How many Zeroes!
1 1 9 6 1 0’s 0 0 0 0 0 0 0 0 … 0 1
Is the number above composite?
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2 solutions
EDIT: Brian below me is correct, this should be to the power of 1962, not 1961. I'll leave the below here as additional insight into these types of problems, but my solution does not work for this problem. (Though see footnote for why it hardly matters)
We can write the number as x 1 9 6 1 + 1 , where x = 1 0 .
Next, we find that this expression is divisible by x + 1 , because 1961 is odd. To see why this is, we can apply long division and find a pattern, which ends up cancelling out any remainder. This implies that the number is divisible by 11.
A footnote: Determining whether a number is prime or not is a very difficult problem in general with no easy solution. On the contrary, we only need to find one out of the countless techniques possible to find a single factor to determine that the number is composite. By the nature of this problem, it becomes easy to assume that the number must be composite, or else it would not be a fair problem to solve.
Wouldn't the number be 1 0 1 9 6 2 + 1 ? We can still factor this, though, since 1 9 6 2 = 3 ∗ 6 5 4 , so
1 0 1 9 6 2 + 1 = ( 1 0 6 5 4 + 1 ) ( 1 0 1 3 0 8 − 1 0 6 5 4 + 1 ) ,
thus ensuring that the given number is composite.
Question: Is it the case that all numbers of the form 1 0 n + 1 for n > 0 , other than 1 1 and 1 0 1 , are composite?
Edit: Apparently 1 0 1 is the largest known prime of this form.
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Nicely done :-
just a program.. def prime(n): count=1 while count<=n: count+=1 if n%count==0: flag='yes' break else: flag='no' return flag print prime((10**1962)+1)
output is..