Minimize This Sum!

Using AM-GM , we have that $\frac { a+b+c+d+e }{ 5 } \ge \sqrt [ 5 ]{ abcde } \\ \frac { a+b+c+d+e }{ 5 } \ge 1\\ \quad a+b+c+d+e \ge 5$

Use your logic to answer this question. We only need to look at two values to determine the logic and the answer. If a < 1, for ab = 1, b must be greater than 1, thereby making the sum of a and b, when a<1 and b>1, greater than the sum of a and b when a = 1, and b = 1. Ex: (1/2) x (2) = 1.

1 x 1 = 1. 1 + 1 = 2. By this example we see that any value of a, b, c, d, or e that is not 1 will yield a sum which is larger than when a, b, c, d, and e all equal 1. The answer is then 5: 1 + 1 + 1 + 1 + 1.

Since a,b,c,d,e are real +ve number and product is 1,that means values of a,b,c,de and e are 1 .therefore sum of all these that is a+b+c+d+e =5 Ans K.K.GARG,India

$\begin{cases}a,b,c,d,e>0\\abcde=1\end{cases}\implies a+b+c+d+e\stackrel{\text{AM-GM}}\ge 5\sqrt{abcde}=5.$

Since $5$ can be reached when $a=b=c=d=e=1$ , the answer is $\boxed{5}$ .

you people make simple questions hard for no reason at all

It was simple really, all you had to do was figure out what value could be multiplied 5 times and still equal one. Which is one, then you plug one into all the other variables, and add them together. It doesn't have to be that complicated

Uhm... But what if it's like this: a, b, c, and d all equal -1, and e equals 1. That would suit the equation and the sum would be -3..?

4 of those can be -1 since a negative times a negative is a positive thus -1×-1×-1×-1×1=1 So -1+-1+-1+-1+1=-3 so really the answer could be -3 or -1 or 5 as an equal amount would have to be negative to remain as a positive one with multiplication.

Answer is 5 as the only way u can get 1 on multipling five digits is twhen each digit is 1 and 1+1+1+1+1 equals 5

I'm pretty sure that "real number" isn't the same as "only integers".

If (a)(b)(c)(d)(e)=1 then, a+b+c+d+e = ? means 1+1+1+1+1=5 so then answer is 5

a×b×c×d×e=1 so the opposite of multiplication is addition,which makes the problem a+b+c+d+e since a×b×c×d×e=1 then the new problem will be 1+1+1+1+1 which equals 5.

5 1+1+1+1+1=5 5 times 1

They multiplied the first equation. And added the second. They all are 1. 1x1=1 as for 1+1=2 so 1x1x1x1x1=1 and 1+1+1+1+1=5.

All integers multiplied together equal one so when each integer is one when added equals five.

First observe the conditions (all are real positive numbers and abcde=1). Real numbers doesn't contain integers .. So we use 1,2,3,4......... Product must be 1...so all letters are substituted by 1. Hence we get abcde=1 and a+b+c+d+e=5.

The answer could also had been 1 because if one though about this logically if the answer is a positive number there only needs to be an odd number of positive factors for the product to become a positive number using this -1•1•1•1•-1 is still equal to positive 1 therefore if you added these numbers together the answer would have been 1

-1x-1x-1x-1x1=1

Minimum = -3

abcde=1. this means a=1,b=1,c=1d=1,e=1 so that a+b+c+d+e=5. (1+1+1+1+1=5)

A b c d and e cannot all be variables for 1, in terms of alegbra, it is not correct.

How about if a,b,c,d=-1and e=1 so -1×-1×-1×-1×1=1 and a+b+c+d+e are -1-1-1-1+1=3

1.0000000000000000000000001 =1

The answer should be -3, if we multiply (-1)×(-1) (-1) (-1)*1=-3 -3 < 5

All the letters = 1 so 1 +1+1+1+1+1 = 5

Perhaps I'm missing something. Given that the integers can all be the same based on the correct answer, couldn't all but one of them be -1, resulting in an answer of -3?

I don't know why but i thought e was the natural number. Maybe the way it was written.

what is AM-GM? somone please explain

Everyone gave such in depth answers, I achieved the answer by realising that one times one is simply one, no matter how many times you times one by one...

What if some of them were negative? E.g. -1x-1x-1x-1x1=1 -1+-1+-1+-1+1=-3 Is there some reason I cannot do this?

Logic.... 1 * 1 * 1 * 1 * 1 = 1

Therefore

1+1+1+1+1=5

If the answer after multiplication will be 1 and they are positive real numbers then the value of a=b=c=d=e=1 so sum is 5

Using Calculus, a minimum value is found by taking the derivative of the equation. The derivative of, a, (variable) is always 1 and derivative of 5(constant) is 0. So 5 variables added together gives you 5 and on the other side the derivative of that constant is 0 so you get a derivative of 5 being the minimum value

Wtf it's so easy guys what's with all this random working out, if it's multiplied and youre answer is 1 and on top of that it's a positive natural number then obviously a b c d e can't be irrational, furthermore if u read the question carefully and think, it doesnt say a b c d e are distinct thus the obvious answer for a b c d e are 1 hence the sum would be 5.

