(i). K product solutions for the multiplication of consecutive positive integer orders. Let's start with a proposition:
The product of k consecutive positive integers must be divisible by k! (factorial of k), i.e., there is a k! | for any non-negative integer m (m+1) (m+2)... (m+k)。
prove:
(m+1) (m+2)... (m+k)
= 1×2×... ×m×(m+1)×(m+2)×...×(m+k) / (1×2×... ×m)
= (m+k)!/m!
= k! [(m+k)!/(m!k!)]
And the number of possible combinations of b elements arbitrarily selected from a element is:
c(a, b) = a!/((a-b)!b!) (b≤a)
令 a = m+k, b = k,则有:
c(m+k, k) = (m+k)!/((m+k-k)!k!) = (m+k)!/(m!k!)
Therefore there are:
(m+1) (m+2)... (m+k) = k!c(m+k, k)
c(m+k, k) i.e., the number of possible combinations of k elements selected arbitrarily from the m+k elements, it must be a positive integer, which means k! Must divide k!c(m+k, k) i.e. divisible (m+1)(m+2)... (m+k)), i.e.,
k! | (m+1) (m+2)... (m+k)
Certified
When in the proposition, k = 5, k! = 5×4×3×2×1 = 120, so we say that the product of 5 consecutive positive integers must be divisible by 120.
The common conclusions of this proposition are:
The product of any 3 consecutive positive integers can be divisible by 6;
The product of any 2 adjacent positive integers can be divisible by 2;
Everyone has probably seen this.
Another proposition associated with conclusion 1 above is that if the sum of the numbers of digits (decimal) of the positive integer m is a multiple of 3, then m can be divisible by 3.
Let,b = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9},Order,
m = ar10r + arkanye₁10r⁻¹ + ... + a₁10¹ + a₀,(ar, archlor|₁, ..., a₁, a₀ ∈ b)
rule:
m = ar(10r-1) + arkanye ₁(10r⁻¹-1) + ... + a₁(10¹-1) + (ar + arkanye ₁ + ... + a₁ + a₀)
And for n = 1, 2, ..., r has:
10ⁿ - 1 = 9×10ⁿ + 9×10ⁿ⁻¹ + ... + 9×10 + 9
Obviously 3 | 10ⁿ - 1, so:
3 | aᵣ(10ʳ-1) + aᵣ₋₁(10ʳ⁻¹-1) + ... + a₁(10¹-1)
Thus just keeping:
3 | aᵣ + aᵣ₋₁ + ... + a₁ + a₀
namely
The sum of the numbers of m is divisible by 3 (i.e. the sum of the numbers of numbers of m is a multiple of 3)
rule
3 | m
(ii). Euclid's prime number infinity order remains solution. In natural numbers integers contain prime numbers 1,2,3,5,7, ... What's the interesting phenomenon of factorial? About 300 BC, Euclid proved the infinity of prime numbers in the Primitives of Geometry.
So what is the interesting phenomenon of the infinite formula multiplication of Euclidean primes? Where is the prime number of the order multiplier tail is zero, can be divided by 5 and 10, so the ancient "Qimen Dunjia" has: five days a yuan Tuesday shuo, ten sons and one A are the same.