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# - How many three-digit numbers are so that they have 2 equal

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- How many three-digit numbers are so that they have 2 equal [#permalink]

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28 Jul 2003, 03:21
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

1.- How many three-digit numbers are so that they have 2 equal digits and the other different from those 2, eg. 211?

2.- There are three sets of numbers from 1 to 8. One number is chosen from each set. How many selections of three numbers sum 16?

3.- 2 two-digit intergers, M and N, have the same digits in reverse order. Which of the following cannot be the sum of M and N?

181 163 121 99 44
[/b]

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28 Jul 2003, 04:20
(1)
we have 10 digits: 0 1 2 3 4 5 6 7 8 9
and three positions _ _ _

Consider variant AAB (there are 3C1=3C2=3 variants)

the first can be 1-9 assuming that 0 is out.
So, the first position can be filled in 9 ways
the second -- the same 9
the third -- the same 9 because one is fixed for the first two.

9*9*9=729

Other variant ABA and BAA will give the same
3*729=2187

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28 Jul 2003, 05:16
stolyar wrote:
(1)
we have 10 digits: 0 1 2 3 4 5 6 7 8 9
and three positions _ _ _

Consider variant AAB (there are 3C1=3C2=3 variants)

the first can be 1-9 assuming that 0 is out.
So, the first position can be filled in 9 ways
the second -- the same 9
the third -- the same 9 because one is fixed for the first two.

9*9*9=729

Other variant ABA and BAA will give the same
3*729=2187

Tovarish,

There are ONLY 900 3-digit number in TOTAL (100 -999). How can your answer be 2187?
_________________

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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

Last edited by AkamaiBrah on 28 Jul 2003, 05:33, edited 1 time in total.

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Re: Digits & numbers... Hard questions [#permalink]

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28 Jul 2003, 05:32
jorgecosano2 wrote:
:?:

1.- How many three-digit numbers are so that they have 2 equal digits and the other different from those 2, eg. 211?

2.- There are three sets of numbers from 1 to 8. One number is chosen from each set. How many selections of three numbers sum 16?

3.- 2 two-digit intergers, M and N, have the same digits in reverse order. Which of the following cannot be the sum of M and N?

181 163 121 99 44
[/b]

(1) assuming the first digit cannot be zero, 243
(2) 42
(3) both A and B as stated in your problem. (I beileve b is a typo and should be 165 -- then the answer is A).
_________________

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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

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Re: Digits & numbers... Hard questions [#permalink]

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28 Jul 2003, 06:13
AkamaiBrah wrote:
jorgecosano2 wrote:
:?:

1.- How many three-digit numbers are so that they have 2 equal digits and the other different from those 2, eg. 211?

2.- There are three sets of numbers from 1 to 8. One number is chosen from each set. How many selections of three numbers sum 16?

3.- 2 two-digit intergers, M and N, have the same digits in reverse order. Which of the following cannot be the sum of M and N?

181 163 121 99 44
[/b]

(1) assuming the first digit cannot be zero, 243
(2) 42
(3) both A and B as stated in your problem. (I beileve b is a typo and should be 165 -- then the answer is A).

Akamaibrah,

(2) I think it should be 21
(3) I think the right answer is 163 - choice (b) since the rest are all divisible by 11. The working is as follows:

Lets assume M has the digits x and y. Therefore, M=10x+y
This gives us N=10y+x
Sum of the two gives M+N=11(x+y) and the answer follows from this.

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Re: Digits & numbers... Hard questions [#permalink]

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28 Jul 2003, 06:18
prashant wrote:
AkamaiBrah wrote:
jorgecosano2 wrote:
:?:

1.- How many three-digit numbers are so that they have 2 equal digits and the other different from those 2, eg. 211?

2.- There are three sets of numbers from 1 to 8. One number is chosen from each set. How many selections of three numbers sum 16?

3.- 2 two-digit intergers, M and N, have the same digits in reverse order. Which of the following cannot be the sum of M and N?

181 163 121 99 44
[/b]

(1) assuming the first digit cannot be zero, 243
(2) 42
(3) both A and B as stated in your problem. (I beileve b is a typo and should be 165 -- then the answer is A).

Akamaibrah,

(2) I think it should be 21
(3) I think the right answer is 163 - choice (b) since the rest are all divisible by 11. The working is as follows:

Lets assume M has the digits x and y. Therefore, M=10x+y
This gives us N=10y+x
Sum of the two gives M+N=11(x+y) and the answer follows from this.

Yes I realize that the number must be divisible by 11. However, NEITHER 181 nor 163 is divisible by 11 (here is a quick check. Subtract the sum of the digits in the odd-positions from the sum of the digits in the even positions. If the answer is 0 or a multiple of 11, then the whole number is divisible by 11). 165 would be. Or 187. So there must be a typo in one of those two. Hence, my answer.
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

Last edited by AkamaiBrah on 28 Jul 2003, 06:31, edited 2 times in total.

