Crazy 8s

Logic Level 2

8 F 8 D + 8 D R 8 G R 8 \large{\begin{array}{cccc} &8& F & 8&D\\ + && 8 &D &R\\ \hline & 8& G & R&8\\ \hline \end{array}}

If D , D, F , F, G , G, and R R are distinct digits satisfying the cryptogram above, and all "8" digits are already shown, what must R R be?


The answer is 3.

View wiki
Jason Dyer by Jason Dyer

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

We know that D+R is either 8 or 18, but for it to be 18 both have to be 9, and that can't be, so D+R=8, and 8+D is either R or 10+R, but from the previous equation we know that R=8-D, not 8+D (unless D=0, but that would mean that R=8, which we know isn't true because all 8's are already shown), so 8+D=10+R. Solving both equations we get that R=3.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...