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How many integers from 101 to 800, inclusive, remains the [#permalink]
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14 Aug 2011, 11:10
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How many integers from 101 to 800, inclusive, remains the value unchanged when the digits were reversed? (A) 50 (B) 60 (C) 70 (D) 80 (E) 90
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Last edited by Bunuel on 18 Feb 2013, 04:24, edited 2 times in total.
Renamed the topic and added OA.



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Re: NS  Integers!! [#permalink]
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14 Aug 2011, 11:38
Was the answer (C)?
If so, I can explain how I got the answer...



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Re: NS  Integers!! [#permalink]
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14 Aug 2011, 11:42
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Take numbers between 101 and 199. Numbers that meet criteria 101,111,121,131,141,151,161,171,181,191. Total 10 number between 101 and 199 Similarly, 10 numbers between 200 and 300, 300 and 400, 400 and 500, 500 and 600, 600 and 700, 700 and 800 for total of 70 numbers. OA C
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Re: NS  Integers!! [#permalink]
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14 Aug 2011, 11:49
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+1 for C. This will be a 3 digit number and it remains same when reversing the digit. So it means 1st and 3rd digit are same. Number of options for 1st and 3rd digit = 7 (1 to 7 inclusive) Number of options for 2nd Digit = 10 (0, 1, to 9)
Total 7 x 10 x 1 = 70 (3rd Digit is already selected when the 1st is selected).



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Re: NS  Integers!! [#permalink]
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14 Aug 2011, 11:56
jamifahad  I did the same way as urs..thanx for the explanation.
timeishere  I liked ur method..thanx a ton.
Kudos to both of you!



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Re: NS  Integers!! [#permalink]
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14 Aug 2011, 12:10
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DeeptiM wrote: jamifahad  I did the same way as urs..thanx for the explanation.
timeishere  I liked ur method..thanx a ton.
Kudos to both of you! Thanks. Press the Kudos if you meant to...



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Re: NS  Integers!! [#permalink]
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14 Aug 2011, 12:29
Yep, I just wrote out the numbers for the first hundred. I dont think there is a quicker or more efficient way to reach the answer.
Last edited by restore on 14 Aug 2011, 16:50, edited 1 time in total.



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Re: NS  Integers!! [#permalink]
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14 Aug 2011, 12:31
restore wrote: Yep, I just wrote out the numbers for the first hundred. I dont think there is a quicker or more efficient way to reach the answer. But what if it is a 4 digit or 5 digit number???



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Re: NS  Integers!! [#permalink]
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14 Aug 2011, 12:34
timeishere wrote: DeeptiM wrote: jamifahad  I did the same way as urs..thanx for the explanation.
timeishere  I liked ur method..thanx a ton.
Kudos to both of you! Thanks. Press the Kudos if you meant to... Ohh now i knw y my kudos arn't reflecting..ooopppsiee thnx for ltng me knw, will do the needful.



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Re: NS  Integers!! [#permalink]
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14 Aug 2011, 12:45
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DeeptiM wrote: timeishere wrote: DeeptiM wrote: jamifahad  I did the same way as urs..thanx for the explanation.
timeishere  I liked ur method..thanx a ton.
Kudos to both of you! Thanks. Press the Kudos if you meant to... Ohh now i knw y my kudos arn't reflecting..ooopppsiee thnx for ltng me knw, will do the needful. I clicked the button..hope you got it this time



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Re: NS  Integers!! [#permalink]
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14 Aug 2011, 16:43
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question is asking for palindrome
first digit possibilities  1 through 7 = 7
8 is not possible here because it would result in a number greater than 8 (i.e 808 , 818..)
second digit possibilities  0 though 9 = 10
third digit is same as first digit
=>total possible number meeting the given conditions = 7 *10 = 70
Answer is C.



