| re. 0
well,
looking at the multiplication tables of the numbers from 1 to 100
the following single zero and double zero split numbers were found:
single zero split:
15 * 7 = 105
18 * 6 = 108
45 * 9 = 405
double zero split:
11 * 91 = 1001
15 * 67 = 1005
18 * 56 = 1008
22 * 91 = 2002
33 * 91 = 3003
44 * 91 = 4004
45 * 89 = 4005
55 * 91 = 5005
66 * 91 = 6006
77 * 91 = 7007
88 * 91 = 8008
99 * 91 = 9009
Enjoy,
Kostas G.
use of calreal was made in the following way:
$assign junk.out pas$output
$ calreal
set up to (100);
multiplication table of ( 91 );
exit(1);
$
$type junk.out
91 * 1 = 91
91 * 2 = 182
91 * 3 = 273
91 * 4 = 364
91 * 5 = 455
91 * 6 = 546
91 * 7 = 637
91 * 8 = 728
91 * 9 = 819
91 * 10 = 910
91 * 11 = 1001
91 * 12 = 1092
91 * 13 = 1183
91 * 14 = 1274
91 * 15 = 1365
91 * 16 = 1456
91 * 17 = 1547
91 * 18 = 1638
91 * 19 = 1729
91 * 20 = 1820
91 * 21 = 1911
91 * 22 = 2002
91 * 23 = 2093
91 * 24 = 2184
91 * 25 = 2275
91 * 26 = 2366
91 * 27 = 2457
91 * 28 = 2548
91 * 29 = 2639
91 * 30 = 2730
91 * 31 = 2821
91 * 32 = 2912
91 * 33 = 3003
91 * 34 = 3094
91 * 35 = 3185
91 * 36 = 3276
91 * 37 = 3367
91 * 38 = 3458
91 * 39 = 3549
91 * 40 = 3640
91 * 41 = 3731
91 * 42 = 3822
91 * 43 = 3913
91 * 44 = 4004
91 * 45 = 4095
91 * 46 = 4186
91 * 47 = 4277
91 * 48 = 4368
91 * 49 = 4459
91 * 50 = 4550
91 * 51 = 4641
91 * 52 = 4732
91 * 53 = 4823
91 * 54 = 4914
91 * 55 = 5005
91 * 56 = 5096
91 * 57 = 5187
91 * 58 = 5278
91 * 59 = 5369
91 * 60 = 5460
91 * 61 = 5551
91 * 62 = 5642
91 * 63 = 5733
91 * 64 = 5824
91 * 65 = 5915
91 * 66 = 6006
91 * 67 = 6097
91 * 68 = 6188
91 * 69 = 6279
91 * 70 = 6370
91 * 71 = 6461
91 * 72 = 6552
91 * 73 = 6643
91 * 74 = 6734
91 * 75 = 6825
91 * 76 = 6916
91 * 77 = 7007
91 * 78 = 7098
91 * 79 = 7189
91 * 80 = 7280
91 * 81 = 7371
91 * 82 = 7462
91 * 83 = 7553
91 * 84 = 7644
91 * 85 = 7735
91 * 86 = 7826
91 * 87 = 7917
91 * 88 = 8008
91 * 89 = 8099
91 * 90 = 8190
91 * 91 = 8281
91 * 92 = 8372
91 * 93 = 8463
91 * 94 = 8554
91 * 95 = 8645
91 * 96 = 8736
91 * 97 = 8827
91 * 98 = 8918
91 * 99 = 9009
91 * 100 = 9100
|
| Anyone for 4-digit integers? The following result was developed with MAPLE:
12[k*88 0's]34 is divisible by 1234, for k=0...
The quotient for k=1 is an 88-digit integer beginning with 9 and ending
with 1. The quotients for larger k are related (k*88)-digit integers, of
the form
9...09...1
9...09...09...1
etc.
Lynn Yarbrough
|
| Here are a few more pairs of numbers including some for N>2 zeros in the
middle.
45 * 9 = 405
18 * 6 = 108
15 * 7 = 105
74 * 946 = 70004
54 * 926 = 50004
82 * 97561 = 8000002
91 * 989011 = 90000001
84 * 952381 = 80000004
78 * 897436 = 70000008
65 * 923077 = 60000005
63 * 952381 = 60000003
35 * 857143 = 30000005
28 * 714286 = 20000008
26 * 769231 = 20000006
21 * 952381 = 20000001
If we have a combination with N zeros, we can generate pairs with any
multiple of N zeros as follows.
For any A,B < 10
If (10A+B)*C = A*10^(N+1) + B (A number with N zeros in
the middle)
Then (10A+B) * ((C-1)*10^N+C) = A*10^(2N+1) + B (A number with 2N zeros in
the middle)
Which can be derived as follows.
(10A+B)*((C-1)*10^N+C) = (10A+B)(C*10^N-10^N+C)
= AC*10^(N+1)-A*10^(N+1)+10AC+BC*10^N-B*10^N+BC
= AC*10^(N+1)+BC*10^N-B*10^N-A*10^(N+1)+10AC+BC
= ((10A+B)*C)*10^N-B*10^N-A*10^(N+1)+(10A+B)*C
= (A*10^(N+1)+B)*10^N-B*10^N-A*10^(N+1)+A*10^(N+1)+B
= A*10^(2N+1)+B*10^N-B*10^N-A*10^(N+1)+A*10^(N+1)+B
= A*10^(2N+1)+B
For example, since 74 * 946 = 70004,
we also know that 74 * 945946 = 70000004
and 74 * 945945946 = 70000000004
and 74 * 945945945946 = 70000000000004, etc.
Also, I believe it is true that if (10*A+B)*C = A*10^(N+1)+B is true for
some A,B,C, and B is even, there exists a D such that (10*A+B)*D/2 will
also have N zeros between it first and last digits, but I haven't proved
that yet.
|
| Almost any number can be zero-split.
Given A*10^w+B, let's try to find k > 0 such that
A*10^w + B | A*10^(w+k) + B, where '|' is read 'divides'.
If we have such a k, then
A*10^w + B | A*10^(w+k) + B - A*10^w - B
A*10^w + B | A*10^w*(10^k-1)
Now, for any integer n, gcd(n,10) = 1 iff there is a k > 0 such that
n | 10^k-1. Let r(n) be the smallest such k > 0 -- it is simply the
number of repeating digits in the decimal expansion of 1/n. The set
of k satisfying n | 10^k-1 is simply the set of all multiples of r(n).
So,
A*10^w + B | A*10^(w+k) + B, for some k > 0 if and only if
gcd( (A*10^w+B) / gcd(a*10^w+B,A*10^w), 10 ) = 1.
Given that there is a solution for k, then the set of k satisfying the
equation are all multiples of r( (A*10^w+B) / gcd(A*10^w+B,A*10^w) ).
Note that if we define gcd(0,m) = m, the above can be better written as:
A*10^w + B | A*10^(w+k) + B, for some k > 0 if and only if
gcd( (A*10^w+B) / gcd(B,A*10^w), 10 ) = 1.
For example, we see that 25 can't be split (between the 2 and the 5),
and that 74 can be split using k = r(37) = 3.
|