Lotsa overthinking here in the comments. The key is the first equation and the axiom that a real product of 1 must have all factors equal 1, making a..e all equal 1. Then simply add them up.

If a=b=c=d=-1 and e=1 => a x b x c x d x e=1 and a+b+c+d+e=-3, min(-3,5)=-3

Guys, Let's consider that a=1 b=-1 c=1 d=1 e=-1 so when multiplying the result is 1 and the addition is 1

I randomly guessed cause I couldn't work it out

Ummmmmm, I just looked at the multiplication part. If all those variables multiplied equals 1, then each variable has to be equal to 1. Adding them together equals 5

If the numbers have to be positive and multiply to be one, then it has to be one. One is the only number that can be multiplied to itself and still have one. Then, 1+1+1+1+1=5

The given equation a.b.c.d.e=1 is only possible if a=b=c=d=e=1 then just add all the five terms you'll get the answer "5".

using the rule of indices, $a^{0} =1$ . therefore, the answer of this question for the sum of all the values a, b, c, d, e can be as little as 0.0000000000000001. However, as the answer required is an integer, the smallest acceptable answer of all the values a, b, c, d, e should be: $1^{0} + 1^{0} + 1^{0} + 1^{0} + 1^{0} = 1+ 1+ 1+ 1+ 1 = 5$

1x1x1x1x1=1, 1+1+1+1+1=5. It's that easy

I get why it is 5.

Nobody ever said that the five numbers had to be different.

Since a,b,c,d,e all are +ve real numbers, a×b×c×d×e=1 iff a=b=c=d=e=1. Hence a+b+c+d+e=1+1+1+1+1=5

If a is 1, b is 2, c is 1/2, d is 3 and e is 1/3, then abcde is 1 and a+b+c+d+e is 5 and 5/6. You can put any guess on the value of abcde and it's always going to be greater than 5. The lowest answer would be 5 based on the assumption that a=b=c=d=e=1.

I just assumed that the only way you can multiply 5 numbers by each other and get 1 is if all the numbers were 1. Then I added to make 5. Surprised I actually got it.

This is my attempt to solve it 1+1.1+1/1.1+1.2+1/1.2= 5,04 That's The nearest I could get to 5 But I realized that we can use a kind of limit. Where any operación we can do will tend to 5, but never tocuh this point

if axbxcxdxe = 1 only when a=b=c=d=e=1, then a+b+c+d+e=1+1+1+1+1=5

1×1×1×1×1=1 so for positive real nos a=1 b=1 c=1 d=1 e=1. So a+b+c+d+e=5

1 times itself is always 1. With that knowledge you just add 1+1+1+1+1=5 Simple logic understanding of basic algebra will help you solve this problem.

We are not given that a,b,c,d,e are distinct s o we can their values as 1

1+0.5+8+1/3+3=5??!?

a b c d e=1 That means all of the letters = 1 So a+b+c+d+e=1+1+1+1+1=5

I didn't use quite as mathmatical methods as most people seemed to have used.

I at first thought that all numbers would be different and would be something like, "0.25x0.2x1x0.2..." etc, because I thought that all the numbers had to be different.

But I couldn't find the 5th term that would enable me to make "a x b x c x d x e = 1".

So I realised the numbers could all be the same and all be 1; thus 1^5 = 1 . Simples. :)

am>=gm (a+b+c+d+e)/5>=9abcde)^(1/5)

ahaha...didn't even got me ten seconds...just think simple...no need for other difficult solutions...

Let : a = 1 ; b = 1 ; c = 1 ; d = 1 ; e = 1

so, 1 x 1 x 1 x 1 x 1 = 1

then : 1 + 1 + 1 + 1 + 1 = 5 that's it!

A b c d e=1 So all are equal to 1so you just need to add all 1 and you get answer

Funny...I just added to get a sum of five. Hmm

You get 1 as an answer in multiplication only by using 1 x 1. Doesn't matter how many times you multiply by 1, so still get 1 as an answer. This means that a, b, c, d and e must all be the number 1.

A+B+C+D+E = 1+1+1+1+1 = 5.

value of a,b,c,d cannot be fractional because in this case there should be a multiplicative inverse too , so 1+1+1+1+1=5

positive real numbers means 1 or above. so minimum value 1. so 1+1+1+1+1=5

Just assume that a,b,c,d,e are 1. That's all.

