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 Decode the double code (Posted on 2011-06-09)
I have multiplied two 3-digit numbers and coded the whole process twice:

In the 1st coding O replaces an odd digit and E the even.
```   OOO
* EOE
OEOE
OOO
OEE
OOOEE ```
In the 2nd coding S replaces a digit smaller than 6 and B a digit bigger than 5.
```   SBB
* SSB
SSSB
SSS
BSB
BBSSB```
Try to get my original numbers(digit 9 does not appear in the multiplication) and all other possible solutions, if any.

 No Solution Yet Submitted by Ady TZIDON Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 computer solutons (spoilers) | Comment 1 of 7

CLS
FOR a = 1 TO 9 STEP 2
FOR b = 1 TO 9 STEP 2
FOR c = 1 TO 9 STEP 2
FOR d = 2 TO 8 STEP 2
FOR e = 1 TO 9 STEP 2
FOR f = 0 TO 8 STEP 2
abc = 100 * a + 10 * b + c
m2 = 100 * d + 10 * e + f
prod = abc * m2
p1 = f * abc
p2 = e * abc
p3 = d * abc
pr\$ = LTRIM\$(STR\$(prod))
IF LEN(pr\$) = 5 THEN
good = 1
FOR i = 1 TO 3
IF INSTR("13579", MID\$(pr\$, i, 1)) = 0 THEN good = 0
NEXT
FOR i = 4 TO 5
IF INSTR("02468", MID\$(pr\$, i, 1)) = 0 THEN good = 0
NEXT
IF p1 < 1000 OR p1 > 9999 THEN good = 0
IF p2 < 100 OR p2 > 999 THEN good = 0
IF p3 < 100 OR p3 > 999 THEN good = 0
IF good THEN
pr1\$ = LTRIM\$(STR\$(p1))
pr2\$ = LTRIM\$(STR\$(p2))
pr3\$ = LTRIM\$(STR\$(p3))
odd\$ = MID\$(pr1\$, 1, 1) + MID\$(pr1\$, 3, 1) + pr2\$ + MID\$(pr3\$, 1, 1)
even\$ = MID\$(pr1\$, 2, 1) + MID\$(pr1\$, 4, 1) + MID\$(pr3\$, 2)
FOR i = 1 TO LEN(odd\$)
IF INSTR("13579", MID\$(odd\$, i, 1)) = 0 THEN good = 0
NEXT
FOR i = 1 TO LEN(even\$)
IF INSTR("02468", MID\$(even\$, i, 1)) = 0 THEN good = 0
NEXT
IF good THEN
PRINT abc: PRINT m2
PRINT p1: PRINT p2: PRINT p3
PRINT pr\$
PRINT
ct = ct + 1
END IF
END IF
END IF

NEXT
NEXT
NEXT
NEXT
NEXT
NEXT
PRINT ct

PRINT

ct = 0
FOR a = 1 TO 5
FOR b = 6 TO 9
FOR c = 6 TO 9
FOR d = 1 TO 5
FOR e = 0 TO 5
FOR f = 6 TO 9
abc = 100 * a + 10 * b + c
m2 = 100 * d + 10 * e + f
prod = abc * m2
p1 = f * abc
p2 = e * abc
p3 = d * abc
pr\$ = LTRIM\$(STR\$(prod))
IF LEN(pr\$) = 5 THEN
good = 1
FOR i = 1 TO 2
IF INSTR("6789", MID\$(pr\$, i, 1)) = 0 THEN good = 0
NEXT
IF INSTR("6789", MID\$(pr\$, 5, 1)) = 0 THEN good = 0
FOR i = 3 TO 4
IF INSTR("012345", MID\$(pr\$, i, 1)) = 0 THEN good = 0
NEXT
IF p1 < 1000 OR p1 > 9999 THEN good = 0
IF p2 < 100 OR p2 > 999 THEN good = 0
IF p3 < 100 OR p3 > 999 THEN good = 0
IF good THEN
pr1\$ = LTRIM\$(STR\$(p1))
pr2\$ = LTRIM\$(STR\$(p2))
pr3\$ = LTRIM\$(STR\$(p3))
big\$ = MID\$(pr3\$, 1, 1) + MID\$(pr3\$, 3, 1) + MID\$(pr1\$, 4, 1)
smal\$ = MID\$(pr1\$, 1, 3) + pr2\$ + MID\$(pr3\$, 2, 1)
FOR i = 1 TO LEN(big\$)
IF INSTR("6789", MID\$(big\$, i, 1)) = 0 THEN good = 0
NEXT
FOR i = 1 TO LEN(smal\$)
IF INSTR("012345", MID\$(smal\$, i, 1)) = 0 THEN good = 0
NEXT
IF good THEN
PRINT abc: PRINT m2
PRINT p1: PRINT p2: PRINT p3
PRINT pr\$
PRINT
ct = ct + 1
END IF
END IF
END IF

NEXT
NEXT
NEXT
NEXT
NEXT
NEXT
PRINT ct

finds (with formatting added by hand):

`  173  216-----   1038 173346-----  37368`
`  177  418-----   1416 177708-----  73986`
`  177  438-----   1416 531708-----  77526`
`  353  214-----   1412 353706-----  75542`
` 4 solutions for part 1`
`  177  437-----   1239 531708-----  77349`
`  177  438-----   1416 531708-----  77526`
` 2 solutions for part 2`

The digit 9 appears in the second solution for part 1 and the first solution for part 2. That leaves three solutions for part 1 and a unique solution for part 2 avoiding the digit 9.

 Posted by Charlie on 2011-06-09 13:59:52
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