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by shashank.ism » Wed Feb 10, 2010 6:12 am
Consider a positive integer Z which when represented in decimal base,does not end in zero. Z and the number obtained by reversing the digits of Z are both multiples of seven. Let the number of such Zs in the set 10 ,11,12... 998,999,1000 be K. Then k is given by which of the following?

a) 13
b) 14
c) 15
d) 16
e) 17

How to solve this problem
Ans E
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by ldoolitt » Wed Feb 10, 2010 12:27 pm
I stared at this for a couple of minutes then decided to start crunching with the divisibility rule.

Obviously there is one number less than 100 that fits, 77.

Above 100 and less than 1000...

I crunched and got to 161 which works (16 - 2 *(1) =14) both ways.
Then got to 168 which works both ways (16- 2 * (8) and 86 - 2 *(1))

Then it was a long haul to 252 which works both ways. Then I noted a pattern that I couldn't explain: For a three digit number if you add the first two digits and the sum is divisible by 7 and add the last two digits and the sum divisible by 7 then the number and its reverse are divisible by 7.

I tested it and sure enough it works. There are 18 numbers between 100-1000 that fit this. Taking out 770 and 700 and adding in 77 I get 17. (e)

I can prove this rule (with some formal arithmetic but I'm not sure how you would derive this on a test...
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by shashank.ism » Mon Feb 15, 2010 12:09 pm
ldoolitt wrote:I stared at this for a couple of minutes then decided to start crunching with the divisibility rule.

Obviously there is one number less than 100 that fits, 77.

Above 100 and less than 1000...

I crunched and got to 161 which works (16 - 2 *(1) =14) both ways.
Then got to 168 which works both ways (16- 2 * (8) and 86 - 2 *(1))

Then it was a long haul to 252 which works both ways. Then I noted a pattern that I couldn't explain: For a three digit number if you add the first two digits and the sum is divisible by 7 and add the last two digits and the sum divisible by 7 then the number and its reverse are divisible by 7.

I tested it and sure enough it works. There are 18 numbers between 100-1000 that fit this. Taking out 770 and 700 and adding in 77 I get 17. (e)

I can prove this rule (with some formal arithmetic but I'm not sure how you would derive this on a test...
you can get divisblity rules at this link. ope this would help u better....
https://en.wikipedia.org/wiki/Divisibility_rule
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by ldoolitt » Mon Feb 15, 2010 1:54 pm
Sorry, still couldn't find a faster way to do it than "first 2 digits sum and last two digits sum are divisible by 7." Is there a faster way?
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by vijay_venky » Mon Feb 15, 2010 9:49 pm
First of all let us take two digit numbers

A number and the reverse of the same should be multiples of 7 and this could be represented as,

10a+b=7X
10b+a=7Y

and 9(a-b)=7(X-Y)

Now either a-b=0 and X-Y=0 (means a=b {77})
or a-b should be a multiple of 7 and X-Y a multiple of 9

And then coming to three digit numbers
100p+10q+r=7M
100r+10q+p=7N

Now, 99(p-r)=7(M-N)
again p-r=0 and M-N=0
Here the first and the last number of the three digit number should be the same. And the number should be a multiple of 7
(means p=r and is satisfied by {161,252,343,434,525,595,616,686,707,777,868,959})
or p-r should be a multiple of 7 and M-N a multiple of 99
p-r a multiple of 7, means the difference between the first and the last number of a three digit number is a multiple of 7 {(168,861),(259,952)}

so a total of 17 numbers
Hope this helps
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by shashank.ism » Mon Feb 15, 2010 9:54 pm
ldoolitt wrote:Sorry, still couldn't find a faster way to do it than "first 2 digits sum and last two digits sum are divisible by 7." Is there a faster way?
No I don't think there is a faster way to check divisiblity by 7 . Though venky has proposed a good solution with a good methodical approach.
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