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

[Z-man]

So - so today's puzzle reads:

Find a six-digit odd number containing no zeros and no repeated digits in which the first digit is four more than the second digit, the third digit is one less than the sixth digit, and the fourth and fifth digits when read as a single number equal the product of the first and sixth digits.

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

Can you find the answer?

-Z

[lendaddy]

628549

[lendaddy]

scratch the second one

[lendaddy]

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

[island911]

Is this in base 10?

[ChemMan]

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

[lendaddy]

Originally Posted by ChemMan

642537

First digit must be four more than second

[Z-man]

Originally Posted by lendaddy

that's the only answer i see, what else did you come up with?

738169

[Amail]

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.

[lendaddy]

Originally Posted by Z-man

738169

"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.

[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

[ChemMan]

Originally Posted by lendaddy

First digit must be four more than second

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

[svandamme]

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

[island911]

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

[krystar]

Originally Posted by ChemMan

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

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

[svandamme]

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

[lendaddy]

Originally Posted by krystar

i think 952637 works

Third digit is not one less than the sixth

[krystar]

oh heh. yea i did 5th.

[krystar]

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

[Z-man]

Originally Posted by lendaddy

"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.

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