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# Of the three digits greater than 700, how many have 2 digit

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Of the three digits greater than 700, how many have 2 digit [#permalink]  20 Nov 2009, 10:07
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55% (medium)

Question Stats:

43% (03:04) correct 56% (01:41) wrong based on 53 sessions
Of the three digits 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

Is there a faster method to find out the answer, because i listed down all the numbers and it took me hardly 1.5 mins.
but what if there are more numbers ?
[Reveal] Spoiler: OA
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Re: Number system (OG 11 ) [#permalink]  20 Nov 2009, 10:28
Expert's post
gurpreet07 wrote:
Of the three digits 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

Is there a faster method to find out the answer, because i listed down all the numbers and it took me hardly 1.5 mins.
but what if there are more numbers ?

Step by step:

The number can have three forms:

XXY meaning that the first digit is the repeated digit;
XYX meaning that the first digit is repeated digit;
and
YXX meaning that the first digit is not repeated digit.

XXY --> the repeated digits can be 7,8 or 9, so 3 possibilities and the third one can take 9 possible values total 3*9=27.

The same with XYX =3*9=27;

YXX --> the non repeated first digit can be 7, 8, 9, so 3 possibilities and the repeated digits can take 9 possible values, total 3*9=27. But this will give us the number 700 as well and we know that the number should be more than 700, hence 27-1=26.

TOTAL=27+27+26=80

Hope it helps.
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Re: Number system (OG 11 ) [#permalink]  20 Nov 2009, 10:43
Bunuel wrote:
gurpreet07 wrote:
Of the three digits 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

Is there a faster method to find out the answer, because i listed down all the numbers and it took me hardly 1.5 mins.
but what if there are more numbers ?

Step by step:

The number can have three forms:

XXY meaning that the first digit is the repeated digit;
XYX meaning that the first digit is repeated digit;
and
YXX meaning that the first digit is not repeated digit.

XXY --> the repeated digits can be 7,8 or 9, so 3 possibilities and the third one can take 9 possible values total 3*9=27.

The same with XYX =3*9=27;

YXX --> the non repeated first digit can be 7, 8, 9, so 3 possibilities and the repeated digits can take 9 possible values, total 3*9=27. But this will give us the number 700 as well and we know that the number should be more than 700, hence 27-1=26.

TOTAL=27+27+26=80

Hope it helps.

thnku....very fast and a efficient method.
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Re: Number system (OG 11 ) [#permalink]  20 Nov 2009, 10:53
4
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Expert's post
This can be solved in another way and maybe it's faster:

Numbers between 700 and 999 =299;

Numbers with all distinct digits = 3*9*8=216;

Numbers with all the same digits more than 700 =3, (777, 888, 999);

Total: 299-216-3=80.
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Re: Number system (OG 11 ) [#permalink]  20 Nov 2009, 11:07
Bunuel wrote:
This can be solved in another way and maybe it's faster:

Numbers between 700 and 999 =299;

Numbers with all distinct digits = 3*9*8=216;

Numbers with all the same digits more than 700 =3, (777, 888, 999);

Total: 299-216-3=80.

Hey Bunuel
thanks man...from where do you get such idea's
just awesome
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Re: Number system (OG 11 ) [#permalink]  27 Mar 2010, 17:29
Bunuel,

I don't understand how there is only 27 numbers based on your calculation. For example, XXY the first number has to be 7, 8 or 9 so 3 possibilities and the next number would also have to be 7, 8 or 9 but the last number could be 9 values but how is that equal to 3*9? Wouldn't that be 3*3*9? I'm don't understand.
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Re: Number system (OG 11 ) [#permalink]  10 Apr 2010, 03:55
Bunuel,

I don't understand how there is only 27 numbers based on your calculation. For example, XXY the first number has to be 7, 8 or 9 so 3 possibilities and the next number would also have to be 7, 8 or 9 but the last number could be 9 values but how is that equal to 3*9? Wouldn't that be 3*3*9? I'm don't understand.

in case of XXY, first digit is 7,8 or 9 then so 3 ways
2nd digit has be same as 1st so only one way
you can also write it as 3*1*9 = 27

Hope it helps
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Re: Number system (OG 11 ) [#permalink]  14 Apr 2010, 02:10
How have you guys become so confident, predicting :

700-999

XXY setup = t occurrences without a single calculation line etc...

Isnt there any mathematical way of solving it.
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Re: Number system (OG 11 ) [#permalink]  14 Apr 2010, 05:02
wow Bunuel ...
I tried to formulate a pattern and reach the answer. But it seems I missed something ...well ....
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Re: Number system (OG 11 ) [#permalink]  14 Apr 2010, 09:24
1
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Expert's post
shrivastavarohit wrote:
How have you guys become so confident, predicting :

700-999

XXY setup = t occurrences without a single calculation line etc...

Isnt there any mathematical way of solving it.

This is how I solved it:

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.
A. all digits are distinct = 3(first digit can have only three values 7,8, or 9)*9*8=216;
C. all three are alike = 3 (777, 888, 999)

So, 299-216-3=80.

Another way to solve this problem is in my first post.

Hope it's clear.
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Re: number property [#permalink]  19 Jun 2010, 21:30
I am getting 78 as the answer. So I went with closest answer 80 (C).

