Download DIVISIBILITY BY (10N+1) THM.1 If a number is divisible 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 subtract N times that same value and
you get a new number divisible also by (10N+1). Also the new number is not divisible by
(10N+1) if the original was not. (Note: if a number has k digits this has the effect of taking the
left k-1 digits and subtracting N times the right digit)
Corollary: If a number is divisible by something it is also divisible by its factors and thus the above
can be used for the factors also. (Thus, for example, by using 21 in the above theorem we could
also use it for 7 or 3)
Examples
Divisibility by11
1991 -> 199-1=198 -> 19-8=11
Thus both 198 and 1991 are divisible by 11 since 11 is divisible by 11.
Divisibility by 11
93973 -> 9397-3=9394 -> 939-4=935 -> 93-5=88 ( you can even go to 8-8=0 which counts as a
multiple of 11) Thus since 88 is divisible by 11 so is 935, 9394, and 93973.
Non-Divisibility by 11
1578=149 -> 14-9=5 since 5 is not divisible by 11 neither is 149 nor 1578.
Non-Divisibility by 21
7653 -> 765-2*3=759 -> 75-2*9=57 -> 5-2*7=-9 which is not divisible by 21, so neither is 57,
759, or 7653.
Divisibility by 31
24304 -> 2430-3*4=2418 -> 241-3*8=217 ->21-3*7=0 which is divisible by 31, thus, so is 217,
2418, and 24304.
Non-Divisibility by 31
56734 -> 5673-3*4=5661 -> 566-3*1=563 -> 56-3*3=47 and this is not divisible by 31 so neither
is 563, 5661, or 56734.
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 subtracted (with the last one
being the one taken leaving only one left) and putting them right to left as the digits of the
number.
Example: 11
Example: 11
8195 -> 819-5
93973 -> 9397-3
= 814 -> 81-4
9394 -> 939-4
= 77
935 -> 93-5
=88
thus 745 *11 = 8915
thus 8543 *11 = 93973
NOTE: The multiple at any step can be found by only taking the digits back to that step.
Example: 21
Example: 31
9492 -> 949-2*2
24304 -> 2430-3*4
=945 -> 94-2*5
2418 -> 241-3*8
=84
217 -> 21-3*7
=0
Thus 452 * 21 = 9492
Thus 784 * 31 = 24304
also 945 = 45 *21
also 2418 = 78 * 31
and 84 = 4 * 21
and 217 = 7 * 31
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 not
divisible by (10N-1) if 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:
52047 -> 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 and Nondivisibiliy by 19
10906 -> 1090+2*6=1102 -> 110+2*2=114 -> 11+2*4=19 which is divisible by 19 and thus 114,
1102, and 10906 are also divisible by 19.
1784 -> 178+2*4=186 -> 18+2*6=30 which is not divisible by 19 and thus neither is 186 nor
1784 divisible by 19.
Examples: Divisibility by 29 and Nondivisibility by 29
18908 -> 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.
DIVISIBILITY BY (10N-1) THEOREM 2. In the above process the multiple of (10N-1) used to
get any step in the original can be found by taking the numbers and taking 9 minus each one times
10 raised to the power= ( step at which it was removed ( in process of level wanted -1). (i.e.
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 1 to these for the number which is the multiple of
(10N-1).
DIVISIBILITY EXAMPLES USING 9
Example:
52047 ->5204+7=5211
9-7=2
(9 minus 4th)
5211 -> 521 +1=522
9-1=8
( 9 minus 2nd)
522 -> 522+2=54
9-2=7
(9 minus 3rd)
54
9-4=5
(9 minus 4th)
Thus (5+1) * 9 = 54
(57+1) * 9 = 522
(578+1) * 9 = 5211
(5782+1) * 9 = 52047
Example:
7074 -> 707+4=711
9-4=5
711 -> 71+1=72
9-1=8
72
9-2=7
Thus
(7+1) * 9 = 72
(78+1) * 9 = 711
(785+1) * 9 = 7074
Example:
40680 -> 4068+0=4068
9-0=9
4068 -> 406+8=414
9-8=1
414 -> 41+4=45
9-4=5
45
9-5=4
Thus
(4+1) * 9 = 45
(45+1) * 9 = 414
(451+1) * 9 = 4068
(4519+1) * 9 = 40680
Example:
90000 -> 9000+0=9000
9-0=9
9000 -> 900+0=900
9-0=9
900 -> 90+0=90
9-0=9
90
9-0=9
Thus
(9+1) * 9 = 90
(99+1) * 9 = 900
(999+1) * 9 = 9000
(9999+1) * 9 = 90000
Example Divisibility by 19
10906 ->
1090+2*6=1102
9-6=3 (9 minus 1st)
1102 ->
110+2*2=114
9-2=7 (9 minus 2nd)
114 ->
11+2*4=19
9-4=5
19
Thus
9-9=0
(0+1) * 19 = 19
(05+1) * 19 = 114
(057+1) * 19 = 1102
(0573+1) * 19 = 10906
Example: Divisibility by 29
18908 -> 1890+3*8=1914
9-8=1
1914 -> 191+3*4=203
9-4=5
203 -> 20+3*3=29
9-3=6
Thus
(6+1) * 29 = 203
(65+1) * 29 = 1914
(651+1) * 29 = 18908
PROOFS
PROOF (Divisibility by 10N+1 Theorem 1)
Let a number be divisible by 10N+1 and call it (10N+1)k. We can write this as
(10N+1)k=10n+c for 0<c<10. Then note that (10N+1)k-c)/10=n is an integer. Now (10N+1)k-c
10
-Nc is also an integer and equals (10N+1)k-c - 10Nc = (10N+1)k-(10n+1)c and note that the top
10
10
10
is divisible by (10N+1) so this number taken by subtracking c, dividing by 10, and subtracting Nc
is again divisible by (10N+1).
Now if some number B is not divisible by (10N+1) then B=10n+c, with 0<c<10, has that
(B-c)/10=n is still an integer. Now B-c -Nc is also an integer and equals B-c - 10Nc = B10
10
10
(10N+1)c is, then B-(10N+1)c is not divisible by (10N+1).
PROOF
(Divisibility by 10N+1 Theorem 2 )
(By induction.) It is true when the multiple of (10N+1) has only one digit since (10N+1)*c=Kc
for some set of digits K. Assume it is true up to multiples of k-1 digits (ie. f1f 2...fk = (bk-1b k-2...b2
b1 * (10N+1) where the b1, b2,..bk-1 are found as above)
Consider a number a1a 2...ak+1 divisible by (10N+1).
Then (A)