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# Of the three-digit integers greater than 600, how many have

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Of the three-digit integers greater than 600, how many have [#permalink]

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11 Feb 2012, 16:04
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Of the three-digit 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?
[Reveal] Spoiler: OA

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11 Feb 2012, 16:10
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enigma123 wrote:
Of the three-digit 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 three-digit integers greater than 600 is 399 (999-600);

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} --> 399-288-4=107.

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Re: Of the three-digit integers greater than 600, how many have [#permalink]

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11 Feb 2012, 16:22
Thanks for very clear and precise explanation Bunuel. Highly appreciate.
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Re: Of the three-digit integers greater than 600, how many have [#permalink]

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04 Oct 2012, 13:05
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 double-checking to make sure.... thanks!
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Re: Of the three-digit integers greater than 600, how many have [#permalink]

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04 Oct 2012, 14:40
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enigma123 wrote:
Of the three-digit 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 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 $$108 - 1= 107.$$

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Re: Of the three-digit integers greater than 600, how many have [#permalink]

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29 Oct 2012, 15:04
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 three-digit integers greater than 600, how many have [#permalink]

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30 Oct 2012, 13:28
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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 three-digit integers greater than 600, how many have [#permalink]

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28 Dec 2013, 08:23
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Re: Of the three-digit integers greater than 600, how many have [#permalink]

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30 Dec 2014, 10:42
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Re: Of the three-digit integers greater than 600, how many have [#permalink]

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30 Dec 2014, 20:54
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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 pattern-match 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").

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

[Reveal] Spoiler:
D

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# Special Offer: Save $75 + GMAT Club Tests 60-point improvement guarantee www.empowergmat.com/ ***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*********************** Senior Manager Status: Math is psycho-logical Joined: 07 Apr 2014 Posts: 443 Location: Netherlands GMAT Date: 02-11-2015 WE: Psychology and Counseling (Other) Followers: 2 Kudos [?]: 104 [0], given: 169 Re: Of the three-digit integers greater than 600, how many have [#permalink] ### Show Tags 10 Jan 2015, 13:09 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... Intern Joined: 18 Aug 2012 Posts: 4 Followers: 0 Kudos [?]: 0 [0], given: 150 Re: Of the three-digit integers greater than 600, how many have [#permalink] ### Show Tags 26 Sep 2015, 08:54 Bunuel wrote: enigma123 wrote: Of the three-digit 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 three-digit integers greater than 600 is 399 (999-600); 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} --> 399-288-4=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 3-digit integer: 3!/2! ==> 4C1*9C1*2C1*(3!/2!) = 4*9*2*3 = 216 216 -1 = 215 (1 is 600) Thank you EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 8015 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Followers: 361 Kudos [?]: 2376 [0], given: 163 Of the three-digit integers greater than 600, how many have [#permalink] ### Show Tags 26 Sep 2015, 13:56 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 3-digit 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 _________________ # Rich Cohen Co-Founder & GMAT Assassin # Special Offer: Save$75 + GMAT Club Tests

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Re: Of the three-digit integers greater than 600, how many have [#permalink]

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23 Jan 2016, 23:37
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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=108-1=107
Re: Of the three-digit integers greater than 600, how many have   [#permalink] 23 Jan 2016, 23:37
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