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three-digit integers

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three-digit integers

by venmic » Sun Nov 13, 2011 10:58 am
Of the three-digit integers greater than 600, how many have
two digits that are equal to each other and the remaining digit
di¤erent from the other two?
(A) 120
(B) 116
(C) 108
(D) 107
(E) 72

Is there a faster way to this - algebrically instead of actually forming sequences

any suggestions


D
Last edited by venmic on Sun Nov 13, 2011 11:21 am, edited 1 time in total.
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by vaibhavgupta » Sun Nov 13, 2011 11:13 am
venmic wrote:Of the three-digit integers greater than 600, how many have
two digits that are equal to each other and the remaining digit
di¤erent from the other two?
(A) 120
(B) 116
(C) 108
(D) 107
(E) 72

Is there a faster way to this - algebrically instead of actually forming sequences

any suggestions
IMO E

whts OA?
If OA is A, IMO B
If OA is B, IMO C
If OA is C, IMO D
If OA is D, IMO E
If OA is E, IMO A

FML!! :/
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by shankar.ashwin » Sun Nov 13, 2011 11:18 am
Looking at the answer choices I would directly pick D

You're interested in numbers of these range (600-700) (700-800) (800-900) and (900-1000)

So say there are 'X' numbers which satisfy the condition asked in the question in the range (600-700). So technically we will have 4X as an answer.

But since question asks us numbers greater than 600, we should not count 600, so the answer choice would be of the form 4X-1. Only D is
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by GMATGuruNY » Sun Nov 13, 2011 12:04 pm
venmic wrote:Of the three-digit integers greater than 600, how many have
two digits that are equal to each other and the remaining digit
di¤erent from the other two?
(A) 120
(B) 116
(C) 108
(D) 107
(E) 72

Is there a faster way to this - algebrically instead of actually forming sequences

any suggestions


D
Good = total - bad.

Total:
Given consecutive integers, the total number = (biggest-smallest) + 1:
999-601+1 = 399.

Bad case 1: all 3 digits are different
Number of options for the hundreds digit = 4. (6,7,8,or 9)
Number of options for the tens digit = 9. (Any digit 0-9 other than the digit already used.)
Number of options for the units digit = 8. (Any digit 0-9 other than the 2 digits already used.)
To combine these options, we multiply:
4*9*8 = 288.

Bad case 2: all 3 digits are the same
Number of options = 4. (666,777,888,999).

Good integers = 399-288-4 = 107.

The correct answer is D.
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by Neo Anderson » Mon Nov 14, 2011 8:26 am
I followed a slightly different approach:-

the solution will comprise of three cases:

1. first two digits same
4*1*9 ways = 36 ways

2. first and last digit same
4*9*1 ways = 36 ways

3. second and third digit same
4*9*1 ways = 36 ways

=> 36*3 = 108; this includes number 600 as well;
as question says above 600, thus answer is 108-1= 107
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