Mensa Calendar Error (aka Z-man has trouble with addition vs. multiplication)

[flipper35]

I guess you have to like math to want to join Mensa. I could do geometry, physics, and aerodynamics but this stuff is not my cup of anything.

[BlueSkyJaunte]

IMHO this isn't really a "brain teaser" or even a puzzle of any kind: it's too easy to brute-force in any programming language you may choose to utilize.

[masraum]

288!!

Wait! Or is it 2?

[Z-man]

Originally Posted by masraum

288!!

Wait! Or is it 2?

I lol'd again.

[BK911]

416287

[BK911]

418369

[GH85Carrera]

8675301 Call Jenny.

[svandamme]

Originally Posted by BK911

416287

The 2nd digit has to be 4 less then the 1st digit

Originally Posted by BK911

418369

The 2nd digit has to be 4 less then the 1st digit
The 3rd has to be 1 less then the 6th

you're getting worse with each try !

[BK911]

Oh crap!
By "4 more" I was multiplying!
Back to grade school math.

[BK911]

Originally Posted by lendaddy

628549

+1

I couldnt find another one.

[svandamme]

Originally Posted by krystar

to translate the puzzle into something more easily readable.

variables A,B,C,D,E in integer space {1,2,3,4,5,6,7,8,9}
variable F in integer space {1,3,5,7,9}

where A<>B<>C<>D<>E<>F (non repeated digits)
A=B+4
C=F-1
D*10+E=A*F



solving....

A=B+4 would limit A to {5,6,7,8,9} and B to {1,2,3,4}
C=F-1 would limit C to {2,4,6,8} and F to {3,5,7,9}

E is basically free to be any digit.
E=A*F-D*10

i think 952637 works

Not really , 3rd digit must b 1 less then the 6th..
your third digit is 5 less then the 6th

[svandamme]

Rule 1
6 digits

Rule 2
odd number

Rule 3
No zeroes

Rule 4
no repeating digits

rule 5
1st digit = 2nd digit +5

Rule 6
3rd digit = 6th digit -1

rule 7
4th-5th= 1st + 6th








1
So taking into account #1 en #2 #3 and #4

the numbers start at 123457


2
Add Rule 5 and it goes up

512347

Anything lower is not possible, because 5 anything lower then 5 can't be on the 1st digit or it would violate rule #5

First 2 digits can only be

51
62
73
84
95

Nothing else is allowed. with rule 5

3
Rule 6

Third digit, must be even, because it's the 6th digit -1, and the 6th digit must be odd(rule #2)
IT also can't be 0 so the last digit cannot be 1

So third one is either 2 4 6 or 8


XX X
51 2
62 4
73 6
84 8
95

4
LAst digit can only be or 3 5 7 9 ,anything else violates rule #2,#3 or #6


XX X xx x
51 2 3
62 4 5
73 8 9
84 6 7
95


5

Start adding up the possible digit 1's with the possible digit 6 to find the possible digits 4 and 5
5+3 = 8 (rule 7 violation, must have 2 digits)
5+5 (rule #4violation)
5+7 = 12
5+9 = 14

6+3 = 9 (rule 7 violation, must have 2 digits)
6+5 = 11 (rule 4 repeating digits)
6+7 = 13
6+9 = 15

7+3 = 10 (rule 3 no zeroes)
7+5 = 12
7+7 (rule #4violation)
7+9 = 15

8+3 = 11
8+5 = 13
8+7 = 15
8+9 = 17


9+3 = 12
9+5 = 14
9+7 = 16
9+9(rule #4violation)


So that leaves only these possibilities for the 4th and 5th digit
11
12
13
14
15
16
17

So this also means #1 is ruled out of all the other digits


XX X xx x
95 2 1 3
62 4 5
73 8 9
84 6 7


6
remaining combinations are

842125
842157
842179
846125
846157
846179
732169
734125
734169
738125
738169
736169
736125
624137
624159
628137
628159
952167
954167
958167

[svandamme]

No calculator , or brute force programming used.
Haven't checked all the results yet...

[svandamme]

CHIT

4th and 5th are the multiplication of 1st and 6th

CRAP

[svandamme]

corrected


Rule 1
6 digits

Rule 2
odd number

Rule 3
No zeroes

Rule 4
no repeating digits

rule 5
1st digit = 2nd digit +4

Rule 6
3rd digit = 6th digit -1

rule 7
4th-5th= 1st + 6th








1
So taking into account #1 en #2 #3 and #4

the numbers start at 123457


2
Add Rule 5 and it goes up

512347

Anything lower is not possible, because 5 anything lower then 5 can't be on the 1st digit or it would violate rule #5

First 2 digits can only be

51
62
73
84
95

Nothing else is allowed. with rule 5

3
Rule 6

Third digit, must be even, because it's the 6th digit -1, and the 6th digit must be odd(rule #2)
IT also can't be 0 so the last digit cannot be 1

So third one is either 2 4 6 or 8


XX X
51 2
62 4
73 6
84 8
95

4
LAst digit can only be or 3 5 7 9 ,anything else violates rule #2,#3 or #6


XX X xx x
51 2 3
62 4 5
73 8 9
84 6 7
95


5

Start multiplying the possible digit 1's with the possible digit 6 to find the possible digits 4 and 5
5x3 = 15 (rule #4violation)
5x5 (rule #4violation)
5x7 = 35 (rule #4violation)
5x9 = 45(rule #4violation)

6x3 = 18
6x5 = 30 (Rule3)
6x7 = 42 (rule #4violation)
6x9 = 54

7x3 = 21
7x5 = 35 (rule #4violation)
7x7 (rule #4violation)
7x9 = 63 (rule #4violation)

8x3 = 24 (rule #4violation)
8x5 = 40 (rule #3)
8x7 = 56
8x9 = 72


9x3 = 27
9x5 = 45(rule #4violation)
9x7 = 63
9x9 (rule #4violation)


So that leaves only these possibilities for the 4th and 5th digit

62x183
62x549

84x567
84x729

95x273
95x637

apply rule 6 (3rd = 6th -1)

622183 (rule #4violation)
628549
846567 (rule #4violation)
848729 (rule #4violation)
952273(rule #4violation)
956637 (rule #4violation)



That leaves only 1

628549

rule le 1
6 digits CHECK

Rule 2
odd number CHECK

Rule 3
No zeroes CHECK

Rule 4
no repeating digits CHECK

rule 5
1st digit = 2nd digit +4 CHECK

Rule 6
3rd digit = 6th digit -1 CHECK

rule 7
4th-5th= 1st + 6th CHECK

[speeder]

Isn't MENSA famous for being full of dip schits? I've heard a lot of funny jokes about it.

[svandamme]

Dunno, i'm to stupid to join.

[flipper35]

8675309

[GH85Carrera]

Good explanation Stijn.

[svandamme]

Originally Posted by flipper35

8675309

troll.