Explaining the Divisibility Rule for 7Date: 02/26/2003 at 10:40:57 From: Abhay Dang Subject: Proving divisibility rule for 7 I found in the Dr. Math FAQ the rule for divisibility of a number by 7. Why do these 'rules' work? http://mathforum.org/k12/mathtips/division.tips.html Please explain (prove) why the following rules work. Rule 1) Divide the number into groups of 3 starting from the right and give these groups alternating +/- signs. Apply the rule again if necessary until you get a three-digit number. The original number is divisible by 7 if this sum is. Eg. for 123456789 => 789 - 456 + 123 Rule 2) To know if a number is a multiple of seven or not, we can use also 3 coefficients (1, 2, 3). We multiply the first number starting from the ones place by 1, then the second from the right by 3, the third by 2, the fourth by -1, the fifth by -3, the sixth by -2, the seventh by 1, and so forth. Example: 348967129356876. 6 + 21 + 16 - 6 - 15 - 6 + 9 + 6 + 2 - 7 - 18 - 18 + 8 + 12 + 6 = 16 means the number is not multiple of seven. So the pattern is as follows: for a number onmlkjihgfedcba, calculate a + 3b + 2c - d - 3e - 2f + g + 3h + 2i - j - 3k - 2l + m + 3n + 2o. Example: 348967129356874. Below each digit let me write its respective figure. 3 4 8 9 6 7 1 2 9 3 5 6 8 7 4 2 3 1 -2 -3 -1 2 3 1 -2 -3 -1 2 3 1 (3*2) + (4*3) + (8*1) + (9*-2) + (6*-3) + (7*-1) + (1*2) + (2*3) + (9*1) + (3*-2) + (5*-3) + (6*-1) + (8*2) + (7*3) + (4*1) = 14 -- a multiple of seven. Kindly explain and prove why these rules work. Best regards Abhay Dang Date: 03/04/2003 at 13:02:51 From: Doctor Peterson Subject: Re: Proving divisibility rule for 7 Hi, Abhay. The key here is that 1000 = 7*143 - 1 Consequently, 1000^2 = (7*143 - 1)^2 = 7^2*143^2 - 2*7*143 + 1 If you think about how higher powers work, you will see that 1000^1 = 7A - 1 1000^2 = 7B + 1 1000^3 = 7C - 1 and so on (where A, B, C, and so on are some whole numbers). This can be better expressed in terms of modular arithmetic, if you are familiar at all with that; 1000^n is congruent (mod 7) to -1 when n is odd, and to +1 when n is even. So looking at the number "abc,def,ghi", and calling the numbers "abc," "def," and "ghi" X, Y, and Z respectively, we can write it as 1000^2 X + 1000 Y + Z which is equal to (7B+1)X + (7A+1)Y + Z = 7(BX + AY) + (X - Y + Z) Do you see that this will be divisible by 7 if and only if X-Y+Z is divisible by 7? The same pattern continues for additional digits. >Rule 2) This is very similar, using the facts that 1 = 1 10 = 7*1 + 3 100 = 7*14 + 2 1000 = 7*143 - 1 Can you fill in the rest of the reasoning? If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 03/05/2003 at 10:01:53 From: Abhay Dang Subject: Thank you (Proving divisibility rule for 7) Thanks a lot. You've been of great help ! |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/