It is proved that 32 cannot be written as the sum of n consecutive natural numbers

The sum of n consecutive natural numbers is.

s=n+(n+1)+(n+2)……n+m)(2n+m)(m+1)/2

If m is an odd number, then 2n+m is an odd number.

If m is even, then m+1 is odd.

Then the sum of n natural numbers must be odd * even or odd * odd.

32 = 2 5 No matter how you distinguish such odd numbers * even numbers, 1 and 32 are not consecutive even numbers, so 32 cannot be written as the sum of n consecutive natural numbers.

Let the three numbers be n-1, n, n+1

The sum of three numbers = n-1+n+n+1=3n

So the sum must be a multiple of 3.

32 3 is not equal to a multiple of 3, so it is proven.

You can just extrapolate the rest, and you won't prove it.

Proof: 1+2+3+4+5+6+7=28

And 32 is in the range of 28 and 36, not the two numbers 28 and 36.

Using the hypothetical method, it is assumed that there are n consecutive natural numbers that are summing and equal to 32. Let the first natural number be x, the second one be x+1, and so on nth is x+n, and add n natural numbers to get n?+(1+2+..

n)=n?+n?1+n) 2=32, because the smallest natural number of x can be taken to 0, so the maximum n can be taken to be 7, and n is equal to 1 to 7 into the above equation, and x is found to be a natural number, which proves that...

First of all, we can find 2013 consecutive natural numbers Minghe, and they search for non-prime numbers, such as 2014!+2,2014!+3,……20141+2014, they are all combinations.

Then shift the 2013 numbers to the left (the first number becomes 2014!).+1,2014!,2014!-1……Until the first number is prime, then the 2013 number satisfies the requirement.

The sum of the three consecutive natural numbers is less than 12, and how many groups of such natural numbers are there before they are brightened.

Three groups. (0,1,2), key leak (1,2,3), (2,3,4).

Well, set a continuous natural number if the dust is dry as k,k+1,., .,k+r-1, then it is proof.

2 n=(k+k+r-1)r 2 is uninteger solved, i.e. .

2k+r-1）r=2^（n+1）

Here 2k+r-1=2 s, r=2 t, then 2k-1=2 s-2 t=2 t[2 (s-t)-1].

This inevitably requires t=0 and r=1, so it is not a certain number to sell wide.

The product of 13 zeros includes at least 13 5s.

There are 11 5 10 15 20 25 30 35 40 45 50 55 There are 11 of them, of which 25 50 each contains 2 5 factors.

So the smallest natural number that comes to the last occurrence is 55

5 10 15 20 25 30 35 40 45 50 55 There are 11 in total, of which 25 50 each contains 2 5 factors.

So the smallest natural number that comes to the last occurrence is 55

1. For a given arbitrary positive integer n>1, the following n consecutive natural numbers are composite numbers (n+1)!+2,(n+1)!+3,..n+1)!+n+1 card.

Ming: Take the k-th number (n+1)!+k+1), 1 k n because (n+1)!=1*2*3*..k+1)..n+1) so (n+1)!+k+1)

At least one factor k+1. Thus for any given positive integer n>1, there are n consecutive composite numbers.

2. The formula is: let k=(n+1)!=2006+1)!=2007!

then k+2=2*(1+3*4*5*....*n+1)) This means that k+2 is divisible by 2.

k+3=3*(1+2*4*5*6*…*n+1)) This means that k+2 is divisible by 3.

k+(n+1)=(n+1)(1+2*3*4*…*n)

Let k = 2007!（k=1*2*3*··2007）

Altogether 2006 counts.

Take k=2007! （k=1*2*3*·· 2007), then k+2 is divisible by 2, k+3 is divisible by 3, ·· k+n is divisible by n, ··

k+2007 is divisible by 2007, for a total of 2006 numbers.

Between 2004 and 2104 there are 3 consecutive natural numbers, the smallest of which is divisible by 3, the largest divisible by 7, and the other divisible by 5. The smallest of these 3 natural numbers is (2049).

A: 2049, 2050, 2051 are eligible, and the smallest one is 2049

According to the drawer principle, there must be at least 2 of the three numbers that are both odd or even (3 numbers in two drawers).

Odd + Odd = Even.

Even number + even number whisper wheel = even number.

Therefore, no matter which type of initiation is the case, there is always an even sum of at least 2 numbers.

Because of the natural number of 3 consecutive companions and the natural number of Mori scribbles, this may be the case.

Odd, even, odd, odd + odd and even this filial number.

Even, odd, even, even + even and even.

So the conclusion holds.