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!