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Re: NS  Integers!! [#permalink]
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14 Aug 2011, 18:14
timeishere wrote: But what if it is a 4 digit or 5 digit number??? How many integers from 10001 to 80008, inclusive, remains the value unchanged when the digits were reversed? 3rd digit can take 10 values 1st 2 digits can take 10x7 = 70 values 80008 is also correct So total number of integers = 10x70+1=701
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Re: NS  Integers!! [#permalink]
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14 Aug 2011, 21:31
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another way to think of it is like this
I made it 100 just to simplify the below
100 < _ _ _ <= 800 which is the same as 101 <= _ _ _ <= 800
so you have a 3 digit number that must be a palindrome.
you have 7 ways to choose the first number (realize that you can't choose the number 8 because there are no palindromes <= 800) , you have 10 ways to choose the second number, but you only have 1 way to choose the 3rd number, because the third number HAS to be the 1st number
7*10*1 = 70
then 5 digit number you can do the following
10000 < _ _ _ _ _ <= 800000
again you have 7 * 10 * 10 = 700
** the other response was for 80008 so that would be another palindrome that's why that answer is 701



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Re: NS  Integers!! [#permalink]
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15 Aug 2011, 07:57
Spidy001 wrote: question is asking for palindrome
first digit possibilities  1 through 7 = 7
8 is not possible here because it would result in a number greater than 8 (i.e 808 , 818..)
second digit possibilities  0 though 9 = 10
third digit is same as first digit
=>total possible number meeting the given conditions = 7 *10 = 70
Answer is C. pinchharmonic wrote: another way to think of it is like this
I made it 100 just to simplify the below
100 < _ _ _ <= 800 which is the same as 101 <= _ _ _ <= 800
so you have a 3 digit number that must be a palindrome.
you have 7 ways to choose the first number (realize that you can't choose the number 8 because there are no palindromes <= 800) , you have 10 ways to choose the second number, but you only have 1 way to choose the 3rd number, because the third number HAS to be the 1st number
7*10*1 = 70
then 5 digit number you can do the following
10000 < _ _ _ _ _ <= 800000
again you have 7 * 10 * 10 = 700
** the other response was for 80008 so that would be another palindrome that's why that answer is 701 Yups, I like the palindrome approach. Let us remove the unnecessary jargon and say this is a simple approach with little logic and common senseKudos for the approach
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Re: NS  Integers!! [#permalink]
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15 Aug 2011, 11:49
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101 111 121 131 141 151 161 171 181 191 202 222 212 232 242 252 262 272 282 292 303 etc 404 505 606 707 so we have always 7 numbers vertically and 10 numbers horizontally. so 7*10 =70 actually the answ is predictable. from the beginning u know that u get a set of seven 3digit numbers (111 ;222; 333 ;*** ;777). so u understand that ur answer choice should be divisible by 7. only 70 is divisible by 7
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Re: NS  Integers!! [#permalink]
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18 Feb 2013, 02:41
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My approach:
Let the no. be N = xyz = 100x + 10y + z .................[1]
Reversed N = N' = zyx = 100z + 10y + x ................ [2]
No. if N = N' then N  N' = 0
or [1]  [2] = 0
i.e. 99 (xz) = 0
or x = z
from 100 to 800 both x & y will take 7 values viz. 1, 2, 3 ...7
However digit y inbetween the three digit no X_Z will take 10 values from 09.
Therefore, total no. satisfying the given condition in the question = 7*10 = 70
To make my sol. more visual
Series of 100 > 1  (09)  1
Series of 200 > 2  (09)  2 . . . . Series of 700 > 7  (09) 7 .... last no. of this series would be 797



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Re: How many integers from 101 to 800, inclusive, remains the [#permalink]
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22 Feb 2013, 22:59
This is how I figured it out: The value will remain same only if all the digits are same OR if the first and the last digits are same. Hence between 101  800 : Total number of nos with same digits is 7 i.e. 111, 222, 333, 444, 555, 666, 777
Now for the first and the last digits as same.
since the range is between 101  800, Out of the 3 digits the first digit can ( or should ) be between 1  7 Total Number of ways in which 7 Numbers can be chosen is 7C1 = 7 ways. This chosen number is same as the 3 rd digit hence the number of ways to chose this same digit is 1 The Middle number can be between 0  9 ( exclusive of the number chosen as the first digit ) Therefore, total number of nos is 9: hence total number of ways to choose a number from 9 : 9C1 = 9
Therefore total number of digits is : 9 * 7 = 63 Total number of same 3 digits : 7 ==> 63 + 7 = 70
Hope I am right and hope this is clear



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Re: How many integers from 101 to 800, inclusive, remains the [#permalink]
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12 Jul 2017, 11:11
10 in an each set of 100 (from 101 to 200  101,111,121,131,141,151,161,171,181,191)
there are 7 sets. so ans = 7X10 =70 (C).




Re: How many integers from 101 to 800, inclusive, remains the
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