a=1; p=0;c=2003%a; if(c==23, goto4, goto6);p=p+1;printp,"-", a; a++;if(a<2000, goto2, 0)
OUTPUT:
n
1 - 30
2 - 33
3 - 36
4 - 44
5 - 45
6 - 55
7 - 60
8 - 66
9 - 90
10 - 99
11 - 110
12 - 132
13 - 165
14 - 180
15 - 198
16 - 220
17 - 330
18 - 396
19 - 495
20 - 660
21 - 990
22 - 1980
+118608 How many natural numbers n have the property that the remainder of dividing 2003 by n is 23?
2003-23=1980
I need to find how many numbers bigger than 23 go into 1980
factor(1980 = ((2^2*3^2)*5)*11
1980 = 11* 5* 3* 3* 2* 2
None of these factors on their own is bigger than 23
Look at pairs.
11*2 =22 which is too small, 11*3 and 11*5 are both fine. So there is 2 (33 and 55)
5 any other than 11 is too small so no more here
Let me look at triples
11*2*2=44 is fine so possible multiples of 11 are fine
11*2*3=66
11*2*5=110
11*3*3=99
11*3*5=165
5 are here
5*2*2=20 which is too small, 5*2*3=30, that is ok. 5*3*3=45, 5*3*2=30 are ok so that is 2 more
3*3*2=18 which is too small, so no more here.
Now quads
11*2*2*3=121 that is ok, all quads with 11 are ok
2*2*3*3=36 which is also ok so all quads are fine.
2*2*3*3=36
2*2*3*5=60
2*2*3*11=132
2*2*5*11=220
2*3*3*5=90
2*3*3*11=198
2*3*5*11=330
3*3*5*11=495
8 of them
groups of 5 factors
leave out, 2,3,5 or 11
4 of them.
6 factors
just 1 of them
2+5+2+8+4+1= 22
------------------------------
33
55
44
66
110
99
165
45
30
2*2*3*3=36
2*2*3*5=60
2*2*3*11=132
2*2*5*11=220
2*3*3*5=90
2*3*3*11=198
2*3*5*11=330
3*3*5*11=495
Plus 5 more
