Raising Squares Part 2

Logic Level 2

$\LARGE \square^\square \ + \ \square^\square \ + \ \square^\square$

Six numbers $1,2,3,4,5,$ and $6$ are to be filled in the square boxes above (without repetition). Of all $6!=720$ possible arrangements, find the minimum value of the resultant number.

Details and Assumptions

As an explicit example, if the six numbers are $2,2,2,3,3,$ and $3,$ then the minimum resultant number is $2^3+2^3+2^3=24.$

See Part 1 and Part 3 .

The answer is 90.

3 solutions

Aravind Raj Swaminathan
Aug 13, 2015

$1^{6}$ $+$ $5^{2}$ $+$ $4^{3}$ $;$ $1$ $+$ $25$ $+$ $64$ $=$ $90$

How do we know that's the minimum?

Chung Kevin - 5 years, 10 months ago

Bcoz there is no repetition of given numbers,we must take aravind Raj's solution

murali mohan - 5 years, 10 months ago

How can we think this method to proceed to get the solution.

Prabhav Bansal - 5 years, 10 months ago

You can't do do anything other than this solution

ANAND THAKRE - 5 years, 8 months ago

To be frank; I used more of logical reasoning. If we have two numbers; take

$X$ $&$ $Y$

$Suppose X$ $>$ $Y$

$Then for sure;$ $X^{Y}$ $<$ $Y^{X}$

Note : This does not apply if 1 or 0 are either of the numbers and/or the first 3 natural numbers are used in a pair. i.e $3^{2}$ $>$ $2^{3}$ .

6 being the last number cannot stay as the base as that would cause the least value of its would be 6. So we take $1^{6}$ as that would be the least value of all.

We have 2,3,4 and 5 left.

Taking a sample space of all;

$2^{3}$ ; $2^{4}$ ; $2^{5}$

$3^{2}$ : $3^{4}$ : $3^{5}$

$4^{2}$ : $4^{3}$ : $4^{5}$

$5^{2}$ : $5^{3}$ : $5^{4}$

From the above sample space; the least value without repeating its value are:

$5^{2}$ $&$ $4^{3}$ .

Adding all:

$1^{6}$ $+$ $5^{2}$ $+$ $4^{3}$ .= $1$ + $25$ + $64$ = $90$

Aravind Raj Swaminathan - 5 years, 10 months ago

I saw your post after I posted my solution. They are pretty much the same. I am sorry.

Shein Phyo - 5 years, 7 months ago

Not a problem at all. :)

Aravind Raj Swaminathan - 5 years, 7 months ago

After starting logically you could have continued doing so rather than using brute force. Let's see if I am right
After sorting out 1 and 6 we are left with 2 3 4 and 5. Let's take 5. What is the minimum number we can get by using 5 as a base and any other as an exponent or vice versa. The obvious solution would be 5^2. That leaves us with 3 and 4 and using your X and Y conclusion it is obvious that 4^3 is smaller than 3^4. If you look at my solution you will find that this will work with any count of numbers.

Zahid Hussain - 1 year, 10 months ago

Jc Mae P
Sep 5, 2015

To ensure that you would be getting the minimum number, take 1 and the maximum value which is 6 and use 1 as a base since in $1^{n}$ whatever value n may be, the result would still be 1.

Now that we have the first addend, the numbers left would be 2,3,4, and 5. In the possible combinations that we can get from those numbers, which would be:

$2^{3}$ $=$ $8$

$2^{4}$ $=$ $16$

$2^{5}$ $=$ $32$

$3^{2}$ $=$ $9$

$3^{4}$ $=$ $81$

$3^{5}$ $=$ $243$

$4^{2}$ $=$ $16$

$4^{3}$ $=$ $64$

$4^{5}$ $=$ $1024$

$5^{2}$ $=$ $25$

$5^{3}$ $=$ $125$

$5^{4}$ $=$ $625$

the numbers that would produce the least value without repetition would be $5^{2}$ and $4^{3}$

Hence:

$1$ $+$ $5^{2}$ $+$ $4^{3}$ $=$ $90$

The answer is 90.

Zahid Hussain
Jul 22, 2019

I think I have got a strategy for this type of problems but would be very grateful if someone could prove or disprove it. It is as follows.
1. Put the numbers in ascending order numbering them 1,2,3, ...n-2, n-1, n
2. Now the numbers to add would be n^1, (n-1)^2, (n-2)^3...
3. One exception is if the first number is 1 then the first term would be 1^n, instead.
4. Another unique case is when there are just 4 numbers and they are 1, 2, 3 and 4. In that case the numbers would be 1^4 and 2^3.
5. I think this would work for any pair of numbers even if they are not consecutive.

Raising Squares Part 2

The answer is 90.

3 solutions

0 pending reports