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How many 3digit integers between 100 and 200 have a digit [#permalink]
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19 Dec 2007, 05:53
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How many 3digit integers between 100 and 200 have a digit that is the average (arithmetic mean) of the other 2 digits? A. 1 B. 7 C. 9 D. 11 E. 19
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Last edited by Bunuel on 14 Jul 2013, 01:11, edited 1 time in total.
Edited the question.



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Since the number is between 100 and 200, the hundreds digit has to be 1.
And since the answer choices are small enough, it won't hurt just to write them out. 102, 111, 120, 123, 132, 135, 147, 153, 159, 174, 195.
Answer D



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let the 3 digit number be xyz, we know X=1.The question leads us to an equation (x+y)/2 = Z and we know we can only use the results which yield whole numbers.Using this equation will lead us to the result which has been already posted earlier.
Cheer!



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Re: number properties [#permalink]
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19 May 2010, 11:08
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einstein10 wrote: How many 3digit integers between 100 and 200 have a digit that is the average (arithmetic mean) of the other 2 digits? a) 1 b) 7 c) 10 d) 11 e) 19
can anyone please explain this? thanks. IMO d, whats the source? Fix the 1_ _ now for every number 1,2,3,.... (x+y)/2 = 1, = 2.... x+y =2 which is only possible when x=1,y=1 or x=2,y=0 or viceversa similarly x+y=4 is only possible when x=1 and y=3....or viceversa, x=3 and y=2 is not possible as one of the x/y has to be 1 102,120 111 123,132 135,153 147,174 159,195
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Re: number properties [#permalink]
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19 May 2010, 19:51
thanks for the explanation. I got this problem from some other forum. some how I was getting 9 numbers.. I was missing first 2 as you pointed out. thanks. kudos for you



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Re: number properties [#permalink]
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19 May 2010, 20:12
Quote: How many 3digit integers between 100 and 200 have a digit that is the average (arithmetic mean) of the other 2 digits? a) 1 b) 7 c) 10 d) 11 e) 19 All of the integers between 100 and 200 are three digit aren't they? Are they trying to confuse us with redundancy? So to the question, in order to be an average of the other two digits we have to take into account that all will include the 1 for the hundreds digit. It helps then for the other number to be odd so that when divided by 2 to take the average we get an integer. 101? no 103 . . . 111 yes, 123 yes, so 132 only because the digits of 123 are transposed, 135, 153, 147, 174, 159, 195. I know I am missing one. I would guess 10 and move on. Can't exactly ask kudos for that mess. Sorry. Should have guessed eleven knowing there would be a pair. 102 and 120. 2+0/2 =1.Skip



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How many 3digit integers between 100 and 200 have a digit [#permalink]
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03 Mar 2011, 11:10
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How many 3digit integers between 100 and 200 have a digit that is the average (arithmetic mean) of the other 2 digits?
A. 1 B. 7 C. 9 D. 11 E. 19



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Re: 3 digit integers [#permalink]
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03 Mar 2011, 11:45
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We know that the first digit (hundreth's place) would be 1 for all such numbers. Now second digit (ten's place) need to be an odd number for the third digit to be an integer arithmetic mean, so it can be 1,3,5,7,9 and the third digit would be 1,2,3,4,5 respectively giving us numbers like 111, 132, 153, 174 and 195. By similar logic, we would have numbers where tens and units digit in above numbers interchanged, so 123, 135, 147 and 159. We also need to worry about cases when the hundred's digit becomes the mean as 1 and in that case, tens and units digit are 0 and 2 and , so we add 102 and 120 to list as well. So, we have total of 11 numbers. Answer D



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Re: 3 digit integers [#permalink]
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16 Jun 2011, 00:21
let the digits be abc c can be 1,2,3,4 only as c>4 means average > 9.Not possible. 1 (1,1);(2,0) 2 (1,3) 3 (1,5) 4 (1,7) now (2,0) can be 2 times so (1,3),(1,5) and (1,7) thus 5+2+2+2 = 11
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Re: How many 3digit integers between 100 and 200 have a digit [#permalink]
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14 Aug 2012, 21:56
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Let the three digit number be xyz X can have only one value that is 1 (as its between 100 and 200) x = (y+z)/2 x is 1 so y + z = 2 so y can be 0 or 1 or 2 and corresponding value of z will be 2, 1, 0 so we have three numbers as of now 102, 111, 120 Now considering different values of y y = (x+z)/2 and x is 1 so x+z will be even only when z will be odd so we can have 5 possible values of z i.e. 1,3,5,7,9 corresponding values of y will be 1,2,3,4,5 so we have following numbers now 111, 123,135,147,159 Now considering different values of z we have z = (x+y) /2 and x=1 => z= (1+y)/2 which is possible only for odd values of y so y can be 1,3,5,7,9 corresponding values of z will be 1,2,3,4,5 So we get following numbers 111,132,153,174,195 So answer will be 102, 111, 120, 123,132, 135,147,153,159,174, 195 So, Answer is 11 Hope it helps!
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Re: How many 3digit integers between 100 and 200 have a digit [#permalink]
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18 Aug 2012, 00:51
Thank you so much. That was very helpful and is now totally clear. What would you say about the difficulty of this problem? How would you rate it?
Thanks



