Multiply together all the prime numbers from 2 up thru n, then add 1 to the result. Show that the final total is never a perfect square.
No doubt lots of you know how easy that one is. So, what’s a better example of a problem that stumps people in spite of being easy?
If you’re not already acquainted with the above example, being told that it’s easy helps you solve it. Can you find a problem where hearing that it’s easy doesn’t help?
This isn’t true. For example for N=4, we have 2 times 3 times 4 plus 1 is 25, which is a perfect square. It isn’t a perfect square of any integer up to N, which may be the proof you were thinking of
ETA: Erroneous answer now spoiler protected, so I don’t confuse other people.
Primes up to N. For N=4, we have 2*3+1 = 7, which is not a perfect square
D’oh. My apologies. I misread the OP and now must slink off in shame.
This one is plainly easy, as long as you’re 75+ years old. How easy is it for younger people? (Pencil and paper allowed, nothing else.)
How many digits in 5 to the 300th power?
A converse is Goldbachs’s conjecture. Every even number can be expressed as the sum ot two primes.
Easy to state and just about anyone can understand it. But still not proved.
In 30 seconds or less, add all the whole numbers from 1 to 100.
The only thing I can think of is that people that old would have worked with logarithm tables enough that they’d have memorized the common log of 5?
That’s easy. You have 1+100, 2+99, 3+98… Fifty pairs of numbers each adding up to 101. 50 x 101 = 5050.
Just about right. If you used logarithms, you remember the base-10 log of 2 is a musical 0.30103, so the log of 5^300 is 300 minus (300 x 0.30103).
In @Andy_L’s defense, n is a whole number; p is a prime.
In expiation, here’s one.
I’m not telling you what base I’m working in - can you if 1001 is a prime number, even with that handicap?
That’s x^3 + 1, where x is the base. Which factors to (x+1)(x^2-x+1). So whatever your base is, 1 plus that is a factor..
OK, here’s an old favorite of mine:
Simplify
(x-a)*(x-b)*(x-c)*...*(x-z)
Heh.
That’s zero, because of the (x-x) factor.
It’s easy if you are Gauss.
No one has yet answered this, so I’ll give my answer. I don’t remember ever seeing this problem before, and the answer wasn’t immediately obvious, but it is relatively straightforward. I’m curious whether my solution is the intended “easy” one.
Let P represent the product of “all the prime numbers from 2 up thru n.”
Suppose P + 1 = k² for some whole number k.
Since P has a factor of 2, P is even, and so P+1 = k² and k are odd.
P = k² – 1 = (k – 1)(k + 1), and since k is odd, both k–1 and k+1 must be even.
But P has only one factor of 2, so we have a contradiction.
and thus 1001 in any base is always divisible by 11
Yeah that’s how the famous story went.
Another answer to the multiply-primes problem is similar to Thudlow_Boink’s. You multiply 2 and a string of odd numbers, so the product is an even number but not a multiple of 4 – so it’s 2 more than a multiple of four, and the final total is 3 more than a multiple of 4, which a square never is. (The square of an odd number 2n + 1 is 4n^2 + 4n + 1.)