Grades 11-12 Video Solutions 2010-11&12 Video Solutions 2010 problem20

Video Transcription

Question number 20. Each symbol here, which I will pronounce as sin, in the pronunciation 1, sin, 2, sin, 3, sin, even sin, 10, can be either mulipplication sin or additionsin.  Let n be the largest possible value of this statement. What is the smallest prime divisor of n? The first task for us is to maximize the value of n.  We have two choices. We can recommend its with additionin, or recommend it with mulplication. So we expect that this operation, given by these two concentric circles,  is multiplication, and it gives a greater value.  And then of course we have to check.  So n to 10 to 9 to 8 to 7 to 3 to 2 to 1 still exists, another name for this is 10! That's the card notation. Then we see that to expect multiplication with additionsin  gives a smaller number, without us doing it for the multiplicative identity. As long as we do not expect the identity 1. And so here finally we see that if we expect  multiplication with additionin then we get a larger number. But in any other case it is not so. So we can say that n is less than here 10 to 9 to 3 to 2 and then to 1, instead of 1  but because 1 is the identity here I can multiply by 1 anyway and it won't expect my number. So what we have is that the largest number is actually 10! plus 1  and now we have to check for divisibility and this is not divisible by any number that divides 10! Why is that so? When we look at this expression  if some number divides both terms then it must divide the first term and it must divide the emptiest so 1 is not prime 2 divide 10!, but not 1 3 divide 10!, but not 1  5 divide 10!, but not 1 7 divide 10!, but not 1 So we always have a problem here with the emptiest term  So we conclude that the narrowest prime term that divides this expression is 10! plus 1 is probably 11 which is the narrowest candidate So actually 10! plus 1 is divisible  of 11 and we can check the divisibility rules or ask a computer to run the problem with factorization but in the problem here we conclude that there is none of the above  11 divides this number but it is not one of the choices so we have to choose E

Video Summary

The task is to find the smallest prime divisor of the expression \( 10! + 1 \). By maximizing \( n \), we find that the largest possible value results from multiplication operations, leading to \( n = 10! + 1 \). The expression \( 10! \) is divisible by smaller primes like 2, 3, 5, and 7. However, \( 10! + 1 \) is not divisible by these primes, leaving 11 as the smallest prime divisor. Therefore, the smallest prime divisor of \( 10! + 1 \) is 11.

Keywords

smallest prime divisor

factorial

10!

prime numbers

mathematics