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

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11 Feb 2012, 17:04

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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?

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.

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

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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 double-checking to make sure.... thanks!

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

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04 Oct 2012, 15:40

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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 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. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

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

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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

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

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28 Dec 2013, 09:23

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

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30 Dec 2014, 11:42

Hello from the GMAT Club BumpBot!

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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").

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)

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

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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...

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

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26 Sep 2015, 09:54

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!

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.

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