A Fairly Simple Equation

Question

I'm new, so I thought I'd start with a fairly simple puzzle.(Simple doesn't necessarily mean easy.)

Can you fit the numbers $2, 3, 4, 5, 6,$ and $10$ into this equation, each replacing one letter, so that the result is $93$ (rounded to the nearest whole number)

Equation: $(A+B)(C-D)(E/F)$

I'm fairly certain there is only one solution. (excluding reversing the order on the addition)

Answer 1

Since rounding is allowed, but the rest are not; I can promise that there are at least 2 solutions:

$(4 + 10)(6 - 2)(5 / 3) = 93$
$(10 + 4)(6 - 2)(5 / 3) = 93$

I was actually curious if there were any further solutions and created a brute force style application in C# to find out. There are actually exactly two solutions.

decimal[] nums = new decimal[] { 2, 3, 4, 5, 6, 10 };
foreach (decimal a in nums)
    foreach (decimal b in nums)
        foreach (decimal c in nums)
            foreach (decimal d in nums)
                foreach (decimal e in nums)
                    foreach (decimal f in nums) {
                        if (a == b || a == c || a == d || a == e || a == f ||
                            b == c || b == d || b == e || b == f ||
                            c == d || c == e || c == f ||
                            d == e || d == f ||
                            e == f) continue;
                            if (Math.Round((a + b) * (c - d) * (e / f)) == 93)
                                Console.WriteLine($"(({a} + {b}) * ({c} - {d}) * ({e} / {f}) = 93");
                    }

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Without doing any of the following, there are no solutions.

• Combining numbers.
• Reusing numbers.
• Rotating numbers.
• Rounding.

When allowing numbers to be combined, there are thousands of solutions given that:

There are 6 numbers, and this means there are 7 basic representations of combinations for each number, for example:
$2, 22, 23, 24, 25, 26, 210$.

However, if you go even further; you can create roughly 36 combinations for each number if you limit yourself to only a single repetition in a single combination.

When rotating numbers is allowed, this number of solutions increases substantially, especially if combination is also allowed. Throw in rounding on top of all of this and you've got yourself a solution celebration similar to the ball drop in Time Square.

Answer 2

I have the answer.

$(10+4)(6-2)(5/3)=93.333$ which when rounded off is 93.

To solve this I tried various ratios of E/F and divided it from 93 and tried to make the nearest integer from the given numbers.

Answer 3

Just to be that guy.

There are technically two unique solutions:
$(10+4)(6−2)(5/3)=93.333$ rounds down to $93$. Found first by Sarthak Rout.
Switching the first two numbers $(4+10)(6−2)(5/3)=93.333$ which rounds down to $93$.