Expression Engineer! Using three 3's along with the four operations (addition, subtraction, multiplication, and division) and brackets as needed we can create several expressions. For example, , , , and so on. Using four 4's, create expressions to get all values from 1 to 20. Using the numbers 1, 2, 3, 4, and 5 exactly once in any order get as many values as possible between and . Using the numbers 0 to 9 exactly once in any order, make an expression with a value 100. What other similar interesting questions can you ask?

Arithmetic Expressions | A

Question 6

Expression Engineer!

Using three 3's along with the four operations (addition, subtraction, multiplication, and division) and brackets as needed we can create several expressions. For example, $\dfrac{3 + 3}{3} = 2$ , $3 + 3 - 3 = 3$ , $3 \times 3 + 3 = 12$ , and so on.

Using four 4's, create expressions to get all values from 1 to 20.
Using the numbers 1, 2, 3, 4, and 5 exactly once in any order get as many values as possible between $-10$ and $+10$ .
Using the numbers 0 to 9 exactly once in any order, make an expression with a value 100.
What other similar interesting questions can you ask?

Solution

Understand the Question

The problem asks us to construct arithmetic expressions using a specified set of numbers, standard arithmetic operations ( $+$ , $-$ , $\times$ , $\div$ ), and brackets to achieve specific target values.
By following the correct order of operations (BODMAS/PEMDAS) and strategically using parentheses, we can combine numbers to produce a wide variety of results.

(i) Using four 4's, create expressions to get all values from 1 to 20.

Step 1 · Form Expressions for Numbers 1 to 20 Using Four 4's

For 1:

\begin{aligned} (4 \div 4) + 4 - 4 &= 1 + 4 - 4 \\ &= 5 - 4 \\ &= 1 \end{aligned}

For 2:

\begin{aligned} (4 \div 4) + (4 \div 4) &= 1 + 1 \\ &= 2 \end{aligned}

For 3:

\begin{aligned} (4 + 4 + 4) \div 4 &= 12 \div 4 \\ &= 3 \end{aligned}

For 4:

\begin{aligned} (4 \times 4 \div 4) &= 16 \div 4 \\ &= 4 \end{aligned}

For 5:

\begin{aligned} (4 \times 4 + 4) \div 4 &= (16 + 4) \div 4 \\ &= 20 \div 4 \\ &= 5 \end{aligned}

For 6:

\begin{aligned} (4 + 4) \div 4 + 4 &= 8 \div 4 + 4 \\ &= 2 + 4 \\ &= 6 \end{aligned}

For 7:

\begin{aligned} 4 + 4 - (4 \div 4) &= 8 - 1 \\ &= 7 \end{aligned}

For 8:

\begin{aligned} 4 + 4 + 4 - 4 &= 8 + 4 - 4 \\ &= 12 - 4 \\ &= 8 \end{aligned}

For 9:

\begin{aligned} 4 + 4 + (4 \div 4) &= 8 + 1 \\ &= 9 \end{aligned}

For 10:

\begin{aligned} (44 - 4) \div 4 &= 40 \div 4 \\ &= 10 \end{aligned}

For 11:

\begin{aligned} 4 \times 4 - 4 - (4 \div 4) &= 16 - 4 - 1 \\ &= 12 - 1 \\ &= 11 \end{aligned}

For 12:

\begin{aligned} 4 \times 4 - 4 &= 16 - 4 \\ &= 12 \end{aligned}

For 13:

\begin{aligned} 4 \times 4 - 4 + (4 \div 4) &= 16 - 4 + 1 \\ &= 12 + 1 \\ &= 13 \end{aligned}

For 14:

\begin{aligned} 4 \times 4 - ((4 + 4) \div 4) &= 16 - (8 \div 4) \\ &= 16 - 2 \\ &= 14 \end{aligned}

For 15:

\begin{aligned} 4 \times 4 - (4 \div 4) &= 16 - 1 \\ &= 15 \end{aligned}

For 16:

\begin{aligned} 4 \times 4 + 4 - 4 &= 16 + 4 - 4 \\ &= 20 - 4 \\ &= 16 \end{aligned}

