Even-Digit Palindromes Divisible by 11Date: 12/08/97 at 14:14:27 From: Bob Subject: Even-Digit Palindromes divisible by 11 When I play with a calculator or computer, I notice that every even- digit palindromic number is divisible by 11: 6006/11 = 546 135531/11 = 12321 64899846/11 = 5899986 Could this be proved as a theorem? Bob Date: 12/08/97 at 16:51:35 From: Doctor Rob Subject: Re: Even-Digit Palindromes divisible by 11 Yes, it can. Even-digit palindromes are of the following form. They are a sum of terms, each consisting of a single digit times a power of 10 times a number which is one greater than an odd power of 10. Example: 64899846 = 60000006 + 4000040 + 800800 + 99000 = 6*10^0*(10^7+1) + 4*10^1*(10^5+1) + 8*10^2*(10^3+1) + 9*10^3*(10^1+1) The key is that 11 always divides evenly into 10^e+1 if e is odd. Prove this by proving that x+1 divides evenly into x^e+1 if e is odd, and then substitute x = 10. The quotient is x^(e-1) - x^(e-2) + ... + x^2 - x + 1 This can also be proved using mathematical induction. Since 11 divides evenly into 10^e+1, it divides evenly into every term in the sum, and thus it divides evenly into the palindrome in question. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 12/08/97 at 23:34:51 From: FTW1 Subject: Re: Even-Digit Palindromes divisible by 11 Thanks Dr. Rob! Bob |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/