9*8=216; Numbers with all the same digits more than 700 =3, (777, 888, 999); Total: 2992163=80.
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Re: Of the three digits greater than 700, how many have two digi
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20 Nov 2009, 11:28
gurpreet07 wrote: Of the three digits greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two
A) 90 B) 82 C) 80 D) 45 E) 36
Is there a faster method to find out the answer, because i listed down all the numbers and it took me hardly 1.5 mins. but what if there are more numbers ? Step by step: The number can have three forms: XXY meaning that the first digit is the repeated digit; XYX meaning that the first digit is repeated digit; and YXX meaning that the first digit is not repeated digit. XXY > the repeated digits can be 7,8 or 9, so 3 possibilities and the third one can take 9 possible values total 3*9=27. The same with XYX =3*9=27; YXX > the non repeated first digit can be 7, 8, 9, so 3 possibilities and the repeated digits can take 9 possible values, total 3*9=27. But this will give us the number 700 as well and we know that the number should be more than 700, hence 271=26. TOTAL=27+27+26=80 Answer: C. Hope it helps.
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Re: Of the three digits greater than 700, how many have two digi
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20 Nov 2009, 11:43
Bunuel wrote: gurpreet07 wrote: Of the three digits greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two
A) 90 B) 82 C) 80 D) 45 E) 36
Is there a faster method to find out the answer, because i listed down all the numbers and it took me hardly 1.5 mins. but what if there are more numbers ? Step by step: The number can have three forms: XXY meaning that the first digit is the repeated digit; XYX meaning that the first digit is repeated digit; and YXX meaning that the first digit is not repeated digit. XXY > the repeated digits can be 7,8 or 9, so 3 possibilities and the third one can take 9 possible values total 3*9=27. The same with XYX =3*9=27; YXX > the non repeated first digit can be 7, 8, 9, so 3 possibilities and the repeated digits can take 9 possible values, total 3*9=27. But this will give us the number 700 as well and we know that the number should be more than 700, hence 271=26. TOTAL=27+27+26=80 Answer: C. Hope it helps. thnku....very fast and a efficient method.



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Re: Of the three digits greater than 700, how many have two digi
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20 Nov 2009, 12:07
Bunuel wrote: This can be solved in another way and maybe it's faster:
Numbers between 700 and 999 =299;
Numbers with all distinct digits = 3*9*8=216;
Numbers with all the same digits more than 700 =3, (777, 888, 999);
Total: 2992163=80. Hey Bunuel thanks man...from where do you get such idea's just awesome



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Re: Of the three digits greater than 700, how many have two digi
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27 Mar 2010, 18:29
Bunuel,
I don't understand how there is only 27 numbers based on your calculation. For example, XXY the first number has to be 7, 8 or 9 so 3 possibilities and the next number would also have to be 7, 8 or 9 but the last number could be 9 values but how is that equal to 3*9? Wouldn't that be 3*3*9? I'm don't understand.



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Re: Of the three digits greater than 700, how many have two digi
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10 Apr 2010, 04:55
adamsmith2010 wrote: Bunuel,
I don't understand how there is only 27 numbers based on your calculation. For example, XXY the first number has to be 7, 8 or 9 so 3 possibilities and the next number would also have to be 7, 8 or 9 but the last number could be 9 values but how is that equal to 3*9? Wouldn't that be 3*3*9? I'm don't understand. Adam in case of XXY, first digit is 7,8 or 9 then so 3 ways 2nd digit has be same as 1st so only one way you can also write it as 3*1*9 = 27 Hope it helps
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Re: Of the three digits greater than 700, how many have two digi
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14 Apr 2010, 03:10
How have you guys become so confident, predicting : 700999 XXY setup = t occurrences without a single calculation line etc... Isnt there any mathematical way of solving it.
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Re: Of the three digits greater than 700, how many have two digi
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14 Apr 2010, 06:02
wow Bunuel ... I tried to formulate a pattern and reach the answer. But it seems I missed something ...well ....



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Re: Of the three digits greater than 700, how many have two digi
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14 Apr 2010, 10:24
shrivastavarohit wrote: How have you guys become so confident, predicting :
700999
XXY setup = t occurrences without a single calculation line etc...
Isnt there any mathematical way of solving it. This is how I solved it: Three digit number can have only following 3 patterns: A. all digits are distinct; B. two are alike and third is different; C. all three digits are alike. We need to calculate B. B=Total  A  C Total numbers from 700 to 999 = 299. A. all digits are distinct = 3(first digit can have only three values 7,8, or 9)*9*8=216; C. all three are alike = 3 (777, 888, 999) So, 2992163=80. Another way to solve this problem is in my first post. Hope it's clear.
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Re: Of the three digits greater than 700, how many have two digi
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19 Jun 2010, 22:30
I am getting 78 as the answer. So I went with closest answer 80 (C).
This is how I arrived @ 78:
For numbers greater than 700 we have following possibilities:
a=> 7xx .. Here x being any digit 09, excluding 7 and 0 {0 is excluded because number should be >700} b=> 7x7 .. Here x can be any digit from 09, excluding 7. {no 3 digits should be same} c=> 77x .. Here x can be any digit from 09, excluding 7 for same reason as above.
a + b + c = 8 + 9 + 9 = 26
Hence same is true with 8xx , 8x8, 88x and 9xx,9x9, 99x ...
So summing up I got 78 ...
Please help me if I am wrong here. I am unable to figure out where did I miss 2 more numbers. Any pointers ?