This is how I arrived @ 78:

For numbers greater than 700 we have following possibilities:

a=> 7xx .. Here x being any digit 0-9, excluding 7 and 0 {0 is excluded because number should be >700}
b=> 7x7 .. Here x can be any digit from 0-9, excluding 7. {no 3 digits should be same}
c=> 77x .. Here x can be any digit from 0-9, excluding 7 for same reason as above.

a + b + c = 8 + 9 + 9 = 26

Hence same is true with 8xx , 8x8, 88x and 9xx,9x9, 99x ...

So summing up I got 78 ...

Please help me if I am wrong here. I am unable to figure out where did I miss 2 more numbers. Any pointers ?
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Re: number property [#permalink]  19 Jun 2010, 22:31
for 700 ==> 711 , 722 , ...... , 799 (except 777) = 8
,707, 717 , 727 , .... , 797 = 9
770, 771 , 772 , ........ , 779 = 9
total = 26

similarly for 800 and 900 series you will have 26 each

so 26 * 3 = 78

and 800 and 900 so 78 + 2 = 80
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Re: Number system (OG 11 ) [#permalink]  20 Jun 2010, 09:45
Bunuel wrote:
gurpreet07 wrote:
Of the three digits 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

Is there a faster method to find out the answer, because i listed down all the numbers and it took me hardly 1.5 mins.
but what if there are more numbers ?

Step by step:

The number can have three forms:

XXY meaning that the first digit is the repeated digit;
XYX meaning that the first digit is repeated digit;
and
YXX meaning that the first digit is not repeated digit.

XXY --> the repeated digits can be 7,8 or 9, so 3 possibilities and the third one can take 9 possible values total 3*9=27.

The same with XYX =3*9=27;

YXX --> the non repeated first digit can be 7, 8, 9, so 3 possibilities and the repeated digits can take 9 possible values, total 3*9=27. But this will give us the number 700 as well and we know that the number should be more than 700, hence 27-1=26.

TOTAL=27+27+26=80

Hope it helps.

thanks! this is an efficient way to solve it. is this method feasible for this kind of questions--fundamental counting?
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Re: Number system (OG 11 ) [#permalink]  20 Jun 2010, 09:59
Expert's post
Bunuel wrote:
Hope it helps.
tt11234 wrote:
thanks! this is an efficient way to solve it. is this method feasible for this kind of questions--fundamental counting?

I don't really understand your question...

Also there is another, easier way to solve this question, shown in my second post:

3-digit numbers more than 700 = 299;

Numbers with all distinct digits = 3*9*8=216 (first digit can take 3 values - 7, 8, 9; second digit can take 9 values and the third digit 8 values);

Numbers with all the same digits more than 700 = 3, (777, 888, 999);

{all} - {all distinct} - {all same} = {two equal} --> 299-216-3=80.

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Re: Number system (OG 11 ) [#permalink]  30 Jun 2010, 16:41
Thanks Bunuel for the effort
but I dont really understand why you choose 299, and what is the meaning of distinct, and why is there 3, 9 , 8?
Pleas explaing that to me
thanks again
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Re: Number system (OG 11 ) [#permalink]  30 Jun 2010, 18:44
xmagedo wrote:
Thanks Bunuel for the effort
but I dont really understand why you choose 299, and what is the meaning of distinct, and why is there 3, 9 , 8?
Pleas explaing that to me
thanks again

The question asks about three-digit integers greater than 700. So, all the numbers in the range 701 to 999, inclusive. There will be 999 - 701 + 1 = 299 such numbers in this range. (Whenever you want to count the number of objects in a range, subtract the smallest number from the greatest and add 1. For example, the number of integers in the range of 1 to 6 is 6 -1 +1 = 6).

"distinct" just means different. So, for example, in the number "799", the final two digits are NOT distinct. Here: 798, they are.
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Re: Number system (OG 11 ) [#permalink]  21 Jan 2012, 19:06

Numbers between 700 and 999 =299;

Numbers with all distinct digits = 3*9*8=216;

Numbers with all the same digits more than 700 =3, (777, 888, 999);

Total: 299-216-3=80.

I've a doubt. Numbers with all distinct digits = 3*9*8=216;

I feel 9 above is not right because if we take one of the 3 digits (7,8 or 9), then we shouldn't take repeated values (how is 9 possible).
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Re: Number system (OG 11 ) [#permalink]  21 Jan 2012, 19:07
Ok, I got it

I forgot to take from 0-9 .. 10 digits

10-1=9
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Re: Of the three digits greater than 700, how many have 2 digit [#permalink]  01 Feb 2014, 11:28
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Re: Of the three digits greater than 700, how many have 2 digit [#permalink]  01 Feb 2014, 19:42
We can fix 2 digits in 3 ways ie 7,8,9 and the third digit can be fixed in 9 ways(10 digits-1 repeating digit)
Total=3*9=27
Since this is possible in 3 different ways(first digit diff from other two,second digit diff from other two and third digit diff from other two) we get 3* 27=81 but we subtract 1 for 700
Ans 80

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Re: Of the three digits greater than 700, how many have 2 digit   [#permalink] 01 Feb 2014, 19:42
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