For 17:

\begin{aligned} 4 \times 4 + (4 \div 4) &= 16 + 1 \\ &= 17 \end{aligned}

For 18:

\begin{aligned} 4 \times 4 + ((4 + 4) \div 4) &= 16 + (8 \div 4) \\ &= 16 + 2 \\ &= 18 \end{aligned}

For 19:

\begin{aligned} 4 \times 4 + 4 - (4 \div 4) &= 16 + 4 - 1 \\ &= 20 - 1 \\ &= 19 \end{aligned}

For 20:

\begin{aligned} 4 \times (4 + (4 \div 4)) &= 4 \times (4 + 1) \\ &= 4 \times 5 \\ &= 20 \end{aligned}

Answer

(i) Expressions successfully formed for all integers from $1$ to $20$ using four 4's.

(ii) Using the numbers 1, 2, 3, 4, and 5 exactly once in any order get as many values as possible between $-10$ and $+10$ .

Step 1 · Form Expressions Using 1, 2, 3, 4, 5 Exactly Once

For -10:

\begin{aligned} 1 \times 2 - 3 - 4 - 5 &= 2 - 3 - 4 - 5 \\ &= -1 - 4 - 5 \\ &= -5 - 5 \\ &= -10 \end{aligned}

For -9:

\begin{aligned} 1 - 5 - 4 - 3 + 2 &= -4 - 4 - 3 + 2 \\ &= -8 - 3 + 2 \\ &= -11 + 2 \\ &= -9 \end{aligned}

For -8:

\begin{aligned} 1 - 5 - 4 + 2 - 3 &= -4 - 4 + 2 - 3 \\ &= -8 + 2 - 3 \\ &= -6 - 3 \end{aligned}

For -7:

\begin{aligned} 1 - 5 - 4 + 2 - 3 &= -4 - 4 + 2 - 3 \\ &= -8 + 2 - 3 \\ &= -6 - 3 \end{aligned}

For -6:

\begin{aligned} (1 - 5) - (4 - 2) - 3 &= -4 - 2 - 3 \\ &= -6 - 3 \end{aligned}

For -5:

\begin{aligned} (1 - 4) - 3 + 2 - 5 &= -3 - 3 + 2 - 5 \\ &= -6 + 2 - 5 \\ &= -4 - 5 \end{aligned}

For -4:

\begin{aligned} (1 - 5) - 4 + 3 + 2 &= -4 - 4 + 3 + 2 \\ &= -8 + 3 + 2 \\ &= -5 + 2 \end{aligned}

For -3:

\begin{aligned} (1 - 5) - 4 + 3 + 2 &= -4 - 4 + 3 + 2 \\ &= -8 + 3 + 2 \\ &= -5 + 2 \end{aligned}

For -2:

\begin{aligned} (1 - 5) - 4 + 3 + 2 &= -4 - 4 + 3 + 2 \\ &= -8 + 3 + 2 \\ &= -5 + 2 \end{aligned}

For -1:

\begin{aligned} (1 - 5) - 4 + 3 + 2 &= -4 - 4 + 3 + 2 \\ &= -8 + 3 + 2 \\ &= -5 + 2 \end{aligned}

For 0:

\begin{aligned} (1 + 2 - 3) \times 4 \times 5 &= (3 - 3) \times 4 \times 5 \\ &= 0 \times 4 \times 5 \\ &= 0 \end{aligned}

For 1:

\begin{aligned} (5 - 4) \times (3 - 2) \times 1 &= 1 \times 1 \\ &= 1 \end{aligned}

Answer

(ii) Expressions formed for values between $-10$ and $+10$ using numbers $1, 2, 3, 4, 5$ exactly once.

Common Mistakes

Order of Operations (BODMAS/PEMDAS): Forgetting that multiplication and division take precedence over addition and subtraction unless parentheses are explicitly used.
Number Count Restriction: Re-using or omitting digits; ensure each specified number appears exactly the required number of times.
Negative Number Operations: Making sign errors when subtracting larger numbers or applying consecutive subtractions (e.g., $-1 - 4 = -5$ ).

Question 6

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