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Re: Digits & numbers... Hard questions [#permalink]

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28 Jul 2003, 06:24
Quote:
Akamaibrah,

(2) I think it should be 21

817
826
835
844
853
862
871
718
727
736
745
754
763
772
781
628
637
646
655
664
673
682

well, i'm up to 22 and there seems to be a few numbers left....
I'll stand by 42 for now.

P.S. If order doesn't matter, then the answer is 9.
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

Last edited by AkamaiBrah on 28 Jul 2003, 06:45, edited 2 times in total.

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28 Jul 2003, 06:31
Agree with you on both....

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28 Jul 2003, 06:35
stolyar wrote:
(1)
we have 10 digits: 0 1 2 3 4 5 6 7 8 9
and three positions _ _ _

Consider variant AAB (there are 3C1=3C2=3 variants)

the first can be 1-9 assuming that 0 is out.
So, the first position can be filled in 9 ways
the second -- the same 9
the third -- the same 9 because one is fixed for the first two.

9*9*9=729

Other variant ABA and BAA will give the same
3*729=2187

my bad, agree. the third 9 is out.

3*9*9*1=243

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28 Jul 2003, 18:51
Question 1: What is the systamatic and faster way to slove this question?

Question 2: My ans is 42. But, how to get hte answer in 2 minutes?

Question 3: I agree that there is a typo in the answer chices. Same
question appaered in a different forum couple of days back with the
same typo?!

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28 Jul 2003, 20:47
Question 1: What is the systamatic and faster way to slove this question?

Question 2: My ans is 42. But, how to get hte answer in 2 minutes?

Question 3: I agree that there is a typo in the answer chices. Same
question appaered in a different forum couple of days back with the
same typo?!

1) There are 10 possible number pairs that can combine with 9 other numbers = 90. Of each of these, there are 3 ways that they can be arranged: AAB ABA BAA. So we now have 90 x 3 = 270. But we need to exclude all of the numbers starting with zero. Since this is a "symmetric" problem, exactly 1/10 of all possible numbers will start with zero so the answer is 270 x 90% = 243.

2) Consider 3 number like 3 dice. Let say we look at the first die. Given that the first die is an 8, the next two must add up to 16. Using the same logic as dice, you can quickly see that there are 7 ways to combine two 8-sided dices to make 8.
If the first die is 7, then the next 2 must add up to 9 and there are 8 ways to do that.
If the first die is 6, then the next 2 must add up to 10 and there are 7 ways to do that. Similarly :
1st = 5, ways to make 11 = 6
1st = 4, ways to make 12 = 5
1st = 3, wyas to make 13 = 4
1st = 2, ways to make 14 = 3
1st = 1, wyas to make 15 = 2
If you spot the pattern and understand how two dice combine, you can quickly fill in this table knowing only the first 3 results.

Hence the total is 7 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 42
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

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30 Jul 2003, 02:17
please... could you give me a systamatic way for 2?

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02 Aug 2003, 04:50
stolyar wrote:
the first can be 1-9 assuming that 0 is out.
So, the first position can be filled in 9 ways
the second -- the same 9
the third -- the same 9 because one is fixed for the first two.

9*9*9=729

my bad, agree. the third 9 is out.

3*9*9*1=243

I guess 243 is not the answer. Considering digits from 1-9, we get 3 x 9 x 8 x 1. When zero is in the last digit, 9 combinations and when zero is in the second digit, 9 more of them. Hence I get 234 as the answer.

When you write 3 x 9 x 9 x 1, you are allowing 0 to occur in all the three places which means it may not be a three digit no. at all.

Bharathi.

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02 Aug 2003, 05:13
bhars18 wrote:
stolyar wrote:
the first can be 1-9 assuming that 0 is out.
So, the first position can be filled in 9 ways
the second -- the same 9
the third -- the same 9 because one is fixed for the first two.

9*9*9=729

my bad, agree. the third 9 is out.

3*9*9*1=243

I guess 243 is not the answer. Considering digits from 1-9, we get 3 x 9 x 8 x 1. When zero is in the last digit, 9 combinations and when zero is in the second digit, 9 more of them. Hence I get 234 as the answer.

When you write 3 x 9 x 9 x 1, you are allowing 0 to occur in all the three places which means it may not be a three digit no. at all.

Bharathi.

the answer is 243. Stolyars method is poor and he was "lucky" to stumble on the correct answer.
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

Kudos [?]: 240 [0], given: 0

02 Aug 2003, 05:13
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