Sandy Wood
Brain Game: Too Easy
by Sandy Wood - August 3, 2009 - 7:30 AM

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Last week, one particular Brain Game was called “too easy” by a handful of players. So we’re going to start Monday with a bang. Remember, you asked for it!
Too Easy

Here is the SOLUTION.
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Comments (10)

  1. got it! that was a fun one.

    posted by grant on 8-3-2009 at 8:47 am

  2. Got it after a little work. I was one of the “handful” last week, and found that this one was a good challenge. Nice job making me wake up on a Monday, Sandy – thanks.

    posted by B on 8-3-2009 at 12:35 pm

  3. I’m happy to hear that, both of you. Thanks!

    posted by Sandy Wood on 8-3-2009 at 12:39 pm

  4. Now I’ve burned up my HDYK brainpower. Must rest until 5pm…

    posted by Seth on 8-3-2009 at 1:46 pm

  5. Woohoo! I got it! I are smrt!

    posted by Jim T on 8-3-2009 at 4:15 pm

  6. Got it. But through trial and error. Was this solvable by algebra?

    posted by SteveO on 8-3-2009 at 6:47 pm

  7. SteveO, there was some algebra/logic that could be used to help you solve it. If you know that T cannot be equal to E, then you know that the hundreds column must be either 19 or 29, but with only 3 or 4 numbers being added together, 19 is the only possibility. That leaves you with H+T equaling either 16, 15, or 14 (15 and 14 are possibilities if you account for either a 1 or 2 being carried over from the 10s column). At least, that’s how I started it.

    posted by Luke on 8-3-2009 at 8:32 pm

  8. Got it! I liked this one. SteveO–here’s how I did it. I started by realizing that T can’t equal E, therefore T+1=E, because with only a 3 in the next column, I could get over 10, but not over 20, so I could only carry over a 1. Given that a=9, I knew I needed H+T+3=19–which wasn’t possible. So with a 1 carried over from the IIAO column, H+T=15, so H and T had to be 8 and 7 (one or the other). That didn’t work, either (because T+1=E, and that number, either 8 or 9, was already used), so I figured there had to be a 2 carried over–easy enough with a 9 in the column. After that, it was easy enough to figure out that I=5 (given what numbers were left and that I had to get that column to be at least 20), and then it was just a question of whether O=1 or O=2, and a quick try of each left O=2.

    So it’s just logic and arithmetic.

    posted by Ryan on 8-4-2009 at 1:41 am

  9. Thank you Sandy for this puzzle.
    I remember when I was younger me and my brother had the Sideways Arithmetic from Wayside School book that started out with these type problems, it was lots of fun.

    posted by Brett on 8-4-2009 at 5:58 am

  10. I am so freaking lost ..

    em such a dumbum .. lool

    Someone please tell me how its done ..

    posted by Dominic on 8-10-2009 at 2:19 am

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