#1**+1 **

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

Guest Aug 12, 2020

#2**+1 **

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

Melody Aug 12, 2020