Divisibility by 11: Proving the Alternating Sum Method

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A number is divisible by 11 if the alternating sum of its digits is divisible by 11. The proof begins with a two-digit number represented as n = a*10 + b, where a is the first digit and b is the second digit. By rewriting n as a*10 + ((b - a) + a), it shows that a*10 + a is always divisible by 11. Therefore, if (b - a) is also divisible by 11, then the entire number n is divisible by 11. This method effectively demonstrates the divisibility rule for 11.
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Homework Statement



what is the test to to see if a number is divisible by 11 and prove it.


The Attempt at a Solution



If the alternating sum of a numbers digits is divisble by 11 then so is the number.


I don't know how to prove it tho.
 
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Do a simple case. Take a two digit number n. First digit a, second digit b. So n=a*10+b. Rewrite this as n=a*10+((b-a)+a)=a*10+a+(b-a). Now a*10+a is always divisible by 11 (why?). So if (b-a) is divisible by 11, then n is.
 
because it equals 11a.

thanks!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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