Try to prove that the 23rd power of 5 minus the 21st power of 5 can be divisible by 600

Try to prove that the 23rd power of 5 minus the 21st power of 5 can be divisible by 600

5^23-5^21
=5^21(5^2-1)
=5^21×24
=5^20×120
=5^19×600
In the natural number of 20, there are () odd numbers, () even numbers, () prime numbers, and () composite numbers, and () in the odd numbers is composite numbers,
() in even numbers is prime
There are (10) odd numbers, (10) even numbers, (8) prime numbers and (11) composite numbers in the natural numbers of 20. Odd numbers: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19 even numbers: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 prime numbers
There are (10) odd numbers, (10) even numbers, (7) prime numbers and (12) composite numbers in the natural numbers from 1 to 20, and (9 and 15) in the odd numbers are composite numbers
There are (10) odd numbers, (10) even numbers, (7) prime numbers and (12) composite numbers in the natural numbers from 1 to 20. The odd numbers (9 and 15) are composite numbers and the even numbers (2) are prime numbers
Hello, among the natural numbers from 1 to 20, there are (10) odd numbers, there are (10) even numbers, there are (8) prime numbers, and there are (11) composite numbers. In odd numbers, (9 and 15) are composite numbers, and in even numbers, (2) are prime numbers.
Prime: 2.3.5.7.11.13.17.19
Total: 4.6.8.9.10.12.14.15.16.18.20
There are 10 odd numbers, 10 even numbers
There are eight prime numbers. In a natural number greater than 1, except 1 and the integer itself, it cannot be divisible by other natural numbers.) ：2,3,5,7,11,13,17,19
There are 11 composite numbers (an integer whose divisor can be divided by other factors besides 1 and itself. Such numbers are called composite numbers.) ：4.6.8.9.10.12.14.15.16.18.20
The odd number is 15
In even numbers, 2 is prime
There are 10 odd numbers, 10 even numbers
There are eight prime numbers. In a natural number greater than 1, except 1 and the integer itself, it cannot be divisible by other natural numbers.) ：2,3,5,7,11,13,17,19
There are 11 composite numbers (an integer whose divisor can be divided by other factors besides 1 and itself. Such numbers are called composite numbers.) ：4.6.8.9.10.12.14.15.16.18.20
In odd numbers, 9 and 15 are combined
Even 2 is prime
It is known that the power of N + 11 of 3 is divisible by 10. It is proved that the power of N + 4 of 3 and the power of M + 2 of 11 can also be divisible by 10
help~
3^(n+4)+11^(m+2)=81*3^n+121*11^m=81*3^n+81*11^m+40*11^m
=81(3^n+11^m)+10(4*11^m)
The power of N can be divided by the first 10 + 11
So the N + 4 power of 3 and the M + 2 power of 11 can also be divisible by 10
prove:
Because, the m-th power of 11 must be 1, and the n-th power of 3 + the m-th power of 11 can be divisible by 10, then the n-th power of 3 must be 9
Then the number of N + 4 power of 3 is 9 times the number of 4 power of 3, and the number of result is 9
The single digit of M + 2 power of 11 is still 1
If the two numbers are added, the number of bits is 0, which is obviously divisible by 10
Let n times of three plus m times of 11 be 10K,
Let the reduction of the proved formula be decomposed again,
The proved formula is 10K + 80 * 3N times + 120 * 11m times = 10p,
It is proved that P is a natural number
In the natural numbers within 20, the numbers that are both even and composite are () and the numbers that are both odd and prime are ()
Among the natural numbers within 20, the numbers that are both even and composite are (4,6,8,10,12,14,16,18,20),
The numbers that are both odd and prime are (3,5,7,11,13,17,19)
Among the natural numbers within 20, the numbers that are both even and combined are (4, 6, 8, 10, 12, 14, 16, 18),
The numbers that are both odd and prime are (3, 5, 7, 11, 13, 17, 19)
I wish you progress in your study, you can ask if you don't understand! thank you!!
Given that m and N are positive integers and M + 3N can be divisible by 11, can m + 3N + 5 be divisible by 11?
Because m + 3N + 5 - (M + 3n) = 3N (35-1) = 242 × 3N = 11 × 22 × 3N, M + 3N + 5 = m + 3N + 11 × 22 × 3N. Since m + 3N can be divisible by 11, M + 3N + 11 × 22 × 3N can also be divisible by 11, so m + 3N + 5 can be divisible by 11
In the range of natural numbers, the smallest prime number is (), the smallest composite number is (), the smallest odd number is (), and the smallest even number is ()（
In the range of natural numbers, the smallest prime number is (2), the smallest composite number is (4), the smallest odd number is (1), and the smallest even number is (0)
Two
Four
One
Two
In the range of natural numbers, the smallest prime number is (2), the smallest composite number is (4), the smallest odd number is (1), and the smallest even number is (0)
Two
Four
One
Zero
Two
Four
One
Zero
Zero
Can the 23rd power of 5 - the 21st power of 5 be divisible by 120?
5^23-5^21
=2^21(5^2-1)
=5^20(5*24)
=120*5^20
It's divisible by 120
Yes, I divided it to 25... In fact, it's 23 power minus 21 power, and the rest is the square of 5~
What is the minimum natural number that is both odd and prime? What is the minimum natural number that is both even and composite?
The minimum natural number that is odd and prime is 3
The minimum natural number that is both an even number and a composite number is 4
25 to the 7th power - 15 to the 12th power can be divisible by 120
25^7-5^12
=(5^2)^7-5^12
=5^14-5^12
=5^12(5^2-1)
=5^12*24
=5^11*5*24
=5^11*120
This number must be a multiple of 120
So it's divisible by 120, and the quotient is 5 ^ 11
Natural numbers include ()
A. Prime, composite B. factors and multiples C. odd and even numbers
According to the analysis, natural number includes odd number and even number (0 is even number)