Of the three-digit integers greater than 700, how many have two digits : GMAT Problem Solving (PS)
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# Of the three-digit integers greater than 700, how many have two digits

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Of the three-digit integers greater than 700, how many have two digits [#permalink]

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07 Sep 2006, 15:31
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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?

(A) 90
(B) 82
(C) 80
(D) 45
(E) 36

OPEN DISCUSSION OF THIS QUESTION IS HERE: of-the-three-digit-integers-greater-than-700-how-many-have-135188.html
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Re: Of the three-digit integers greater than 700, how many have two digits [#permalink]

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07 Sep 2006, 16:35
Rayn wrote:
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?

A) 90
B) 82
C) 80
D) 45
E) 36

Explain your reasoning. Also, can PR or CR be used for this problem...how so?

I get none of the above answers...

The numbers are:

707, 717, 727,...797 - 9 numbers (excluding 777)
711, 722, 733, ...799 - 8 numbers (excluding 700 and 777)

808, 818, 828,...898 - 9 numbers
800, 811, 822,...899 - 9 numbers

909, 919, 929, ...989 - 9 numbers
900, 911, 922, ...988 - 9 numbers

Totals to 53..
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Re: Of the three-digit integers greater than 700, how many have two digits [#permalink]

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07 Sep 2006, 19:59
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We will count only from 700-799

7XX where X is any digit other than 7. = 9
7X7 where X is any digit other than 7 = 9
77X where X is any digit other than 7 = 9
Total form 700-799 = 27
Similarly there are 27 from 800-899 and 27 from 900-999.

Total = 27+27+27 = 81
but we have to exclude 700.
So answer = 81-1 = 80
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Re: Of the three-digit integers greater than 700, how many have two digits [#permalink]

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09 Jan 2015, 18:06
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Re: Of the three-digit integers greater than 700, how many have two digits [#permalink]

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10 Jan 2015, 04:51
Rayn wrote:
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?

(A) 90
(B) 82
(C) 80
(D) 45
(E) 36

Three digit number can have only following 3 patterns:
A. all digits are distinct;
B. two digits are alike and third is different;
C. all three digits are alike.

We need to calculate B. B=Total - A - C

Total numbers from 700 to 999 = 299 (3-digit numbers greater than 700);
A. all digits are distinct = 3*9*8=216 (first digit can have only three values 7, 8, or 9);
C. all three are alike = 3 (777, 888, 999).

So, 299-216-3=80.

OPEN DISCUSSION OF THIS QUESTION IS HERE: of-the-three-digit-integers-greater-than-700-how-many-have-135188.html
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Re: Of the three-digit integers greater than 700, how many have two digits   [#permalink] 10 Jan 2015, 04:51
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