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)

  • $\begingroup$ Welcome to Puzzling.SE! This is a great puzzle, are we allowed to reuse numbers to make additional numbers? Such as $(23 + 3)(45 - 5)(610 / 10)$? I just want to ensure we know the rules of this. Also, I would assume that the pairing of parenthesis is implying multiplication as with traditional math? $\endgroup$ Sep 21, 2018 at 16:45
  • $\begingroup$ You can't combine or reuse numbers, and the pairs of parentheses is multiplication. $\endgroup$ Sep 21, 2018 at 16:55

3 Answers 3


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");

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.


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.

  • $\begingroup$ This was the first correct answer. Should have been selected. Oh well. $\endgroup$
    – Robert S.
    Sep 21, 2018 at 20:45
  • $\begingroup$ @RobertS. This was a good answer by SarthakRout for sure; however, as with all SE websites, the first answer (nor the selected answer) is ever guaranteed to be the best answer. Who knows, some one could come in a year or two from now and post a better answer that none of us thought about. $\endgroup$ Sep 21, 2018 at 21:35

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 $93$.

  • 1
    $\begingroup$ I counted those as the same XD $\endgroup$ Sep 21, 2018 at 17:08

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