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DIVISIBILITY BY (10N-1) Thm. 1 If a number is divisible by (10N-1) then you can take the number and subtract the ones digit then divide by 10 and add N times the value subtracted off back again and you get a new number divisible also by (10N-1). Also the new number is no0t divisible by (10N-1) uf the original was not. (Note: if a number has k digits this has the effect of taking the left k-1 digits and adding N times the right digit) Example Divisibility by 9: 52947 -> 5204+7=5211 -> 521+1=522 -> 52+2=54. Thus, since 54 is divisible by 9 so is 522, 5211, and 52047. Example Divisibility by 9: 7074 -> 707+4=711 -> 71+1=72 (you can even keep going to 7+2=9) Thus since 72 is divisible by 9 so is 711 and 7074. Example: 789 -> 78+9=87 which is not divisible by 9 so neither is 789. Examples: Divisibility by 19 10906 -> 1890+3*8=1914 -> 191+3*4=203 -> 20-3*3=29 which is divisible by 29 and thus so is 203, 1914, and 18908 divisible by 29. 1543 -> 154=3*3=163 -> 16+3*3=25 is not divisible by 29 so neither is 163 nor 1543 divisible by 29. PROOF: Let a number be divisible by (10N-1) and call it (10N-1)k. We can write this as (10N- 1)k=10n+c with 0<c<10. Then note that ((10N-1)k-c)/10=n is an integer. Now (10N-1) k-c +Nc 10 is also an integer and equals (10N-1) k-c + 10Nc = (10N-1) k+ (10N-1)c and note that the top is 10 10 10 divisible by (10N-1) so this new number taken by subtracting c dividing that by 10 and adding Nc back is again divisible by (10N-1). Now if some number B is not divisible by (10N-1) then B=10n+c is such that (B-c)/10=n is still an integer. Now B-c +Nc is also an integer and equals B-c + 10Nc = B+(10N-1)c but since B is not 10 10 10 10 divisible by (10N-1) and (10N-1)c is, then B-(10N-1)c is not divisible by (10N-1). DIVISIBILITY BY (10N-1) THEOREM 2. In the above process the multiple of (10N-1) used to get the original value can be found by taking the numbers and taking 10 minus the first one then 9 minus each subsequent one times 10 raised to the power equal to the number of the step minus one at which it was removed. (Ie. 10^1 for the second taken off, 10^2 for the third taken off etc.) Do this until the last one taken occurs when there is only two digits left. Add these for the number which is the multiple of (10N-1). Example: 52047 -> 5204+7=5211 10-7=3 (10 minus first number) 5211 -> 521+1=522 9-1=8 522 -> 9-2=7 522+2=54 54 9-4=5 5783 thus 5783 * 9 = 52047 (9 minus all other numbers) Example: 40680 -> 707+4=711 10-4=6 711 -> 71+1=72 9-1=8 (times 10) 72 9-2=7 (times 10^2) 786 thus 786 * 9 = 7074 Example: 40680 -> 4068+0=4068 10-0=10 4068 -> 406+8=414 9-8=1 414 -> 41+4=45 9-5=4 4520 thus 786 * 9 =40680 Example: 90000 -> 9000+0=9000 10-0=10 9000 -> 900+0+900 9-0=9 900 -> 90+0=90 9-0=9 10000 thus 10000 * 9 = 90000 Example: Divisibility by 19 10906-> 1090+2*6=1102 1102 -> 110+2*2=114 10-6=4 9-2=7 (10 minus first number) (9 minus all other numbers) 114 -> 11+2*4=19 9-4=5 19 9-9=0 574 thus 574 * 19 = 10906 Example: Divisibility by 29 18908 -> 1890+3*8=1914 10-8=2 1914 -> 191+3*4=203 9-4=5 (times 10) 203 -> 20+3*3=29 9-3=6 (Times 10^2) 29 9-9=0 (Times 10^3) 0652 thus 652 * 29 = 18908 NOTE: The numbers can be found at each preceding step by only taking the multiple back to that level and adding 1. Example: Divisibility by 29 18908 -> 1890+3*8=1914 10-8=2 1914 -> 191+3*4=203 9-4=5 (times 10) 203 -> 20+3*3=29 9-3=6 (Times 10^2) 29 9-9=0 0652 thus 652 * 29 = 18908 (Times 10^3) Example: Divisibility by 19 10906-> 1090+2*6=1102 10-6=4 1102 -> 110+2*2=114 9-2=7 114 -> 11+2*4=19 9-4=5 19 9-9=0 (10 minus first number) (9 minus all other numbers) 574 thus 574 * 19 = 10906 but also (57+1) *19=1102 (5+1) *19=114 PROOF: (By induction.) It can be shown true for all numbers with multiples of 1 digit. (Note this can be shown by the fact that the a1 a2 . . . an 9*a = b1 b2 . . . bk and 10-bk = a). Assume it is true for numbers which are (10N-1) times numbers with k-1 digits. (A) Consider a1 a2 . . . ak+1 +Nak+1 = f1f2...fk (By thm. 1 above) 10 Which (also by thm. 1 above) is divisible by (10N-1). Now this either is (10N-1) times a number that is k-1 digits or will reduce to one that is. Thus we will show it true for those which have f1f2...fk which are (10N-1) times a nuber with k-1 digits, then by the fact that the other numbers with k+1 digits reduce to these we can use an identical process on those. Thus by the induction step (B) f1f2...fk = (bk-1b k-2...b2b 1) * (10N-1) for some b1b 2,...,bk-1 found as stated in the theorem. Since the element b1 was the difference from 10 for the last number and we need it to be the difference from 9 if the theorem is to hold, we must subtract 1 from bk-1b k-2 ...b1-1. Also since ak+1 is the first number added on we need (10-ak+1 ) as the additional digit on the right. Thus we must show (10 * (bk-1bk-2...b1-1) + (10-ak+1) ) * (10N-1) = a1a2...ak+1 Now (10 * (bk-1b k-2...b1-1) + (10-ak+1) ) * (10N-1)= (10 * (f1f 2...fk -1) + (10-ak+1) ) * (10N-1) 10N-1 (By (B) above) = 10 * (f1f 2...fk - (10N-1)) + (10N-1) * (10-ak+1) But f1f 2 ...fk= a1 a2 ...ak+1-ak+1 + Nak+1 10 =a1a 2...ak+1 + (10N-1)ak+1 (By (A) above) 10 Thus by substituting we have: 10 * (a1a 2...ak+1 + (10N-1) ak+1 - (10N-1)) 10 +100N-10-(10N-1)ak+1 = a1a 2...ak+1 + (10N-1)ak+1-100N+10+100N-10-(10N-1)ak+1 = a1a 2...ak+1 Thus it works for numbers with multiples of k digits that reduce to numbers which are (10N-1) times numbers with k-1 digits. Now since all other numbers with multiples of k digits will reduce to these by the above process, then by doing an identical proof on them we get that it works for all numbers with multiples of k digits.