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# Proof Practice-I

$$\text {Proof the following}$$

$$\bullet$$ Given any positive integer $$\displaystyle k$$ ,prove that there are $$\displaystyle k$$ consecutive integers that are all composite.

Note by Anik Mandal
2 years, 10 months ago

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One line proof. $$(k+1)! + i$$ where $$2 \leq i \leq k+1$$

- 2 years, 10 months ago

Sorry,I didn't get it.Please explain it clearly@Siddhartha Srivastava

- 2 years, 10 months ago

Expanding on what Dinesh Chavan said. $$(k+1)! = (k+1)*k*(k-1)...*4*3*2$$. So for each $$2 \leq i \leq k+1$$, we see that $$i|(k+1)!$$. Also, it is trivially true that $$i|i$$. Therefore, $$i| (k+1)! + i$$. Since $$i$$ divides the number and it is greater than one, the number must be composite.

- 2 years, 10 months ago

If u know factorial notation then notice that for each $i$ that same $i$ from factorial expansion will come out common and the expression will be composite.

- 2 years, 10 months ago