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Counting integers

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Counting integers

by ReenaMo » Sat Aug 23, 2008 10:45 am
Of the three-digit integers that are greater 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 arrived at the right answer, 80 (C). However, if I employed that method on the actual test, I'd still be on number 11 at the end of the 75 minutes! :)

Can you explain the formulas/quick tricks we can use during the test on these types of problems (of even greater difficulty)? I saw a few others posted that were kind of similar, but none really explain what the actual formula is.

Thanks so much!
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by pranavc » Sat Aug 23, 2008 10:49 am
I landed up with something significantly smaller. How did you end up with 80? Please do let me know.
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by parallel_chase » Sat Aug 23, 2008 11:28 am
Here is the answer.

For first digit we have 3 letters to play with (7,8,9)
Next two digits we can have any letter (0,1,2,3,4,5,6,7,8,9)

CASE I (ABB)
for first digit we have (7,8,9) = 3 ways
for second digit we have ((0,1,2,3,4,5,6,7,8,9) except for 1 letter (7,8,9) which we have used at first digit = 9 ways
third we only 1 way because we want the same digit as second = 1 way
3*9*1 = 27

Similarly the other two cases can be followed in conjunction with above method.
CASE II (BAB)
3*9*1 = 27

CASE III (BBA)
3*1*9 =27

27+27+27 = 81

subtract 1 because we want the number to be greater than 700, the above combination include 700

81-1=80

Let me know if you have any doubts.
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by pranavc » Sat Aug 23, 2008 11:53 am
Makes perfect sense. Thanks a lot.
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by ReenaMo » Sat Aug 23, 2008 12:16 pm
Great - thanks!
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by ankijain » Sat Aug 23, 2008 12:31 pm
Yea nice one.. i was thinking 54... misssed the BBA part....:P
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by rahul.s » Thu Jan 28, 2010 4:31 am
parallel_chase wrote:Here is the answer.

For first digit we have 3 letters to play with (7,8,9)
Next two digits we can have any letter (0,1,2,3,4,5,6,7,8,9)

CASE I (ABB)
for first digit we have (7,8,9) = 3 ways
for second digit we have ((0,1,2,3,4,5,6,7,8,9) except for 1 letter (7,8,9) which we have used at first digit = 9 ways
third we only 1 way because we want the same digit as second = 1 way
3*9*1 = 27

Similarly the other two cases can be followed in conjunction with above method.
CASE II (BAB)
3*9*1 = 27

CASE III (BBA)
3*1*9 =27

27+27+27 = 81

subtract 1 because we want the number to be greater than 700, the above combination include 700

81-1=80

Let me know if you have any doubts.
amazing post! thank you.
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by rajman41 » Tue Feb 09, 2010 11:39 pm
THanks a lot.
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by hala0987 » Thu Apr 01, 2010 10:18 am
Guys...I actually did it in the EXCEL sheet and i got did this in the following way...i got 80 even though the book did not pay attention to the NOT EQUAL part of -999 , SO we're doing from 701-998 and also the 999,888,777 and 700 are considered 3 digits equal to each other and the last one is just the limitation of the question (more than 700)...Look at this XL Sheet please and tell me what you think...

That's why i trust the method which involves probability better but then i worry that im missing something in it so i start doubting and trying to write down the actual numbers...


[/img]
Attachments
Probability-Q-701-998,2EqualDigits.xls
(14.5 KiB) Downloaded 127 times
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by inmate » Tue Oct 05, 2010 12:45 pm
I am on this problem and trying to understand it.

"In three-digit integers, there are three pairs of digits that can be the same while the other digit is different: tens and ones, hundreds and tens, and hundreds and ones." What does tens and ones, hundreds and tens, and hundreds and ones mean?
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by armand_h » Wed Jul 25, 2012 9:11 am
You don't need to count.
Here is another way to think of it.
The number of 2 similar digits between 700 and 999 is 3 times the number of similar digits between 700 and 799
So the answer is in the form of 3N-1 (the -1 is because we have to exclude 700)
Which means if you add 1 to the answer it will be a multiple of 3
The only choice that verifies this condition is 80 (80+1=3*27)
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