Tuesday puzzle

Re: Tuesday puzzle

If a number cannot be repeated for it to work the individual digits are between 0 and 9 unless it is a trick and the the number is 0000, and the 1st and last numbers have to be even (1 through 9 when multiplied by 4) and the last (D) has to be multiplied by four so it must be formed from a number lower than 3 or it kicks to a double digit. So two and eight are the only possibilities since 1 and 7 are odd numbers and A and D must be even. So A is 2 and D is eight (I think). The middle two have to be odd numbers so that when multiplied by four the last(D)is eight--using numbers not already used. If it is a odd number the only possibility for C is one or 7 but doing the long hand it has to be 2 or lower and 2 is already used and the end number is 8. That leaves 1. Any of the other numbers not already used do not work. That leaves the second number (B) and it has to be 7 because it is the only number when multiplied by four that provides an eight and not already used. There has to be a formula for this but by my logic I arrive at 2718. It works but if numbers in the answer can be repeated I'm dead.

2718 doesn't work. Sorry. My numbers

or logic is screwed up. It worked in my scribbles but I need to go back.

not quite

2718 X 4= 10872.

Re: Tuesday puzzle

A=2, B=1, C=7, D=8 2178 x 4 = 8712

Galen wins

You'd think after all these years I'd be able to keep straight thought encompasing addition and multiplying four numbers to test. Even then I bollixed the logic. So much for my chances at MIT and CalTech (and about 100 others).

;-) congrats Galen