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

So - so today's puzzle reads:

I came up with two possible solutions, but the Mensa answer says there is only one.

Can you find the answer?

-Z

628549

scratch the second one

That's the only answer I see, what else did you come up with?

Is this in base 10?

642537
or
624537 if they mean the 4th digit must come before the 5th in the product sequence.

First digit must be four more than second

738169

But I thought the fourth and fifth digits when read as a single number equal the product of the first and sixth digits. The product of 7 and 9 is not 16.

"the fourth and fifth digits when read as a single number equal the product of the first and sixth digits"

You're adding, "product" means multiply.

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

I was assuming the 1st digit was in the 10^0 place, 2nd digit was in the 10^1 place etc.

What do you mean with "2 numbers , read as a single digit" ????

"when read as a single number " 12, for example.

ahh true there's an ambiguity i assumed first digit is 10^5 place.

ok, now i get it
well, what island said : 628549 seems right to me

Third digit is not one less than the sixth

oh heh. yea i did 5th.

the real thing is. not only do u have to find 1 case. u have to then prove that there's none other than 1 case proving it....i need to do more formal math....and i'm out of brain cells now

Yep. Oppsy. Never could get my addition and mutiplication straight! :eek: