Math problem - 17-digit number

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RikiMartin

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A 17-digit number is reversed and the two numbers are added. Prove that the sum has at least one even digit.
 
RikiMartin said:
A 17-digit number is reversed and the two numbers are added. Prove that the sum has at least one even digit.
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In particular, if you can't show any work, please at least tell us what you have learned that might apply. If this is for a course, what topics have been covered recently? What facts do you have on which to base a proof?
 
Think about the digit in the very middle. What does that get added to? So what parity does it have?
Otis said:
Hi Riki. What two numbers are added? Please check to ensure that you've provided us with all of the given information. Thank you!

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[MATH]u = d_{16} d_{15} \dots d_1 d_0 \\ u_r = d_0 d_1 \dots d_{15} d_{16}\\ U = u + u_r [/MATH]
It looks reminiscent of a pigeon hole problem but I'm not sure.
 
Thanks, Romsek! I'd mistakenly thought 'two digits' instead of two 17-digit numbers.

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Partial proof. The 8th digits of both numbers are the same so adding them will yield a number that ends in an even digit. Now if you add the 6th digits (from the right) and there is no carry over then you are done. Now what happens if the you do have a carry over?

Another thing I would think of that goes along with Romsek's thought is that the sum of the digits of the sum of the two numbers must be a multiple of 9.
 
Jomo said:
Partial proof. The 8th digits of both numbers are the same so adding them will yield a number that ends in an even digit. Now if you add the 6th digits (from the right) and there is no carry over then you are done. Now what happens if the you do have a carry over?

Another thing I would think of that goes along with Romsek's thought is that the sum of the digits of the sum of the two numbers must be a multiple of 9.
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I think it's only the difference that has to be a multiple of 9. But my first thought was to use casting out nines: the digit sum has to be even, which tells you something in the case where all digits are odd. I haven't had time to delve into it, though.

@RikiMartin: any thoughts yet? Or any background on the problem?
 
Dr.Peterson said:
I think it's only the difference that has to be a multiple of 9. But my first thought was to use casting out nines: the digit sum has to be even, which tells you something in the case where all digits are odd. I haven't had time to delve into it, though.

@RikiMartin: any thoughts yet? Or any background on the problem?
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Yes the difference is a multiply of 9. Corner time for 9 minutes. Thanks for pointing that out.
 
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