Cryptarithm solution for DOUBLE TROUBLE
A friend of Nicole's shared this challenge that they gave his kid in school. The kid is in second grade and going to a gifted and talented program. Seems harder than it is when you sit down, it requires very good understanding on how addition works - which sounds easy, but...
You have 9 letters to solve, and each one can have a value between 0 and 9, so 10 values. No number can be repeated, so that becomes permutations of 10 over 9, i.e. the number of ways you can assign one of 10 values to 9 items, without repetition allowed, and with saying that every time you change the order it is a different arrangement. That is technically P(10,9) = 10! which is over 3 million combinations! Luckily they give us some breaks, including the fact that T and D cannot be 0, along with a few other clues that reveal themselves as you attempt to solve it.
I was trying to do it in my head at first, but that did not wok. I had to get old school pen and paper, and work through it.
011/020/02025 9:24 AM