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Re: How many 3digit integers between 100 and 200 have a digit [#permalink]
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18 Aug 2012, 02:15
welcome, actually the problem was bit lengthy. It was not tough though. So i guess 600700 is the right tag for the problem.
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nevergiveup wrote: Since the number is between 100 and 200, the hundreds digit has to be 1. And since the answer choices are small enough, it won't hurt just to write them out. 102, 111, 120, 123, 132, 135, 147, 153, 159, 174, 195.
Answer D Is there another approach for this question? How do you pick the tenths and units digit? It isn't too clear to me. The first part with the hundreds digit was fine but then the average stuff wasn't that clear Thanks Cheers J



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Re: How many 3digit integers between 100 and 200 have a digit [#permalink]
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For jlgdr: The first digit is obviously 1. A) Let assume that the third digit is the mean of the first and second digits: for the second, we have to choose an odd number to result in an even one when added to 1(first digit), since the digits must be whole numbers(and odd+even results in odd and cannot be divided by 2 to get the other digit). >>> We have 5 options for the second digit. B) The other way around is true if assuming the second digit to be the mean. >>> Another 5 options. NOTE: The 111 is the number that we counted twice in scenario A and scenario B.>>> So far we have 5+51=9 options. C) Lastly, assuming that 1 (first digit) is the mean, we reach 102,120, and 111(counted before). 2 more options >>>> we have 9+2=11 choices. Answer is D. P.S. This approach is not preferable unless you master the process. For me, it took less time than counting. I am not good at counting though! Hope it helps...



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Re: How many 3digit integers between 100 and 200 have a digit [#permalink]
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29 Dec 2013, 14:27
panacia wrote: For jlgdr: The first digit is obviously 1. A) Let assume that the third digit is the mean of the first and second digits: for the second, we have to choose an odd number to result in an even one when added to 1(first digit), since the digits must be whole numbers(and odd+even results in odd and cannot be divided by 2 to get the other digit). >>> We have 5 options for the second digit. B) The other way around is true if assuming the second digit to be the mean. >>> Another 5 options. NOTE: The 111 is the number that we counted twice in scenario A and scenario B.>>> So far we have 5+51=9 options. C) Lastly, assuming that 1 (first digit) is the mean, we reach 102,120, and 111(counted before). 2 more options >>>> we have 9+2=11 choices. Answer is D. P.S. This approach is not preferable unless you master the process. For me, it took less time than counting. I am not good at counting though! Hope it helps... Yeah me neither, I prefer these conceptual approaches or method of counting (combinatorics) rather than listing options by process of brainstorming Kudos to you!! Cheers J



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Re: How many 3digit integers between 100 and 200 have a digit [#permalink]
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21 Apr 2014, 07:37
nktdotgupta wrote: Let the three digit number be xyz
X can have only one value that is 1 (as its between 100 and 200) x = (y+z)/2 x is 1 so y + z = 2 so y can be 0 or 1 or 2 and corresponding value of z will be 2, 1, 0 so we have three numbers as of now 102, 111, 120
Now considering different values of y y = (x+z)/2 and x is 1 so x+z will be even only when z will be odd so we can have 5 possible values of z i.e. 1,3,5,7,9
corresponding values of y will be 1,2,3,4,5
so we have following numbers now 111, 123,135,147,159
Now considering different values of z we have z = (x+y) /2 and x=1 => z= (1+y)/2
which is possible only for odd values of y so y can be 1,3,5,7,9 corresponding values of z will be 1,2,3,4,5 So we get following numbers 111,132,153,174,195
So answer will be 102, 111, 120, 123,132, 135,147,153,159,174, 195
So, Answer is 11
Hope it helps! Is there any way to know that we counted two numbers twice without actually writing them down?



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