Current location - Training Enrollment Network - Mathematics courses - Mathematical power table
Mathematical power table
First of all, a b stands for the b power of a.

I don't know what you have learned, but there are many ways to solve it, which is convenient for you.

Method 1: find the law.

The fifth power of any number, the unit is the unit of the original number.

The following is a list of 1 to the 5th power.

0、0、0、0、0

1、 1、 1、 1、 1

2、4、8、6、2

3、9、7、 1、3

4、6、4、6、4

5、5、5、5、5

6、6、6、6、6

7、9、3、 1、7

8、4、2、6、8

9、 1、9、 1、9

That is to say, any integer power can be the single digit period of a number of 4 (of course, 4 is not necessarily the minimum positive period).

1997 ÷ 4 = 499... 1 and 1949 ÷ 4 = 487... 1

Therefore, the single digits of n1997 and n1949 are the same as those of n digits.

So the bits of N 1997-N 1949 are all zeros, which can be divisible by 10. Method 2: Fermat's Little Theorem

Note that 10=2×5, and 2 and 5 are prime numbers.

So,

① For any number that is coprime with 5,

n^4≡ 1(mod? 5)

Therefore,

n^ 1997≡(n^499)^4×n≡ 1×n≡n(mod? 5)

n^ 1949≡(n^487)^4×n≡ 1×n≡n(mod? 5)

Therefore, n1997-n1949 ≡ n-n ≡ 0 (mod? 5)

(2) For any number that is not coprime with 5,

Both:

n^ 1997-n^ 1949=5p-5q=5×(p-q)≡0(mod? 5)

To sum up, that is to say, whether n and 5 are coprime or not, n 1997-n 1949 is a multiple of 5.

On the other hand, N 1997 and N 1949 are odd and even, so N 1997-N 1949 must be even, that is, a multiple of 2.

Therefore, N 1997-N 1949 must be a common multiple of 2 and 5, so it is a multiple of 10.

Prove completion. Method 3: Residual classes

n^ 1997-n^ 1949

=n^ 1949×(n^48- 1)

=n^ 1949×(n^24+ 1)×(n^24- 1)

=n^ 1949×(n^24+ 1)×(n^ 12+ 1)×(n^ 12- 1)

=n^ 1949×(n^24+ 1)×(n^ 12+ 1)×(n^6+ 1)×(n^6- 1)

=n^ 1949×(n^24+ 1)×(n^ 12+ 1)×(n^6+ 1)×(n^3+ 1)×(n^3- 1)

(1) Note that all natural numbers can be divided into five categories according to the remainder of ÷5.

5-{0}、5-{ 1}、5-{2}、5-{3}、5-{4}

(In fact, what is the remainder of ÷5 is assigned to the corresponding tag category. )

When n∈5-{0}, n 1949 ∈ 5-{0}, so the product is also 5-{0}, which is a multiple of 5;

When n∈5-{ 1}, (n 3-1) ∈ 5-{ 0}, the product is also 5-{0}, which is a multiple of 5;

When n∈5-{2}, (n 6+1) ∈ 5-{ 0}, so the product is also 5-{0}, which is a multiple of 5;

When n∈5-{3}, (n 6+1) ∈ 5-{ 0}, so the product is also 5-{0}, which is a multiple of 5;

When n∈5-{4}, (n 3+1) ∈ 5-{ 0}, so the product is also 5-{0}, which is a multiple of 5;

That is to say, regardless of n∈5-{? }, the product is a multiple of 5.

(2) Note that all natural numbers can be divided into two categories according to their ? 2 remainder.

2-{0}、2-{ 1}

(In fact, no matter what the remainder is, it is assigned to the corresponding tag category, that is, parity).

When n∈2-{0}, n 1949 ∈ 2-{0}, so the product is also 2-{0}, which is a multiple of 2;

When n∈2-{ 1}, (n 3-1) ∈ 2-{ 0}, the product is also 2-{0}, which is a multiple of 2;

That is to say, regardless of n∈2-{? }, the product is a multiple of 2.

To sum up, we take n, and the product is a common multiple of 2 and 5, so it is a multiple of 10. The economic mathematics team will answer for you!