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

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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

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seekmba 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 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.
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Mikko wrote:
seekmba 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


Guys, i need help, it is 81 not 80.

abc is the 3 digit integer greater than 700:

Then a can be 7, 8, 9: we have 3 choices for a
- if b=a, c#a=b: 1 choice for b and 10-1=9 choices for c: 3*1*9=27
- if b=c#a: 9 choices for b, and 1 choice for c: 3*9*1=27
- if a=c: 1 choice for c, 9 choises for b (b is different fr a): 3*1*9=27

Then we have 27+27+27=81, i cant not reason why? plz help.


The red part could give you 700 but we are told that numbers must be more than 700, so subtract 1 --> 81-1=80.

Hope it's clear.
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New post 22 Oct 2010, 18:43
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total 3 digit numbers greater than 700 are 299 --- A

total 3 digit numbers greater than 700 and having all digit different = 3* 9 * 8 = 216 --- B

(first position can take 3 values, 2nd position can take 9 values and 3rd place can take 8 values)

total 3 digit numbers greater than 700 and having all digit same = 3 ---C

Hence total 3 digit numbers greater than 700 and having only 2 digit same = A-B-C = 80 :option c

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Hey Bunnel, your answer is super but beyond my grasping power to do by self and less than 2 mins.

So I had a fairly different approach:
For a 3-digit number we have a digit in the tens, hundreds and thousands place.
So counting the number of digits when tens and hundreds are same but not thousands:
i.e: 711. 722...766,788,799,800 = 8+9+9 = 26
So counting the number of digits when tens and thousands are same but not hundreds:
i.e: 707,717,...767,787,... = 9+9+9 = 27
So counting the number of digits when hundreds and thousands are same but not tens:
i.e: 770,771,772...776,778... = 9+9+9 = 27

Total it up to 80.
Answer C
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Re: Of the three digit integers [#permalink]

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Bunuel wrote:
seekmba 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 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.


Bunuel -
Your method of pattern recognition is by far the best and quickest approach to problems such as these. Thanks for the help.
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Re: Of the three digit integers [#permalink]

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hfbamafan wrote:
Bunuel wrote:
seekmba 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 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.


Could you elaborate on part A just a bit more please.

Thanks


Sure.

A. all digits are distinct, are numbers like 701, 702, ...987, so all numbers from 701 to 999, inclusive which have all three different digits.

Now, the first digit can take 3 values 7, 8, or 9.
The second digit can take 9 values: all 10 digits minus the one we used for the first digit.
Similarly the third digit can take 8 values: all 10 digits minus the one we used for the first digit and the one we used for the second digit.

So, for A. there are 3*9*8=216 numbers.
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Re: Of the three digit integers [#permalink]

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New post 27 Sep 2010, 23:00
seekmba 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


Guys, i need help, it is 81 not 80.

abc is the 3 digit integer greater than 700:

Then a can be 7, 8, 9: we have 3 choices for a
- if b=a, c#a=b: 1 choice for b and 10-1=9 choices for c: 3*1*9=27
- if b=c#a: 9 choices for b, and 1 choice for c: 3*9*1=27
- if a=c: 1 choice for c, 9 choises for b (b is different fr a): 3*1*9=27

Then we have 27+27+27=81, i cant not reason why? plz help.
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Re: Of the three digit integers [#permalink]

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New post 29 Sep 2010, 00:07
Nice Post Thanks +1 to Bunuel for the explaination
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Re: Of the three digit integers [#permalink]

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New post 18 Nov 2011, 02:00
Bunuel wrote:
seekmba 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 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.



New approach. Thanks alot
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Re: Of the three digit integers [#permalink]

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New post 09 Dec 2011, 12:03
Bunuel wrote:
seekmba 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 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.



Hats off, this is a brilliant way to solve!
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Re: Of the three digit integers greater than 700, how many have [#permalink]

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New post 09 Dec 2011, 22:19
ABC: for 700 series, PQR: for 800 and XYZ:900
A, P, and X are reserved for 7, 8 and 9 respectively.
for each serise, there are three cases:
1. tens digit is same as hundreds digits, but unit digit is different.
2. Unit digit is same as hundreds digits, but tens digit is different.
3 .Both unit and tens digit are same but not = hundreds digit

Therefore, we have:
for 700 series: (1C1*1C1*9C1)+ (1C1*9C1*1C1) +(1C1*8C1*1C1)=9+9+8=26
for 800 series: (1C1*1C1*9C1)+ (1C1*9C1*1C1) +(1C1*9C1*1C1)=9+9+9=27
for 900 series: (1C1*1C1*9C1)+ (1C1*9C1*1C1) +(1C1*9C1*1C1)=9+9+9=27

Total: 26+27+27=80 ways.
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Re: Of the three digit integers greater than 700, how many have [#permalink]

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New post 13 Dec 2011, 01:38
superb explained by bunel
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Re: Of the three digit integers greater than 700, how many have [#permalink]

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New post 27 Mar 2012, 00:32
Simple and sweet explanation. Thanks Bunuel!
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Re: Of the three digit integers greater than 700, how many have [#permalink]

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New post 27 Mar 2012, 04:55
Saw a lot of ways to solve this. But Bunuel, you are awesome.
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Re: Of the three digit integers greater than 700, how many have [#permalink]

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New post 27 Apr 2012, 11:42
Hi bunuel,

Nice method.


Can you please post links of similar problems so that i can tune this area?

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

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New post 27 Apr 2012, 12:18
Thanks Bunel, nice explaination 1+
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Re: Of the three digit integers [#permalink]

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New post 27 Apr 2012, 22:27
Bunuel wrote:
seekmba 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 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.


Could you elaborate on part A just a bit more please.

Thanks
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Re: Of the three digit integers   [#permalink] 27 Apr 2012, 22:27
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