57 of 100: Curious Calculator

Start from 1 and construct backwards. Going one step forward from 6 shows that we must get one of 12, 11, 2, or -1 in backward construction. It can't be done with three steps. But with four, 6 appears and there are seven ways to do it. Thus the minimum required button presses is $\boxed{4}$ .

All the possible solutions that require only $\boxed{4}$ steps are:

• $6 \boxed{\times 2} \to 12 \boxed{\times 2} \to 24 \boxed{\div 3} \to 8 \boxed{-7} \to 1$
• $6 \boxed{\times 2} \to 12 \boxed{\div 3} \to 4 \boxed{\times 2} \to 8 \boxed{-7} \to 1$
• $6 \boxed{+5} \to 11 \boxed{-7} \to 4 \boxed{\times 2} \to 8 \boxed{-7} \to 1$
• $6 \boxed{\div 3} \to 2 \boxed{\times 2} \to 4 \boxed{\times 2} \to 8 \boxed{-7} \to 1$
• $6 \boxed{-7} \to -1 \boxed{\times 2} \to -2 \boxed{\times 2} \to -4 \boxed{+5} \to 1$
• $6 \boxed{-7} \to -1 \boxed{\times 2} \to -2 \boxed{+5} \to 3 \boxed{\div 3} \to 1$
• $6 \boxed{-7} \to -1 \boxed{+5} \to 4 \boxed{\times 2} \to 8 \boxed{-7} \to 1$

First step 6 x2=12

Second step 12/3 = 4

Third step 4 x2= 8

Fourth step 8 - 7 =1

6-7=-1

-1*2=-2

-2+5=3

3/3=1

The fastest way to get to 1 is done in 4 steps :
1) 6 ÷ 3 = 2
2) 2 × 2 = 4
3) 4 × 2 = 8
4) 8 - 7 = 1

You can use Python to write a program that uses recursion to find out the solution! :)

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17

def solution(steps, number):
    if number == 1:
        return steps
    if steps > 5:
        return steps

    Min = 10000000 

    if number%3 == 0:
        Min = min(solution(steps + 1, number/3), Min)
    Min = min(solution(steps + 1, number + 5), Min)
    Min = min(solution(steps + 1, number - 7), Min)
    Min = min(solution(steps + 1, number*2), Min)

    return Min

print solution(0,6)

All the 7 ways to get 4. There are 256 possible button push combinations (consider each button a digit in base 4, count all 4 digit numerals in base 4, $4^4 = 2^ 8 = 256$ ). Examine the 16 two button pushes (conveniently arranged in a 4x4 table). Find that 4 occurs 4 tines, 24 once, and -2 once. Find also that these are the only combinations that lead to a solution, and that none are 1 nor yield 1 with one more button therefore no 2 or 3 button solutions, four pushes is the minimum.

These yield at total of 7 solutions. 4 can be finished off one way, x2, -7 yielding 4 independent solutions. 24 can be finished one way ÷3, -7. -2 can be finished two ways, +5, ÷3, and x2, +5.

Two ways to finish off -2 after two pushes.

-1,-2,3,1 (-7,x2,+5,÷3) uses each button once. Arguably prettiest.

-1,-2,-4,1 (-7,x2,x2,+5) no divide.

One way to get 24 and finish it off.

12, 24, 8, 1 ( x2, x2, ÷3, -7) no add.

Four ways to get 4 in two pushes.

12, 4, 8, 1 ( x2, ÷3, x2, -7) no add.

-1,4,8,1 (-7,+5, x2, -7) no divide.

11, 4, 8, 1 (+5, -7, x2, -7) no divide.

2, 4, 8, 1 ( ÷3, x2, x2, -7) no add.

