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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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+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).

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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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

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 sense

Kudos for the approach
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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 3-digit 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: 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

Re: How many integers from 101 to 800, inclusive, remains the [#permalink]

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17 Aug 2014, 03:45

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Re: How many integers from 101 to 800, inclusive, remains the [#permalink]

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07 Sep 2015, 20:05

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