Integer all

We have $\dfrac{n+k}{k}$ is an integer if and only if $\dfrac{n}{k}$ is an integer.

All of numbers are integers if $n$ is multiple of $1, 2, 3, 4$ and $5$ .

Since $lcm(1,2,3,4,5)=60$ , $n$ must has form as $60m$ where $m$ is an integer.

So, the minimum value of positive integer $n$ is $60$ .

We can express the given expressions as, $\frac{n}{1} + 1, \frac{n}{2} + 1, \frac{n}{3} + 1, \frac{n}{4} + 1, \frac{n}{5} + 1,$ .

This means we are finding the first positive integer that divides $1, 2, 3, 4, 5.$

This is the Least Common Multiple(LCM) of the set, and for this set it is 60.

Let's go divide and conquer as following;

1. $\frac{n+1}{1}$ is an integer for all integral values of $n$ which does not reduce the set of possibilities.

2. $\frac{n+5}{5}$ is an integer for $n=5, 10, 15, ...$ .

3. $\frac{n+2}{2}$ and $\frac{n+4}{4}$ are integer iff $n$ is even. Implies that $n$ belongs to ${10, 20, 30, . . .}$

4. $\frac{n+3}{3}$ is an integer iff $n$ is divisible by $3$ . Implies that $n$ belongs to ${30, 60, . . .}$

For $n=30$ , $\frac{30+4}{4}$ is not an integer.

For $n=60$ , $\frac{60+1}{1}=61$ , $\frac{60+2}{2}=31$ , $\frac{60+3}{3}=21$ , $\frac{60+4}{4}=16$ and $\frac{60+5}{5}=13$ are all integer.

Therefore $60$ is the smallest possible integer after zero such that all the given fractions are an integer.

What we have: $\frac{n+m}{m}$ Lets split up the fraction $=\frac{n}{m} + \frac{m}{m} =\frac{n}{m} + 1$ Now the solution looks more clean right ?

Since $m = (1,2,3,4,5)$ and we want to know the lower $n$ that divided by $m$ give us a integer we just take the lcm from $1,2,3,4,5$ and it's $60$

Q.E.S

We must find a number that is exactly divisible by all denominators to get integers(a multiple of the denominator is added, so the numerator becomes exactly divisible by denominator). LCM is 60, so that's the answer

This might make a fun anecdote, I used an intuitive approach. I'm a graphic designer. When doing textual layout, it is common to divide a page into a grid of equal width columns (with margins inbetween so things don't bump into each other). The number of horizontal cells is often $12$ . Because $12$ is a nice number since it's divisible by $1, 2, 3, 4, 6$ and $12$ . This makes it convenient to align text columns of different widths onto a page with easy to manage fractions. That alone makes $12$ one of my favorite integers. So when searching for a least common multiple of small numbers, I often think of $12$ , I know the least common multiple for $1$ to $4$ is $12$ . The $5$ was in the way though, but $12*5$ solves that. Is there anything smaller? Apparently not.

I'm slightly jealous of countries on the imperial system sometimes -- then again, not really...

Every given ratio can be expressed in the form n/m +1.

m={1,2,3,4,5}

So it represents that n must be that number which is multiple of m But we want the number n which is just after zero and satisfy the condition, clearly it is asking the lcm(1,2,3,4,5) = 60 So n takes the value=60∆

∆ is a whole number

The LCM of the denominators is the answer since n is divisible by all of the denominators , as a multiple of the denominator is being added to n every time . Hence the answer is 60

n must have a factor common to 2, 3, 4, 5 so that when they are added to i and then the whole is divided by i they yield a integer as the denominator is a factor of the numerator; where i =1,2,3,4,5. Therefore lcm of 2,3,4,5 is 60 and so n =60.

N has to be divisible by 1,2,3,4,5, hence it has to be it's LCM. Which is equal to 60

Take lcm of the denominators

Think of LCM (lowest common multiple) so therefore 60 would be the smallest number that can be divided by 1,2,3,4,5.

$\dfrac{n+x}{x}$ $=\dfrac{n}{x}+1$ Thus, we can conclude that in order for $\dfrac{n+x}{x}$ to be an integer, $x$ must be divisible by $n$

In the above problem, using the aforementioned rule, we were told that $n$ must be divisible by 1, 2, 3, 4, and 5, ergo, $n$ will be the lowest common multiple of 1, 2, 3, 4, and 5, which is 60.

Therefore, the answer is 60.

All are equal to 1. Therefore we have to find a soln that will give us perfect integer , therefore we have to find LCM of denominators therefore we can find a common dividend and it will give an integer LCM of 1,2,3,4,5 is 60 therefore it is the answer