Integer chemistry

Its obvious. Only those numbers n which are not prime can be written as $a \times b$ , where both a and b are not equal to 1(prime factors of n). Thus multiplication can be used to decrease the number of 1's required, as shown in the example of 6 in the question.

Thus, $15=3 \times 5$ , will give least number of 1's. But $3=3 \times 1$ and $5=5 \times 1$ are themselves prime. Therefore, 3 requires 3 *1's, as addition $3=1+1+1$ will give them least number of 1's, whereas multiplication increases number of 1's as $3=(1+1)\times 1+1$ or $3=1 \times 1+1+1$ , both give * 4 1's.

Similarly $5=1+1+1+1+1$ requires *5 * 1's. Hence total number of 1's $=3+5=\boxed{8}$ .

Find the prime factors of the number 15.

We find "prime" factors because they cannot further be reduced. So they can only be formed by addition of "1's" .

Moreover, this gives us the fewest no. of "1's" because these are the lowest possible numbers which can be multiplied to form the given number. Hence we get the lowest no. of "1's".

Here its 3 and 5.

So, 3x5=(1+1+1)x(1+1+1+1+1)=15

Total no. of "1's" used is 8 ANS

For complexity of 6, the shortest way is the prime fatcors= 3 * 2=(1+1+1) * (1+1). Similarly in the case of fifteen the prime factors are-3 * 5 or (1+1+1) * (1+1+1+1+1).Therefore it consists of 8 ones(1) and therefore the answer is 8.

15 is the product of 5 and 3 ...and it cant be simplified anymore as both of them are prime numbers yet..so 5's complexity is 5 and 3's is 3 .Adding them it sums to 8 , the answer. Easy stuff .all you got to do is to simplify it and try to find the maximum factors although i didn't need to do anything such here....

the only numbers by which 15 is divisible is 3 and 5. so adding 3 and 5 we get 8, which is the answer to the given question.

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

Basically, if we factorize the number, we will get the product of the minimum numbers to make that number . . . So, when we factorize 15, we get it as 3*5 . . . so we will need 8 = (5+3) 1s.

Since 6=(1+1+1)*(1+1),Hence 6 is a product of 3 and 2. Here the sum of 1's is 5.

Similarly, 15 is a product of 3 and 5, Hence 15=(1+1+1)*(1+1+1+1+1). Here the sum of 1's is 8.

15= 3 5= (1 1 1) (1+1+1+1+1)

(1+1+1+1+1)*(1+1+1) =5 * 3 =15 so complicity is 5+3=8

15=(1+1+1)*(1+1+1+1+1)

8 is the obvious answer because the only simple way to get 15 is either (5) (3) or (3) (5). So 5+3 is nothing but 8. So Ans=8.

3 5 (1+1+1) (1+1+1+1+1) = 15

$Let\quad 6\quad =\quad 3\quad *\quad 2\quad \{ a\quad product\quad of\quad primes\} \\ \qquad \qquad (1+1+1)\quad *\quad (1+1)\\ \therefore Complexity:\quad 3+2\quad =\quad 5\\ Let\quad 15\quad =\quad 3\quad *\quad 5\quad \{ a\quad product\quad of\quad primes\} \\ \qquad \qquad =(1+1+1)*(1+1+1+1+1)\\ \therefore Complexity:\quad 3+5\quad =\quad 8\\ \\ \Longrightarrow \quad n\quad =\quad a_{ 1 }*\cdots *a_{ i }\quad \{ where\quad a_{ 1 }...a_{ i }\quad are\quad prime\} \\ Complexity\quad :=\quad a_{ 1 }+...+a_{ i }$

if we look closely 6=2*3

similarly 15 shlould be equaly to 3*5 as per given pattern it wil be

15=(1+1+1)*(1+1+1+1+1) hence complexity = 8

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

15=(1+1+1)*(1+1+1+1+1)

so, the number 15 has complexity 1+1+1+1+1+1+1+1=8

according to fundamental theorem of Arithmetics "Any integer greater than 1 is either a prime number, or can be written as a unique product of prime numbers (ignoring the order)".

(1+1+1) (1+1+1+1+1) which means 3 5=15 where 3,5 are primes and there is no way you can make 3 and 5 from any product combination

The complexity of number 15 is 8.

15 = (1 + 1 + 1) * (1 + 1 + 1 + 1 + 1)

the complexity of number 15 can be written as 15=(1+1+1) (1+1+1+1+1)

15 = 1+1+1+1+1+1+1+1+1+1+1+1+1+1+1

= (1+1+1) * (1+1+1+1+1)

so, we can say the no. 15 has complexity 8.

we break 15 into its prime factors by prime factorisation method. We get that there is no other way to get 15 than to multiply 3 with 15... 3 can be broken as (1+1+1) and 5 can be broken as (1+1+1+1+1) so we get the complexity 3ones + 5ones= 8

for this we we can do it by finding the biggest 2 numbers and adding it in this case the number is 15 we can split it into 3*5 and adding it =3+5=8 we can check this by (1+1+1)(1+1+1+1+1)

As number 6 has complexity 5 (1+1) (1+1+1) just like that 15 has complexity of 8 (1+1+1) (1+1+1+1+1)

Well, I saw that the ways can be as follows :---

$15=1+1+1+1+1+1+1+1+1+1+1+1+1+1+1$

$15=(1+1+1)*(1+1+1+1+1)=(1+1+1+1+1)*(1+1+1)$ and,

$15=((1+1)*(1+1)+1)*(1+1+1)$ , etc.

Then, I counted number of 1's in each of the formations and saw that 15 can be created at the very least by using 8 1's. Actually, truth be told, I just used trial and error method to solve this.

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

As you can see in the example they have given the complexity of 6 is 5 because it can be solved by using 5 ones, which is (3) (2) = (1+1+1) (1+1). similarly 15 = (3) (5) = (1+1+1) (1+1+1+1+1). as we know the lowest number of ones that we can use is only when the multiples of 15 will be multiplied. and then only you will get lowest value of complexity. so this is how we get complexity= 8 for 15.

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

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

11=1+(1+1+1+1+1)*(1+1)

number 6 is equal to minimum value of multiplier 2 and 3.2+3=5 same as case of 15 is equal to minimum value of multiplier 3 and 5 .3+5=8 so ans is 8