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madcowudub
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OG question. 17 Jul 2007, 02:40
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This a question in the OG.

Of the three-digit integers greather 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

i see the answer and explanation in the OG but still don't understand. please help. thanks



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GK_Gmat
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OG question. 17 Jul 2007, 02:52
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madcowudub
This a question in the OG.

Of the three-digit integers greather 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

i see the answer and explanation in the OG but still don't understand. please help. thanks
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Counting the numbers greater than 700 and less than 800, that have 2 digits that are equal and one digit different, I get 17.

We can multiply 17 by 3 (for 700, 800 and 900) to get the digits greater than 700 and less than 1000 with 2 digits that are equal and one digit different to give us 51. Since we haven't counted the numbers 800 and 900 (which have two 0's same and one digit different) we need to add 2 to 51 to get an answer of 53.

This is not even in the answer choices!! So obviously I'm going very wrong somewhere.
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[email protected]
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OG question. 17 Jul 2007, 02:52
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Is it 80.

If correct, I will explain.
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Finding Perdition
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OG question. 17 Jul 2007, 03:08
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Is it 80.

If correct, I will explain.
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How did you get 80?

FP
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OG question. 17 Jul 2007, 03:20
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Finding Perdition
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Is it 80.

If correct, I will explain.

How did you get 80?

FP
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First we count the numbers with the pattern, 77x , 88x and 99x .
Such numbers will be 9+9+9 = 27

Count the numbers with the pattern, 7x7 , 8x8 and 9x9 .
Such numbers will be 9+9+9 = 27

Count the numbers with pattern 7xx, 8xx and 9xx
Such numbers will be 8+8+8 = 24

Add two more numbers 700 and 800.
So numbers = 27+27+24+2 = 80
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madcowudub
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OG question. 17 Jul 2007, 06:56
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great reply! much better than the OG reply....is there any other way that would be faster than counting? such as using (numbers 701-999)-(numbers that don't contain two 7s)?
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OG question. 17 Jul 2007, 10:30
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Let me try a faster way

Lets find 7xx. We get 9 ways. So 8xx and 9xx will have 9 ways each so 27
Remember x is not equal to 7, 8, 9 respectively

Lets find 77x We get 9 ways. Similary 88x and 99x so total is 27
Remember x is not equal to 7, 8, 9 respectively

Let find 7x7. We get 9 Similarly for 8x8 and 9x9. we get total of 27
Remember x is not equal to 7, 8, 9 respectively

Now rememeber to do away with 700 in step as the stem says > 700.

So it is 81-1 = 80



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
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