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Of the threedigit integers greater than 600, how many have
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11 Feb 2012, 17:04
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Of the threedigit integers greater than 600, how many have two digits that are equal to each other and the remaining digit different from the other two? (A) 120 (B) 116 (C) 108 (D) 107 (E) 72
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Re: Of the threedigit integers greater than 600, how many have
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11 Feb 2012, 17:10
enigma123 wrote: Of the threedigit integers greater than 600, how many have two digits that are equal to each other and the remaining digit different from the other two? (A) 120 (B) 116 (C) 108 (D) 107 (E) 72
I always struggle to solve these. What is the concept behind solving these questions? # of threedigit integers greater than 600 is 399 (999600); Numbers with all distinct digits: 4*9*8=288. First digit can take 4 values: 6, 7, 8 or 9, second digit can take 9 values (10 minus the one we used for the first digit) and third digit can tale 8 values (10 minus 2 digits we've already used for the first two digits); Numbers greater than 600 which have all alike digits: 4 (666, 777, 888, 999); {Total}{all distinct}{all alike}={two alike, one different} > 3992884=107. Answer: D.
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Re: Of the threedigit integers greater than 600, how many have
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11 Feb 2012, 17:22
Thanks for very clear and precise explanation Bunuel. Highly appreciate.
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Re: Of the threedigit integers greater than 600, how many have
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04 Oct 2012, 14:05
If the problem asks for numbers greater than 600, don't we have to start counting from 601 forward? I think we don't, but just doublechecking to make sure.... thanks!



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Re: Of the threedigit integers greater than 600, how many have
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04 Oct 2012, 15:40
enigma123 wrote: Of the threedigit integers greater than 600, how many have two digits that are equal to each other and the remaining digit different from the other two? (A) 120 (B) 116 (C) 108 (D) 107 (E) 72
I always struggle to solve these. What is the concept behind solving these questions? The question is about counting/combinations. The numbers can be of the form \(AAB, \,ABA,\) or \(ABB\) where \(A\) and \(B\) are different digits and \(A\geq6\). There will be \(4\cdot9\cdot3=108\) possibilities, but we have to subtract \(1\) for the number \(600\) (which is obtained for \(A = 6\) and \(B = 0\)). Therefore, total number of possibilities \(108  1= 107.\) Answer D.
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Re: Of the threedigit integers greater than 600, how many have
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29 Oct 2012, 16:04
Bunuel, thanks for the explanation. Mine looks more complicated, but could you please check if my solution is correct.
1) yxx  first digit can have 4 values, second  9, third  1 = 36 2) yyx  first digit  4, second  1, third  9 = 36 3) yxy  first digit  4, second 9, third  1 = 36
1)+2)+3) = 108  1*= 107 1*  we should exclude 600



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Re: Of the threedigit integers greater than 600, how many have
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30 Oct 2012, 14:28
Ev,
what you did was basically the same as what evajager did.
Let me clarify.
Three digit number: 6xy
First digit: either 6, 7, 8, or 9
So out of 4 possible digits, you choose 1. (4C1)
Either 2nd or 3rd digit must match the first digit > Only 1 possibility so (1C1)
Then the remaining last digit (opposite of the you chose above) can have 9 remaining digits to choose from. So (9C1).
Of course, out of 3 available spots, we have to pick 2 of them to be the same (3C2).
So multiply them together:
= # of ways to arrange a pair within 3 slots * [ digit #1 * pair digit * remaining digit] = (3C2) [ (4C1) * (1C1) * (9C1) = 36 * 3 = 108
Since the question technically asks for numbers >600, the exact value of 600 (which was included) should not be included.
So we subtract 108  1 = 107



