+11912 which of the following is a prime no. ...
43!+2 , 43!+3, 43!+4...................43!+43.
pls explain in detail. thank you!
None of them is prime !! They are 43 consecutive composite numbers by definition, since all the added numbers[2, 3, 4, 5..........43] are already factors of 43!
+129852 None are prime....as the guest said....
To see this more clearly....note that we can write
43! + n where n is an integer that varies from 2 to 43
Note that in every case, n will divide both terms evenly because n will be a factor of both terms
Thus
43! + n will be divisible by 2,3,4,5,6......41, 42, 43
So.....it can't be prime because it is diviible by some integer other that 1 (and itself)
+11912 CPhill...i have no idea about your answer......shouldnt n vary from 1 to 43?
*Note that in every case, n will divide both terms evenly because n will be a factor of both terms*
what do you mean by that statement..
i dont understand anything after that either....pls could you elaborate.....thank you!
It works like this: Take a smaller number such as 10!. I hope you know what 10!(ten factorial) means. It means this: 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 =3,628,800. Notice that 2, 3, 4, 5, 6, 7, 8, 9, 10 are ALL factors of their product of 3,628,800. This means that 3,628,800 will divide any one these factors EVENLY. So, for example, 3,628,800 / 7 =518,400. OK?. Now, if you add another 7 to 10!, you will have:3,628,800 + 7 =3,628,807. Now, if you divide this new number by 7, you will get the following:
3,628,807 / 7 =518,401.... So, now you can see 10!+7 will divide 7 EVENLY again, and the original quotient of 518,400 is gone up to 518,401. And the 1 at the end of 518,401 simply means we have 1 extra 7 because we deliberately added to 10!. And this applies to all numbers from 2 to 10 when added to 10!. I hope this is clearer now.
