My Math Forum divisibility by 3

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 April 7th, 2014, 11:17 AM #2 Newbie   Joined: Apr 2014 From: Pleven Posts: 2 Thanks: 0 Found the answer. If our number is abcd then we have 1000a + 100b + 10c + d. Whe are gonna subtract dcba from it which is 1000d + 100c + 10b +a 1000a + 100b + 10c + d -(1000d + 100c + 10b +a)=1000a + 100b + 10c + d - 1000d - 100c - 10b - a =999a + 90b - 90c - 999d= 9(111a + 10b - 10c - 111d)
 April 7th, 2014, 11:35 AM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs You will find such a difference is in fact always divisible by 9. Consider the $n$-digit decimal number: $\displaystyle N=\sum_{k=0}^n\left(a_k10^k \right)$ And so the difference D you describe is: $\displaystyle D=\sum_{k=0}^n\left(a_k10^k \right)-\sum_{k=0}^n\left(a_{k}10^{n-k} \right)$ $\displaystyle D=\sum_{k=0}^n\left(a_k\left(10^k-10^{n-k} \right) \right)$ $\displaystyle D=\sum_{k=0}^n\left(a_k10^{n-k}\left(10^{2k-n}-1 \right) \right)$ A power of ten less 1 is always divisible by 9, and we see that each term has such a factor. Thanks from agentredlum, Olinguito and nicksona
 April 7th, 2014, 11:40 AM #4 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 An integer is divisible by 3 IF AND ONLY IF the sum of the digits is divisible by 3. The sum of the digits of an integer is called the digital root , example $$DR(142536) = 1+ 4+ 2+ 5 +3 +6 = 21$$ The $DR$ is iterative. $$DR( 142536) = DR(21) = 2 + 1 = DR(3)$$ 21 is divisible by 3 , that makes 142536 divisible by 3 and furthermore , those digits in any order would form a number divisible by 3. Iteration helps when the number has many digits , you apply the $DR$ procedure until the answer becomes obvious , technically until you get a single digit ... that single digit is then the true $DR$. Thanks from MarkFL and nicksona
 April 7th, 2014, 04:31 PM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra That doesn't quite answer the question, since not every number has the digital root 3 and we want to know about the digital root of the difference between two numbers containing the same digits. You need to prove that $DR(a - b) \equiv DR(a) - DR(b) \mod 3$.
 April 7th, 2014, 09:27 PM #6 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 The $DR$ of any positive integer is a single digit from 1 to 9 ($DR(0) = 0$) So it is quite easy to observe that 3 and 9 are the only contestants that work for the relation $$DR(a - b) = DR(a) - DR(b)$$ by eliminating the other contestants with counterexamles. I make no claim that what follows is a rigorous proof. Counterexample to contestant 2 $$2 = DR(3 - 1) = DR(30) - DR(01)$$ $$30 - 01 = 29$$ NOT divsible by 2 Counterexample to contestant 4 $$4 = DR(5 - 1) = DR(50) - DR(01)$$ $$50 - 01 = 49$$ NOT divsible by 4 Counterexample to contestant 5 $$5 = DR(6 - 1) = DR(60) -DR(01)$$ $$60 - 01 = 59$$ NOT divsible by 5 Counterexample to contestant 6 $$6 = DR(7 - 1) = DR(70) -DR(01)$$ $$70 - 01 = 69$$ NOT divsible by 6 (but divisible by 3 ) Continue like this to eliminate contestants 7 and 8. Now , if we try to eliminate contestants 3 and 9 using this procedure we find it impossible to construct a counterexamle. IF the statement for 3 , 9 does not hold , THEN there should exst a counterexamle. Since a counterexamle is impossible then the statement for 3 , 9 must hold. Also , If $DR(a - b) = DR(a) - DR(b) = 0 , 3 , 6 , 9$ Then divisibility by 3 is gauranteed , no matter what positive integer values for a , b Finally , an answer to his question totally (If i have not misunderstood) Since he reverses digits and subtracts we can write $$DR(a - b) = DR(a) - DR(a) = 0$$ And 0 is divisible by 3 We do not need to know the numerical value of any DR , which is a bit surprising. Note $DR(a) = DR(b)$ 3 is really quite special IMHO. As MarkFL has pointed out , the OP procedure gaurantees that the number obtained is divisible by 9 BUT as far as $DR$ is concerned we need to throw out $DR(a-b) = 3 , 6$ If i made mistakes i apologize. Thanks from MarkFL and nicksona
 April 7th, 2014, 09:36 PM #7 Senior Member     Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra In general … Lemma: Take a number and swap any two of its digits. Then the difference between the old and new numbers is divisible by 9. Proof: Similar to the one by MarkFL. Corollary: Take a number and permute its digits in any way. Then the difference between the old and new numbers is divisible by 9. Proof: Follows from the fact that any permutation is a product of transpositions. For example: $$321-123\,=\,198\,=\,9\times22$$ $$321-132\,=\,189\,=\,9\times21$$ $$321-213\,=\,108\,=\,9\times12$$ $$321-231\,=\,90\,=\,9\times10$$ $$321-312\,=\,9\,=\,9\times1$$ $$321-321\,=\,0\,=\,9\times0$$ Thanks from MarkFL, agentredlum and nicksona

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