All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Find my number (Posted on 2011-12-19)
My number:
1. Is a 3-digit number.
2. All its digits are prime digits.
3. A number formed by its 1st 2 digits is prime.
4. So is the number formed by its last 2 digits.
5. It is the smallest number satisfying the above conditions.

Find it.

Rem: The digits in my number do not have to be distinct.

Comments: ( Back to comment list | You must be logged in to post comments.)
 re(3): No problem (spoiler), if composite | Comment 4 of 10 |
(In reply to re(2): No problem (spoiler), if composite by Dej Mar)

573 doesn't work as 57=3*19.

373 is the only one of the four remaining that is itself prime, per Ady's original intent, and so is not only the smallest, but also the only number satisfying that intent.

10   dim Prd(4)
20   Prd(1)=2
21   Prd(2)=3
22   Prd(3)=5
23   Prd(4)=7
40   for A=1 to 4:D1=Prd(A)
50   for B=1 to 4:D2=Prd(B)
60   for C=1 to 4:D3=Prd(C)
100        Pr1=10*D1+D2:Prwhole=100*D1+10*D2+D3
110        if prmdiv(Pr1)=Pr1                             then
120          :Pr2=10*D2+D3
130          :if prmdiv(Pr2)=Pr2 then
140             :print Prwhole
150   next
160   next
170   next
OK
run
237
373
537
737
OK

10   dim Prd(4)
20   Prd(1)=2
21   Prd(2)=3
22   Prd(3)=5
23   Prd(4)=7
40   for A=1 to 4:D1=Prd(A)
50   for B=1 to 4:D2=Prd(B)
60   for C=1 to 4:D3=Prd(C)
100        Pr1=10*D1+D2:Prwhole=100*D1+10*D2+D3
110        if prmdiv(Pr1)=Pr1 and prmdiv(Prwhole)=Prwhole then
120          :Pr2=10*D2+D3
130          :if prmdiv(Pr2)=Pr2 then
140             :print Prwhole
150   next
160   next
170   next
OK
run
373
OK

 Posted by Charlie on 2011-12-20 10:20:35

 Search: Search body:
Forums (0)