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OG 11- Quant Sample- Question 11 [#permalink]
19 Mar 2007, 16:10

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Hello,

Would anybody have a formula/explanation for this:
"Of the three-digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two?"

This isn't a formula or a mathematical explanation but I would just go with the primitive method (I won't be able to come up with a formula in 2 minutes anyway... )

8xx where x = 0~7, 9 - 9 of them
8x8 where x = 0~7, 9 - 9 of them
88x where x = 0~7, 9 - 9 of them
total 27

since it would be the same for the others in 700s and 900s

27 x 3 = 81

however, since it has to be greater than 700, subtract the case 7xx where x = 0.

81 - 1 = 80

What is the OA?
(I might have missed out on something...)

Could you explain this more? I am not sure to get it

8xx where x = 0~7, 9 - 9 of them 8x8 where x = 0~7, 9 - 9 of them 88x where x = 0~7, 9 - 9 of them

8xx where x = 0~7, 9 - 9 of them -> Keep the 8 fixed and make the ten's and unit's digits the same. For example, 800, 811, 822, etc. However, you cannot have 8xx where x = 8 since then, there would be 3 (not 2) identical numbers.

8x8 where x = 0~7, 9 - 9 of them -> Keep the hundred's and unit's digits fixed and rotate the middle digits.

88x where x = 0~7, 9 - 9 of them -> Keep the first two digits fixed and rotate the last digits.

so every change in the tens digit, there are 2 possibilities, except for 77x where there are 10 possibilities, so therefore there are 26 possibilities for the 700's, same of 800's and the 900's, so

so every change in the tens digit, there are 2 possibilities, except for 77x where there are 10 possibilities, so therefore there are 26 possibilities for the 700's, same of 800's and the 900's, so

26*3=78

Hi Tuneman,

You shouldn't include 700 cuz it said that the number is greater than 700. More important, 77x does not have 10 possibilities because 777 would have three identical numbers instead of two.

This isn't a formula or a mathematical explanation but I would just go with the primitive method (I won't be able to come up with a formula in 2 minutes anyway... )

8xx where x = 0~7, 9 - 9 of them 8x8 where x = 0~7, 9 - 9 of them 88x where x = 0~7, 9 - 9 of them total 27

since it would be the same for the others in 700s and 900s

27 x 3 = 81

however, since it has to be greater than 700, subtract the case 7xx where x = 0.

81 - 1 = 80

What is the OA? (I might have missed out on something...)

This might not be what you wanted..but...

thanks. very helpful _________________

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson