Thus, the smallest base for which these alphametics hold true is 14. If b = 13, O = 2, O = 3, O = 4, O = 5, or O = 6. The Confetti Teacher Want to teach your students how to solve Logic Puzzles Or need a time filler These 5 brain teasers will challenge your students to think outside the box Answer keys included Subjects: Back to School, End of Year, Other (Math) Grades: K - 2 nd Types: Worksheets, Activities, Centers 2.00 4.Also, O must be less than one-half of the base, since 1 + O + O = T does not exceed the base, and O + 0 = R does not exceed the base.Therefore 2O = R, 2O + I = T, R + l = T, R is even, T is odd, and F = I.O + O = R, but O + O = T therefore, in ONE+ ONE= TWO, a 1 must be carried over from N + N.

Some “truths” we can determine without regard for the base:

Since each letter in the alphametic represents a unique and different “digit” and there are eight different letters used (O, N, E, T, W, F, U, R), the positive integral base must be equal to, or greater than, eight.

Setting the first term (x 2 + x -5) to zero yields: Setting the second term (x 2 – x – 4) to zero gives:īut neither answer satisfies the original equation. Square again to get: x 4 – 10x 2 + x + 2 = 0 Square both sides (just remember that this process may give extraneous solutions – we will test each for validity later) to get: There are many YouTube videos that show the setup and solution, here is a good one: Did you figure out this issue’s answers? If not, we’re here to help! Check out the solutions below.

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