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# Hard Maths question watch

1. 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
2. (Original post by CraigBackner)

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.

Spoiler:
Show

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

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Updated: January 16, 2017
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