If a x b x c x d x e = 1, then all the variables have to equal 1. The only way for a number to be by multiplied by another number and equal 1 is 1x1=1. If each variable is 1, then a + b + c + d + e = 1+1+1+1+1 = 5. Also note, there is nothing written in the problem that says that the variables cannot be the same number.

the lowest possible positive number is 1, so i supposed all are 1, it could not be 0 coz after multiplication.. the answer would be zero.. so then supposing all lowest possible, i.e. 1 we get the ans 5 after addition

Now at first I was confused because even though it said the minimum number, it didn't exactly say it had to be an integer. So there are an infinite number of possible minimum outcomes. But I went based off the assumption that all numbers are equal to 1. I was making the same point to myself as to what another person said that two numbers could be reciprocals of each other and just one constant be equal to one.

thanks to Cauchy and AM-GM inequality

It'sreally very easy. a x b x c x d x e =1. They are all whole numbers. No decimals. The only way this is possible is for a,b,c,d,and e to equal 1. Plug in any other number, and it doesn't work. Since they all equal 1, simple arithmetic: 1 + 1 + 1 + 1 + 1 =5

If we multiply any no. By one then the answer is that no. A b c d e=1 So logically all veriable are 1 A+b+c+d+e=5. So,, 1+1+1+1+1=5 So answer. Is 5

First, think, if the value of product of "n" numbers is 1, then all the numbers got to be 1 or fractions like 1 (1/5) 1 (1/5) 1. But, solving these fractions would result in a number, which is greater than 5. So, 1+1+1+1+1=5 gotta be the answer fellas!

I know this will stupid but if in the multiplication problem it said (a)(b)(c)(d)(e)=1 that says each number is 1 so when added instead I got 5. maybe some people just thought to much into an easy problem.

Kinda a lazy way to solve the problem but just assume every letter equals 1 lol.

let the all numbers is 1 the 1/1 1/1 1/1 1/1 1/1=1 now , a+b+c+d+e = 1+1+1+1+1/1=5/1 the answer is 5

am>=gm (a+b+c+d+e)/5>=1 a+b+c+d+e>5

Ok, this one is not stark logical, i make two unproved assumptions, but that's how i got a 5 on first attempt:

I assumed you can get a $1$ from multyplying terms if pairs of them are co-inverse $b = 1/a, d = 1/b$ and the remaining term $e=1.$

In this case you get an expression:

$a + 1/a + b + 1/b + 1$

$(2a+1)/a+(2b+1)/b + 1 = 5 + (a+b)/ab$

$(2ab+b+2ab+a+ab)/ab$

$(5ab+a+b)/ab$

$5 + (a+b)/ab$

and from there i assumed $a$ and $b$ can be as small as 0+, which makes the relation 0+

Its quite simple.. 1 is a prime number while multiplying each variable must have the same value=1.. When u add all variable its equal to 5

I got 5 as the answer too, but then I thought of something: what if a=100 and b=1/100; c,d, and e are all 1. The sum would then be greater than 5, no? Someone correct me if I'm wrong.

We know a b c d e are +ve real no.s ... so we know we have two real no.s and their multiplicative inverse amongst these 5 no.s... Now we are left with only one remaining no. But it should be 1 because the product of the (no. , multiplicative inverse) pair is 1 and the actual product of 5 no.s is 1...

So say we have a, 1/a, b, 1/b, 1 as our 5 given no.s... and we want to find out min. Value of a+1/a+b+1/b+1.... and we know min. Value of a +1/a is 2... so we have... 2+2+1 which equals 5.

It never states that the integers must be distinct, so you can automatically assume that all the numbers equal 1 if the product itself is one.

This solution is not valid because when working in algebra all letters that are different require different numbers. In this problem, however, to reach a minimum of 5, a, b, c, d and e must have a numerical value of 1. In algebra, if a number is the same, the pronumeral is the same as well. Therefore, this solution is invalid and the problem needs to be rewritten to account for this.

here a,b,c,d,e are all positive real number in other word (a,b,,c,d,e)>0. in order to ; a x b x c x d x e=1 the value of a,b,c,d & e will be 1 and therefore a+b+c+d+e=1+1+1+1+1=5

AM>/=GM (a+b+c+d+e)/5 >/=(abcde)^(1/5)=1 so, a+b+c+d+e>/=1

a x b x c x d x e = 1, if a=1, b= 1, c=1, d= -1 and e=-1, so: 1 x 1 x 1 x (-1) x (-1)= 1! a + b + c + d + e = 1 + 1 +.1 + (-1) + (-1) = 3 + (-2) = 1 Why 1 is an wrong answer?

Why is it AM >= GM ???

Look at a x b x c x d x e=1, meaning all of them equals 1, so when you add a + b+ c + d + e the answer is obviously 5. simple as that .

It never said that the numbers had to be distinctively different. It did say, however, that they had to be positive and real.. meaning no fractions (so you cant multiply by 3 and then 1/3, etc) The only option remains for each of the variables to be equal to 1. 1+1+1+1+1=5.