# Maximizing the common value of both sides of an equation

Using all the integers from 1 to 10 inclusive (each used exactly once), along with any number of the four arithmetic operations (add, subtract, multiply and divide) and parentheses, construct an equation with the largest (and identical) number on both sides.

Clarifications:

$$1+2+3+4+5+6=10 \times (9+8+7)$$
is not a valid equation because the value on the left hand side (21) is not equal to the value on the right hand side (240).

$$(2-1) \times (4-3)=(10-9) \times (8-7) \times (6-5)$$
is a valid equation because the value on both sides equals $$1$$. However the common value of $$1$$ is not the largest that can be achieved.

You can’t use anything other than what is explicitly allowed. So, for example, you can’t use square roots, concatenation, decimal points, bases other than 10, ...

• I assume powers are not allowed either? Commented Aug 31 at 17:13
• @quarague We could go much higher if they were: for example, $2^{9^{4 + 7}} = 8^{3^{1 \times 5 + 6 + 10}}$, where both sides have nearly ten billion digits. Commented Aug 31 at 18:04
• @quarague Your assumption is correct; powers are not allowed. Commented Aug 31 at 19:10

We can do

1920

which equals

6x5x(9x7+1) = 3x10x2x4x8

• That's a great answer! Surely this is optimal. Commented Aug 30 at 11:48
• Why is this optimal? Commented Aug 30 at 17:43
• @ThomasL Without proof, it is a good candidate because: 1) The fastest way to grow the value with the allowed operands is to us nothing but multiplication. 2) This solution uses almost entirely multiplication. 3) The one operand that isn't multiplication is addition - the second-fastest way to grow the value - and it's used for +1 which is actually better than x1 because that doesn't increase the value at all. Commented Aug 30 at 18:16
• After seeing this I did a quick check to see if there were any other ways to write $10!=a*b$ where $b-a$ is a divisor of $a$, because that might allow one to insert the $+1$ on the left hand side of $a=b$ to make both sides equal. This is the only solution of that kind, and there seems to be no way to get anything larger. Commented Aug 30 at 18:47
• The maximum you can get with 1-10 and the 4 basic symbols is by adding 1 to 2 (because adding 1 to anything else is a smaller percentage increase) and then multiplying everything, giving 5,443,200. We need two sides which are equal, so the upper bound for this puzzle is the square root, which is 2333.06.... So 1920 is getting pretty close to that upper bound. Commented Sep 2 at 14:56

I can do

1440

With

$$4*5*8*9 = (7+2-1)*3*6*10$$

• Oh lol, I actually did this by doing 7+1=8 and then thought "well, now there's nothing I can do about that 2!". Nice answer, +1 ;) Commented Aug 30 at 9:50
• Instead of ROT13(fhogenpgvat bar sebz n snpgbe ba gur euf) you could ROT13(nqq bar gb n snpgbe ba gur yrsg). Commented Aug 30 at 10:18
• @Jaap Scherphuis Indeed this is an even better solution (higher score)! I won't post it since this is not my idea... Commented Aug 30 at 11:18

This is a small improvement on franck vivien's answer. If you upvote this, please also upvote theirs.

The equation

$$4∗5∗(8+1)∗9=(7+2)∗3∗6∗10$$

has value

$$1620$$

I can do

$$900$$

with

$$(2 \times 5) \times (8+1) \times (4+6) = 10 \times 9 \times (7+3)$$

• I like how your answer is of the form aba=aba. Commented Aug 30 at 9:53