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Re: Of the threedigit integers greater than 600, how many have
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30 Dec 2014, 21:54
Hi All, In these types of questions, the real issue is thoroughness  make sure that you're not "missing" any of the possibilities and make sure that you're not "counting" a possibility that should NOT be counted (or accidentally counting a possibility more than once). Your ability to patternmatch will help speed you up. With the limitations posed by this question, we COULD break the numbers down into smaller groups and then total up all of these smaller numbers (it's a slightly longer way to do things, but if you don't immediately see the more complex calculations, you can still get to the correct answer with a bit of "hand math"). Let's start with making the first 2 digits the same... 66_ 77_ 88_ 99_ Since the third digit has to be DIFFERENT from the matching pair, we have 9 options for each of the 4 groups above (you CAN'T count 666, 777, 888 or 999  the numbers don't fit the restrictions). Total of this group = 36 Next, let's make the first and third digits the same... 6_6 7_7 8_8 9_9 Here, we have a similar situation to the one we had above; we have 9 options for each of the 4 groups (you CAN'T count 666, 777, 888 or 999). Total of this group = 36 Finally, let's make the second and third digits the same (I'll refer to those digits with the variable X)... 6XX 7XX 8XX 9XX In this grouping, we have 1 "catch"  X can be any digit, BUT the number 600 is NOT permissible, since the prompt tells us for numbers GREATER THAN 600. So, 6XX has 8 possibilities (you CAN'T count 600 or 666) 7XX, 8XX and 9XX have 9 possibilities each (you CAN'T count 777, 888 or 999) Total of this group = 35 Overall total = 36 + 36 + 35 = 107 Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: Of the threedigit integers greater than 600, how many have
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10 Jan 2015, 14:09
Many very good approaches here! Good job everyone! I did it in the amature's way: So, I started calculating for 600, like this: 606 611 616 622  633  644  655  677  688  699 626  636  646  656  676  686  696 So, from here I only did 17*4= 68. By 4 because we are interested in 600, 700, 800 and 900. Then I added the numbers after 660, that I had left out so that I wouldn't get confused: 660, 661, 662, 663, 664, 665, 667, 668, 669. This is 9*4= 36. Adding 68+36= 104. I saw that I was missing 3 from answer D. So, I realised that I didn't add the 600 (not allowed), 700, 800, 900. So, 104+3 = 107. It took a little more that 2 minutes, but again less than 2 1/2... You can't learn everything at the same time (unfortunately), so this solution kept me satisfied...



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Re: Of the threedigit integers greater than 600, how many have
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26 Sep 2015, 09:54
Bunuel wrote: enigma123 wrote: Of the threedigit integers greater than 600, how many have two digits that are equal to each other and the remaining digit different from the other two? (A) 120 (B) 116 (C) 108 (D) 107 (E) 72
I always struggle to solve these. What is the concept behind solving these questions? # of threedigit integers greater than 600 is 399 (999600); Numbers with all distinct digits: 4*9*8=288. First digit can take 4 values: 6, 7, 8 or 9, second digit can take 9 values (10 minus the one we used for the first digit) and third digit can tale 8 values (10 minus 2 digits we've already used for the first two digits); Numbers greater than 600 which have all alike digits: 4 (666, 777, 888, 999); {Total}{all distinct}{all alike}={two alike, one different} > 3992884=107. Answer: D. Hi Bunuel, Can you please tell me what is wrong with my approach? Here is the way that i did: The hundreds digit can be 6,7,8 or 9 => 4C1 Pick one more value for either tens or unit digit: 9C1 Choose 1 of 2 values above to be the repeated digits: 2C1 The permutation of the 3digit integer: 3!/2! ==> 4C1*9C1*2C1*(3!/2!) = 4*9*2*3 = 216 216 1 = 215 (1 is 600) Thank you



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Re: Of the threedigit integers greater than 600, how many have
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26 Sep 2015, 14:56
Hi vihavivi, The big error in your calculation is in how you deal with the 'permutation'... You described it in this way: "The permutation of the 3digit integer: 3!/2!" However, this calculation includes options that are NOT allowed. For example, if you have the digits 4, 4 and 6, you could have 3 possible values: 446, 464 and 644. However, two of those values (446 and 464) are NOT greater than 600, so they should not be included. GMAT assassins aren't born, they're made, Rich
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Re: Of the threedigit integers greater than 600, how many have
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24 Jan 2016, 00:37
It works well for me, hope it helps u too:
600 700 800 900 abc 9 9 9 9 abc 8 9 9 9 abc 9 9 9 9
9*12=1081=107



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Re: Of the threedigit integers greater than 600, how many have
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25 Aug 2017, 03:18
I used bunuel's approach here.
Numbers between 600 & 999 =399. Numbers with distinct digits =6*9*8 =288 Numbers with all 3 digits alike =4
So Numbers with 2 digits alike =399(288+4) = 107



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Re: Of the threedigit integers greater than 600, how many have
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23 Sep 2019, 08:14
enigma123 wrote: Of the threedigit integers greater than 600, how many have two digits that are equal to each other and the remaining digit different from the other two?
(A) 120 (B) 116 (C) 108 (D) 107 (E) 72 SOLUTION 1 (CASES) \([011:different.pair]…4(6,7,8,9)•9•1•arrangements(3!/2!)=36•3=1081(#600)=107\) Answer (C) SOLUTION 2 (TOTALNOT) \([XXX:any.600999]…4•10•10=4001(#600)=399\) \([000:all.different]…4•9•8=4•72=288\) \([333:triple]…4•1•1=4\) \(totalnot=3992884=107\) Answer (C)




Re: Of the threedigit integers greater than 600, how many have
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