Download DIVISIBILITY BY (10N-1)

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.