Look at this situation: 1 1 1 1 = 6.
We can make this a correct equality by inserting different number operations. For instance
(1 + 1 + 1)! × 1 = 6
Note that !, pronounced factorial, works this way:
2! = 2 × 1
3! = 3 × 2 × 1
4! = 4 × 3 × 2 × 1, etc.
Which of the following can you complete by inserting various common number operations?
(i) 2 2 2 2 = 6
(ii) 3 3 3 3 = 6
(iii) 4 4 4 4 = 6
(iv) 5 5 5 5 = 6
(iv) 6 6 6 6 = 6
(vi) 7 7 7 7 = 6
Solution 1:
There are many possibilities for some of these numbers. We will only give one answer here.
(i) (2 × 2 × 2) – 2 = 6
(ii) 3 + 3 + 3 – 3 = 6
(iii) (4 + √4) × (4 ÷ 4) = 6
(iv) [√(5 × 5)] + (5 ÷ 5) = 6
(iv) √(6 × 6) × (6 ÷ 6) = 6
(vi) [(7 + 7 + 7) ÷ 7]! = 6
(Actually the method of (vi) will work for all of these problems.)
Extension:
1. Can you do this with four 8s and four 9s?
2. How many ways can you find to produce each of these numbers?
3. What numbers other than 6 can you get using four digits?