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Of the three-digit integers greater than 700, how many have two digits

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

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New post 06 Oct 2019, 02:28
Bunuel wrote:
SOLUTION

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.

Answer: C.


Hi Bunuel, I have a question here.

Total numbers from 700 to 999 = 299 (3-digit numbers greater than 700);

I thought the question said it needs to be greater than 700 so shouldn't it be 299-1 (as 700 cannot be one of the number).
Any response will be very much appreciated!

Thank you JC
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Re: Of the three-digit integers greater than 700, how many have two digits  [#permalink]

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New post 06 Oct 2019, 05:08
We don't even have to solve this question. Just use the logic!!

Because of the symmetry, 3 digit integers that have two digits that are equal to each other and the remaining digit different from the other two from 700-799 is equal to that are from 800-899 or 900-999

Hence total 3 digit numbers greater than 700 have two digits that are equal to each other and the remaining digit different from the other two
= 3k-1 (as we have to exclude 700)

Only option C is in the form of 3k-1

Bunuel 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

Diagnostic Test
Question: 11
Page: 22
Difficulty: 650
Manager
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Re: Of the three-digit integers greater than 700, how many have two digits  [#permalink]

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New post 06 Oct 2019, 07:40
Bunuel 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

Diagnostic Test
Question: 11
Page: 22
Difficulty: 650


According to the question
3 digits will be AAB,ABA,ABB
start with 9
AAB- For A - 1 option(9), For A-1 option, For B-9 options. Total=9
In the same way for ABA and ABB
So total=27
For 8
in the same way
So total=27
case with 7 is a bit different as question say >700
AAB-For A-1(option-7),For A-1 option, For B-9 options. So Total= 9
ABA- For A-1 option, For B-9 options, For A-1 option
ABB- For A- 1 option, For B-8 Options, For B- 8 options( You can not include two zeroes as it is mentioned that >700)
So total=9+9+8=26
So required number is= 27+27+26
80:)
C:)
GMAT Club Bot
Re: Of the three-digit integers greater than 700, how many have two digits   [#permalink] 06 Oct 2019, 07:40

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