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

 Who is afraid of 7 (or 23) ? (Posted on 2010-11-08)
No matter how we permute the 3 digits 1,2 4 , none of the 6 three- digit numbers (124,142,214,241,412,421) is a multiple of 7.

Clearly, out of 84 possible triplets created by choosing distinct non-zero digits
there must be several like this, i.e. generating a 6-pack of three-digit numbers,
non divisible by 7.

1. List those triplets.
2. Similar question re:4 digits and divisibility by 23.

 No Solution Yet Submitted by Ady TZIDON No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 computer solution | Comment 2 of 5 |

Part 1:

DEFDBL A-Z
CLS
FOR a = 1 TO 7
FOR b = a + 1 TO 8
FOR c = b + 1 TO 9
n1 = a * 100 + b * 10 + c
n2 = a * 100 + c * 10 + b
n3 = b * 100 + a * 10 + c
n4 = b * 100 + c * 10 + a
n5 = c * 100 + b * 10 + a
n6 = c * 100 + a * 10 + b
IF n1 MOD 7 > 0 THEN
IF n2 MOD 7 > 0 THEN
IF n3 MOD 7 > 0 THEN
IF n4 MOD 7 > 0 THEN
IF n5 MOD 7 > 0 THEN
IF n6 MOD 7 > 0 THEN
PRINT a; b; c
END IF
END IF
END IF
END IF
END IF
END IF
NEXT
NEXT
NEXT

finds

1  2  4
1  2  5
1  2  9
1  3  6
1  3  8
1  4  6
1  4  8
1  4  9
1  5  9
1  6  7
1  7  8
2  3  4
2  4  8
2  5  6
2  5  7
2  5  8
2  6  9
2  7  9
2  8  9
3  4  5
3  4  7
3  4  9
3  5  6
3  6  8
4  6  8
4  8  9
5  6  9
5  7  9
5  8  9
6  7  8

Part 2:

DECLARE SUB permute (a\$)
DEFDBL A-Z
CLS
FOR a = 1 TO 6
FOR b = a + 1 TO 7
FOR c = b + 1 TO 8
FOR d = c + 1 TO 9

n\$ = LTRIM\$(STR\$(a)) + LTRIM\$(STR\$(b)) + LTRIM\$(STR\$(c)) + LTRIM\$(STR\$(d))
h\$ = n\$
good = 1
DO
n = VAL(n\$)
IF n MOD 23 = 0 THEN good = 0: EXIT DO
permute n\$
LOOP UNTIL n\$ = h\$
IF good THEN PRINT a; b; c; d

NEXT
NEXT
NEXT
NEXT

(Listing for permute shown elsewhere on Perplexus.)

finding:

1  2  3  4
1  2  3  5
1  2  3  6
1  2  3  7
1  2  4  9
1  2  6  7
1  2  6  8
1  2  7  8
1  3  4  9
1  3  5  6
1  3  5  8
1  3  7  8
1  3  8  9
1  4  5  8
1  4  8  9
1  5  6  7
1  5  6  9
1  5  7  9
1  6  8  9
2  3  4  9
2  3  5  7
2  3  5  9
2  3  7  9
2  4  5  9
2  4  6  7
2  4  6  8
2  5  6  9
2  5  8  9
2  6  8  9
2  7  8  9
3  4  5  6
3  4  5  7
3  4  6  7
3  5  6  8
3  5  7  8
3  7  8  9
4  5  6  7
4  5  6  9
4  5  7  8
4  5  8  9

If the second part had stuck with 7, instead of 23, we'd have the more exclusive:

1  2  3  8
1  3  8  9
2  4  6  9

 Posted by Charlie on 2010-11-08 16:22:21

 Search: Search body:
Forums (0)