When n is an integer, we try to explain that (2n + 1) (2n + 1) is a multiple of 8 (2n + 1) (2n + 1) - 1 is a multiple of 8

When n is an integer, we try to explain that (2n + 1) (2n + 1) is a multiple of 8 (2n + 1) (2n + 1) - 1 is a multiple of 8

4n2+4n+1-1=4n2+4n=4n(n+1)
n. N + 1 has an even number, so it's a multiple of 8

(2) Is the square of any odd number minus 1 a multiple of 8

It must be
Prove: (2n + 1) ^ 2-1 = (2n + 1 + 1) (2n + 1-1)
=4n(n+1)
So n (n + 1) is even. So 4N (n + 1) is a multiple of 8. So the square of any odd number minus 1 must be a multiple of 8

Is the square of any odd number minus 1 a multiple of 8

Let the odd number be 2N-1
(2n-1)²-1=4n²-4n+1-1=4n(n-1)
N and N-1 are two adjacent positive integers, so one of them is even
So n (n-1) is a multiple of 2
So 4N (n-1) is a multiple of 4 * 2 = 8
So the square of any odd number minus 1 is a multiple of 8

From the natural numbers of 1-2001, what are the numbers that can be divided by 2 but not by 3 or 7?

572
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The product, sum and difference of two nonzero natural numbers must have a number divisible by 3
Right or wrong? Why?

Because there are only three cases, that is, the remainder is 0, 1 and 2. Because these two are non-zero natural numbers, so (1) when one of these two numbers is a multiple of 3 (the remainder of 3 is 0), the product of these two numbers can be divided by 3; (2) when the remainder of these two numbers divided by 3 is different (except 0, the remainder is 0)

The product of any three continuous natural numbers must be divisible by 2 and 3 at the same time. It is best to give an example

1、2、3

The sum of two can be divided by three natural products at the same time
Give reasons

I'll explain to you on the next 1.2 floor that the product of the multiple of 2 and the multiple of 3 is the multiple of 6. The multiple of 6 multiplied by any number can also be the multiple of 2 and 3, so it must be divisible by 2 and 3 at the same time

Find the sum of squares of numbers divisible by 37 in 1000

The quotient is 27
So it's 1 & # 178; × 37 & # 178; + 2 & # 178; × 37 & # 178; + +22²×37²
=37²×(1²+2²+…… +22²)
=1369×22×(22+1)×(2×22+1)/6
=5195355

The sum of squares of numbers between 10 and 150 divisible by 3 and 7
Hurry, hurry!

21^2+(21*2)^2+(21*3)^2+(21*4)^2+(21*5)^2+(21*6)^2+(21*7)^2
=21^2*(1+4+9+16+25+36+49)
=21^2*140
=61740

Let the sum of squares of the digits of a four digit positive integer be 50. How many such four digit positive integers are there,

The square of no more than 50 is 1 49 16 25 36 49
50＝1+49＝36+9+4+1＝25+16+9
therefore
Class I: 17000: C32 * 2 = 6
Class 2: 1236: a44 = 24
Class III: 5430:3 * A33 = 18
Summary = 6 + 24 + 18 = 48