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Re: Of the three digits greater than 700, how many have two digi
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19 Jun 2010, 23:31
for 700 ==> 711 , 722 , ...... , 799 (except 777) = 8 ,707, 717 , 727 , .... , 797 = 9 770, 771 , 772 , ........ , 779 = 9 total = 26
similarly for 800 and 900 series you will have 26 each
so 26 * 3 = 78
and 800 and 900 so 78 + 2 = 80



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Re: Of the three digits greater than 700, how many have two digi
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20 Jun 2010, 10:45
Bunuel wrote: gurpreet07 wrote: Of the three digits greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two
A) 90 B) 82 C) 80 D) 45 E) 36
Is there a faster method to find out the answer, because i listed down all the numbers and it took me hardly 1.5 mins. but what if there are more numbers ? Step by step: The number can have three forms: XXY meaning that the first digit is the repeated digit; XYX meaning that the first digit is repeated digit; and YXX meaning that the first digit is not repeated digit. XXY > the repeated digits can be 7,8 or 9, so 3 possibilities and the third one can take 9 possible values total 3*9=27. The same with XYX =3*9=27; YXX > the non repeated first digit can be 7, 8, 9, so 3 possibilities and the repeated digits can take 9 possible values, total 3*9=27. But this will give us the number 700 as well and we know that the number should be more than 700, hence 271=26. TOTAL=27+27+26=80 Answer: C. Hope it helps. thanks! this is an efficient way to solve it. is this method feasible for this kind of questionsfundamental counting?



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Re: Of the three digits greater than 700, how many have two digi
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20 Jun 2010, 10:59
Bunuel wrote: Hope it helps. tt11234 wrote: thanks! this is an efficient way to solve it. is this method feasible for this kind of questionsfundamental counting?
I don't really understand your question... Also there is another, easier way to solve this question, shown in my second post: 3digit numbers more than 700 = 299; Numbers with all distinct digits = 3*9*8=216 (first digit can take 3 values  7, 8, 9; second digit can take 9 values and the third digit 8 values); Numbers with all the same digits more than 700 = 3, (777, 888, 999); {all}  {all distinct}  {all same} = {two equal} > 2992163=80. Answer: C.
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Re: Of the three digits greater than 700, how many have two digi
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30 Jun 2010, 17:41
Thanks Bunuel for the effort but I dont really understand why you choose 299, and what is the meaning of distinct, and why is there 3, 9 , 8? Pleas explaing that to me thanks again



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Re: Of the three digits greater than 700, how many have two digi
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30 Jun 2010, 19:44
xmagedo wrote: Thanks Bunuel for the effort but I dont really understand why you choose 299, and what is the meaning of distinct, and why is there 3, 9 , 8? Pleas explaing that to me thanks again The question asks about threedigit integers greater than 700. So, all the numbers in the range 701 to 999, inclusive. There will be 999  701 + 1 = 299 such numbers in this range. (Whenever you want to count the number of objects in a range, subtract the smallest number from the greatest and add 1. For example, the number of integers in the range of 1 to 6 is 6 1 +1 = 6). "distinct" just means different. So, for example, in the number "799", the final two digits are NOT distinct. Here: 798, they are.



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Re: Of the three digits greater than 700, how many have two digi
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21 Jan 2012, 20:06
This is the answer posted:
Numbers between 700 and 999 =299;
Numbers with all distinct digits = 3*9*8=216;
Numbers with all the same digits more than 700 =3, (777, 888, 999);
Total: 2992163=80.
I've a doubt. Numbers with all distinct digits = 3*9*8=216;
I feel 9 above is not right because if we take one of the 3 digits (7,8 or 9), then we shouldn't take repeated values (how is 9 possible).



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Re: Of the three digits greater than 700, how many have two digi
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21 Jan 2012, 20:07
Ok, I got it I forgot to take from 09 .. 10 digits 101=9



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Re: Of the three digits greater than 700, how many have two digi
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01 Feb 2014, 20:42
We can fix 2 digits in 3 ways ie 7,8,9 and the third digit can be fixed in 9 ways(10 digits1 repeating digit) Total=3*9=27 Since this is possible in 3 different ways(first digit diff from other two,second digit diff from other two and third digit diff from other two) we get 3* 27=81 but we subtract 1 for 700 Ans 80
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Re: Of the three digits greater than 700, how many have two digi
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16 Jun 2016, 11:55
Repeating the explanation from above, the sequence of nos. should be XXY, XYX, YXX. As we need 3 digit nos. greater that 700, the 100th place in can be either of 7, 8 or 9 (3 ways to select 100th place digit)
Once 100th place if fixed, say units place is same as hundred place (XYX). this means that the tens place can be picked in 9 ways (all digits from 09 except the place in hundred place) and unit place can be picked in 1 way only (same as hundred place). So total ways for XYX is 3*9*1=27
We need to repeat this for XXY format. This would again generate 3*1*9 ways = 27
We need to repeat this now with YXX which will give 3.9.1 ways = 27. But we need to exclude 700 from this as we need nos greater than 700. This this combination will give 271=26 ways
Total ways = 27+27+26=80




Re: Of the three digits greater than 700, how many have two digi
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