Quinary Quantity Part 4

Logic Level 2

$\large \square \ + \ \square \ - \ \square \ \times \ \square \ \div \ \square$

What is the minimum possible value if you fill in the boxes with one of each $1, 2, 3, 4$ and $5?$

Note: You cannot use parentheses and must evaluate the expression using proper order of operations.

Like this problem? Try these versions: Part 1 , Part 2 , Part 3 .

The answer is -15.000.

2 solutions

Chew-Seong Cheong
Aug 11, 2015

For $S = A + B - C \times D \div E$ to be minimum, $C \times D \div E$ must be maximum, which is $4 \times 5 \div 1$ and $S_{min} = 2+3 - 4 \times 5 \div 1 = \boxed{-15}$ .

This combination is much easier to see than parts 1, 2, and 3 of these problems.

Scott Ripperda - 5 years, 9 months ago

Can you prove that it is not possible to obtain any smaller value?

Owen Leong - 5 years, 8 months ago

You can write a short program to calculate all the 5! = 120 cases. I think I did with Python.

Chew-Seong Cheong - 5 years, 8 months ago

Hazard Sharp
Aug 12, 2015

$S = A + B - C \times D ÷ E$

Notice that we are finding the Minimum, not the Smallest number.

Since we are subtracting $C \times D ÷ E$ we need to maximize this part to get the minimum answer.

So, C and D should be the largest numbers, 4 and 5, and E should be the smallest, 1.

A and B get the leftovers 2 and 3.

$S_{min} = 2 + 3 - 4 × 5 ÷ 1$

$-15 = 5 - 20$

Quinary Quantity Part 4

The answer is -15.000.

2 solutions

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