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    how would you solve this without using trial and error?

    I've tried representing it using algebra but it doesn't get me anywhere

    I represented a 3 digit number as 100a+10b+c and created this inequality by reversing it and adding it together:

    100a+10b+c+100c+10b+a<100

    simplifying it, i get: 101a+20b+101c<100

    where do i go from here?

    any help will be appreciated
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    (Original post by CraigBackner)
    Name:  AQA 90 Questions.pdf - Google Chrome 16_01_2017 20_51_39.png
Views: 21
Size:  27.8 KB
    how would you solve this without using trial and error?

    I've tried representing it using algebra but it doesn't get me anywhere

    I represented a 3 digit number as 100a+10b+c and created this inequality by reversing it and adding it together:

    100a+10b+c+100c+10b+a<100

    simplifying it, i get: 101a+20b+101c<100

    where do i go from here?

    any help will be appreciated
    You don't really need algebra here.

    We're looking for the biggest number so start with the 900s : we need the reversed number to be less than 100 otherwise the total would exceed 1000. But that's not possible since you can only use the digits 1 - 9. Therefore the number we're looking for cannot begin with a 9.

    Now think about the 800s:

    Spoiler:
    Show

    The reversed number can't exceed 200 so must begin with a 1. How does that help?
 
 
 
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