Blank Space Part 1

Logic Level 3

$\large 2 \, \square \, 2 \, \square \, 2 = A$

There are $4^2= 16$ ways in which we can fill the squares with $+ , - ,\times, \div$ .

How many distinct positive integer values can $A$ take?

Note :
You are not allowed to use parenthesis.
Obey the order of operations.

The answer is 5.

2 solutions

Nihar Mahajan
Jul 15, 2015

$\large {2+2+2=6 \\ 2+2-2=2 \\ 2+2\times 2=6 \\ 2+2 \div 2=3 \\ 2-2+2=2 \\ 2-2-2=-2 \\ 2-2\times 2=-2 \\ 2-2\div 2=1 \\ 2\times 2+2=6 \\ 2\times 2-2=2 \\ 2\times 2 \times 2=8 \\ 2\times 2\div 2=2 \\ 2\div 2+2=3 \\ 2\div 2-2=-1 \\ 2\div 2\times 2=2 \\ 2\div 2\div 2= 0.5}$

Hence acceptable values are $(1,2,3,6,8)$ that are $\boxed{5}$ in total.

Is there any mathematical way

Department 8 - 5 years, 11 months ago

This is the mathematical way of "checking cases". As you might have realized, there isn't a simple way of reducing the number of cases to check, and we should just run through all possibilities.

Chung Kevin - 5 years, 11 months ago

Alexander Weddle
Jul 24, 2015

4 operations taken 2 at a time minus 1 for the case 2 divided by 2 divided by 2.

4!/2! - 1, or 5, values

Note: For this solution, one must first realize that only performing division first $\large {2 \div 2}$ would result in a number not divisible by 2 in a second division operation.

Blank Space Part 1

The answer is 5.

2 solutions

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