Digit Dilemmas



1. Two other digits name the number of segments they contain: 5 and 6

2. You can discard segment combinations 1, 5, and 6. As shown here, you need only shapes 2, 3, and 4 to make all the digits.


3. The smallest number of changes is just one segment, which occurs between 5 and 6 and between 8 and 9. The largest number of changes is five segments, which occur between 1 and 2 and between 6 and 7.


4. The following order, found by Nick Baxter, requires just 16 segment changes, the minimum amount possible. The numbers below the digits are the number of segment changes required for each transition.






When One Plus One doesn't equal Two





1. TWO + FIVE = 7

2. THREE + SEVEN = 10

3. FIVE + SIX + ZERO = 11

4. NINE + TWO + ZERO = 11

5. SIX + SEVEN + ZERO + TWO = 15

6. ZERO + ONE + TWO + SEVEN + NINE = 19

7. ZERO + ONE + TWO + THREE + FOUR + FIVE + SIX + SEVEN + EIGHT = 36

8.



Cross Purposes



While you can solve these through educated guessing, you can also tackle them algebraically. Write the first problem as (10 + a)/(10a + b) = 1/b, which becomes a = 9b/(10 - b). A few substitutions reveal that a = 9 and b = 5, or a = 6 and b = 4.




This is the fraction with the smallest numbers that cancels to 8/4. To solve, reason that since the numerator must be twice the denominator, the first number in the denominator must be 4. The second number in the numerator must also be 4. The second number in the denominator, then, must be half that, or 2. Continue working left to right, carrying numbers as necessary. Another approach is to treat all the letters as a single variable: (8 x 10¹⁸ + a)/(10a + 4) = 8/4.

Try working out the answers for similar fractions that reduce to 6/3, 4/2, and 2/1. The answers are closely related.





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