Maximizing Ones and Twos

Logic Level 2

You have six numbers: three 1's and three 2's as shown below.

$1 \quad 2 \quad 1 \quad 2 \quad 1 \quad 2$

Using only addition (+), multiplication ( $\times$ ), and parentheses, make the expression above as large as possible without rearranging the digits or concatenating digits.

The answer is 27.

6 solutions

Atul Shivam
Jan 11, 2016

I did $(2+1) \times (2+1) \times (2+1)$ which yields an answer of $\boxed{27}.$

Before this question was re-written/clarified, I concatenated the digits and came up with a very large number. Afterward I came up with the expected solution. Nice puzzle, too bad I misunderstood and got it wrong. Thanks for fixing it! ☺

Jeff Carter - 5 years, 5 months ago

I did 121*212=25652 which is probably the largest as its not mentioned that we have to insert operators between all!

Prince Loomba - 5 years, 5 months ago

you should apply ur logic that why those gaps between digits are there

Atul Shivam - 5 years, 5 months ago

lol it's also not mentioned to insert operation in middle, I think $121212$ is itself probably the largest number, and if u have to insert any symbol than $211×221= 46631$ is the largest one{ it is also not mentioned in the question that you can't rearrange the digits as per ur statement }

Atul Shivam - 5 years, 5 months ago

@Atul Shivam My question is, how did you know that your answer is the largest of them?

John Michael Gogola - 5 years, 5 months ago

I have found no any other combination bigger than $27$ and lill bit of analysis

Atul Shivam - 5 years, 5 months ago

UPDATE

Using Calculus( or the formula -b/2a for parabolic functions)

In this situation, we can get a maximum of 3 numbers to use as a factor of multiplication

We can get:

(1+2) (1+2) (1+2)

We can get 2 numbers by using one of these combinations (or similars):

(1+2+1) (2+1+2) (1+2+1+2) (1+2)

And we can obviously get 1 number:

1+2+1+2+1+2 = 9

*It's kinda impossible to get 4 numbers 'cause one of these numbers would be 1, and multiply by one wouldn't change anything.

let's work with the situation we have 2 numbers. Let's call them x and y

we know that x + y has to be 9 what's the largest value of xy?

since x+y = 9 --> y= 9 - x. Using substitution we have( ? means a real number):

xy = ? x(9-x) = ? 9x - x² = ?

Derivatin' we must get: 9 - 2x = 0 (since the derivate of a real number) --> x= 4.5 (and y = 4.5)

Since we must have real numbers, the closest combination is 4 and 5(or 5 and 4, however) We also know that (9x - x²) is a parabolic function, so how closer the numbers are from the vertex, the bigger they are(in this case). So we have the fact that:

(1+2+1)*(2+1+2) will be the largest IF we can only have 2 numbers(and the result would be 20)

BUT there's another possibility, the case (1+2) (1+2) (1+2)

And this result will be 27. Which is bigger than 20, and this will be the largest result.

BONUS

We can also use the formula: -b/2a to found x=4.5 will be the largest, and using logic u can assume that x=4 and y = 5 will be the largest if we only had 2 numbers to multiply.

Victor Veras - 5 years, 4 months ago

Eduard Stoleru
Jan 13, 2016

Since 1 cannot increase the result in a multiplication, to get the largest result we have use it only in summation. Also, 2 is getting a larger result if it used in a multiplication than in a summation (only exceptions are $2 X 1$ and $2 X 1$ ). Therefore, we have to avoid multiplication between 1 and 2, that is why we need to use brackets to separate them :). Finally, the expression $(2 + 1) X (2 + 1) X (2 + 1) = 27$ gets the largest result.

Luca Ng
Jul 22, 2020

(2 + 1)^3 should be the answer. Putting 1 and 2 in addition then multiplying them will maximize the answer.

Daniel Ellesar
Jan 18, 2016

Can anyone prove that 27 is the largest?

Using Calculus (or the formula -b/2a for parabolic functions)

In this situation, we can get a maximum of 3 numbers to use as a factor of multiplication

We can get:

(1+2) (1+2) (1+2)

We can get 2 numbers by using one of these combinations (or similars):

(1+2+1)*(2+1+2)

(1+2+1+2)*(1+2)

And we can obviously get 1 number:

1+2+1+2+1+2 = 9

*It's kinda impossible to get 4 numbers 'cause one of these numbers would be 1, and multiply by one wouldn't change anything.

let's work with the situation we have 2 numbers. Let's call them x and y

we know that x + y has to be 9

what's the largest value of xy?

since x+y = 9 --> y= 9 - x. Using substitution we have( ? means a real number):

xy = ?

x(9-x) = ?

9x - x² = ?

Derivatin' we must get: 9 - 2x = 0 (since the derivate of a real number is 0) --> x= 4.5 (and y = 4.5)

(1+2+1)*(2+1+2) will be the largest IF we can only have 2 numbers(and the result would be 20)

BUT there's another possibility, the case (1+2) (1+2) (1+2)

And this result will be 27. Which is bigger than 20, and this will be the largest result.

BONUS

We can also use the formula: -b/2a to found x=4.5 will be the largest, and using logic u can assume that x=4 and y = 5 will be the largest if we only had 2 numbers to multiply.

P.s: Sorry about grammar, i'm not an english speaker. Thx

Victor Veras - 5 years, 4 months ago

Henny Lim
Jan 13, 2016

Actually I answered 27 on the third attempt.

First, I got 20 (I was stuck after getting sum of 1-2-1 and 2-1-2, then I multiply them)

Second I got 24 (I added 1-2-1 then multiply it with 2, then multiply it again with 1+2)

At last I realized that I should add 1+2 and multiply all of them, to get 27.

Intuitively I think there is no possible result which is larger than 27. But don't know how to prove it. Since powering, factorials, concatenating, or any other signs are prohibited. If not so, there will be so much possibilities.

Actually, you can prove it by trying all the possible expressions amd tabulate them. That would take around 9 combinations. Useless, just saying though.

Alex Diola - 5 years, 4 months ago

Karthick Shiva
Jan 11, 2016

2 + 1 = 3

3 x 3 x 3 = 27

Maximizing Ones and Twos

The answer is 27.

6 solutions

UPDATE

BONUS

BONUS

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