Three ways but all 4 steps.

\[\begin{array} {} 6 & \boxed{\times 2} = 12 & \boxed{\div 3} = 4 & \boxed{\times 2} = 8 & \boxed{-7} = 1 \\ 6 & \boxed{+ 5} = 11 & \boxed{-7} = 4 & \boxed{\times 2} = 8 & \boxed{-7} = 1 \\ 6 & \boxed{-7} = -1 & \boxed{+5} = 4 & \boxed{\times 2} = 8 & \boxed{-7} = 1 \end{array} \]

1) 6 - 7 = -1 2) -1 * 2 = -2 3) -2 * 2 = -4 4) -4 + 5 = 1

6+5+5=16/2+8-7=1

6-7= -1 -1x2=-2 -2x2=-4 -4+5=1

6x2=12 12x2=24 24/3=8 8-7=1

Subtract 7 from 6, leaving you with -1.

Multiply -1 by 2, leaving you with -2.

Add 5 to -2, leaving you with 3.

Divide 3 by 3, leaving you with 1.

6 - 7 = -1; -1 x 2 = -2; -2 + 5 = 3; 3 / 3 = 1 Ans

First step $6-7=-1$

Second step $-1× 2=-2$

Third step $-2+5=3$

Fourth step $3÷3=\boxed{1}$

6 - 7 = -1 => -1 * 2 = -2 => -2 * 2 = -4 => -4 + 5 = 1

• 6 -7 = -1

-1 x 2 = -2

-2 +5 = 3

3/3 = 1

6 + 5 = 11, 11 - 7 = 4. 4 x 2 = 8, 8 - 7 = 1, therefore there are 4 button presses.

6-7 = -1, -1x2 = -2, -2+5 = 3, 3/3 = 1

One solution : 6 2 2/3 - 7 = 24/3 - 7 = 1 . This requires 4 steps. To prove that we cannot get a value of 1 with less than 4 steps, 1 brute force approach is to calculate all the 16 combinations of 2 calculator operations , check if 1 is amongst any of them. If not, for each of the 16 values above, calculate if any of the calculator buttons yield a value of 1. If not, we are done. The above is not that complex since values would be repeated and many can be discarded at a glance

6-7+5x2 4 steps!

All solutions for 4 or 5 steps, without permutations of sums/substractions and multiplications/divisions ( +5 –7 $\times$ 2 –7 will be shown, but not –7 +5 $\times$ 2 –7)

×2  ×2  /3  -7                                                                                                                                                                                                           
+5  -7  ×2  -7
-7  ×2  ×2  +5
-7  ×2  +5  /3

×2  +5  -7  -7  /3
×2  /3  +5  /3  /3
×2  -7  ×2  -7  /3
×2  -7  -7  ×2  +5
+5  ×2  -7  -7  -7
+5  +5  -7  /3  /3
/3  -7  -7  /3  +5
-7  ×2  +5  +5  -7

It is simple, very simple. You have to just do the following: 6 + 5 (one) - 7 (two) * 2 (three) - 7 (four). In this way, it could be done by pressing just four buttons.

Here is one solution: 6 ( - 7 ) then ( * 2 ) then ( + 5 ) and finally ( / 3 ) to get 1. This accounts for a total of 4 steps.

6 - 7 = -1 -1 x 2 = -2 -2 x 2 = -4 -4 + 5 = 1

To get to $1$ with two operations, you need to get to $\frac{1}{2} = 0.5$ , $1-5 = -4$ , $1*3 = 3$ , or $1+7 = 8$ in a single step from $6$ —clearly not possible. To get to $1$ in three operations, you need to be able to get to any of those four numbers in two steps, but the only ones that look like they can possibly be reached in two steps are $1*3 = 3$ and $1+7 = 8$ . So see if you can get to one step away from either of those two in only one step from $6$ —the numbers that are one step away from those are $\frac{3}{2}$ , $4$ , $3$ , $-2$ , $24$ , $9$ , $10$ , $15$ . You can't get to any of these numbers with only one operation. So sequences involving four button presses are the shortest sequences possible. (Look at other solutions to find those sequences.)

I have found six solutions of length (so far). To find them all I need to develop an algorithm that enumerates all the 256 possible strings of the four characters and tests them all. This should not be too hard.

http://www.codeskulptor.org/#user43 n263Hpy25L 0.py

It takes 4 steps. One possible way is:

6 divided by 3 = 2

multiplied by 2 = 4

multiplied by 2 again = 8

minus 7 = 1