If n is a positive integer, define n! = n * (n-1) * (n-2) * 3 * 2 * 1, let m = 1! + 4! + +2006! + 2007!, then the sum of the last two digits of M is:
It's four
RELATED INFORMATIONS
- 1. Find the sum of the last two digits of M = 1! + 2! + 3! +. + 2003! + 2004
- 2. How many natural numbers can be taken from 1, 2, 3, 2004, 2005 at most, and the difference between each two numbers is not equal to 4
- 3. How many natural numbers can be taken from 1.2.3.2004, so that the difference between each two numbers is not equal to 51005
- 4. All natural numbers are integers, and all integers are natural numbers______ (judge right or wrong)
- 5. Please use n for three consecutive even numbers (n is a natural number), and their sum is. Use M for two consecutive odd numbers (M is a natural number), and their sum is The two questions are irrelevant If there are three consecutive even numbers, whether there must be two consecutive odd numbers satisfying the upper number filling method, if there must be, please explain the filling method: if not, please explore and give the conditions that the number in the middle of the three consecutive even numbers must exist
- 6. Is 0 a natural number? What is the concept of natural number?
- 7. Define an operation "*" for any non-zero natural number n, which satisfies the following operation properties: (1) 1 * 1 = 1; (2) (n + 1) * 1 = 3 (n * 1). Try to find the algebraic expression of n * 1 with respect to n
- 8. Define an operation "*": for natural number n, it satisfies the following operation properties: (I) 1 * 1 = 1, (II) (n + 1) * 1 = n * 1 + 1, then n * 1 equals () A. nB. n+1C. n-1D. n2
- 9. There is an operation program, when a ⊕ B = n (n is a constant), define (a + 1) ⊕ B = n + 1, a ⊕ (B + 1) = n-2, now known 1 ⊕ 1 = 2, then 2010 ⊕ 2010=______ .
- 10. When a ♁ B = n (n is a constant), we can get (a + 1) ♁ B = n + 1, a ♁ (B + 1) = n + 2, then (a + 2) ♁ (B + 2)= (can you talk about the process? Thank you. Thank you very much. The sooner the better.)
- 11. If n is a positive integer, define n! =n×(n-1)×(n-2)×… X 3 × 2 × 1, let m = 1! +2! +3! +4! +… +2003!+2004! Then the sum of the last two digits of M is______ .
- 12. Given that the polynomial x ^ 2 + ax-9 can be factorized in the range of integers, then the value of natural number a is 8, why should - 8 be abandoned?
- 13. If x ^ 2 + ax-9 = (x + m) (x + n) m, n is a positive integer, then the value of natural number a is? A.0, - 8.8b. Plus or minus 8 c.0, - 8 d.0,8 d.0 If x ^ 2 + ax-9 = (x + m) (x + n) n is an integer, then there is the value of natural number a? A. 0, - 8.8 B. plus or minus 8 C. - 8 d.0,8
- 14. Given that the natural numbers x and y satisfy the equation x & sup2; - Y & sup2; = 13, find the values of X and y On factorization X & sup2; - Y & sup2 is the square of x minus the square of Y
- 15. (X-5) (x + 6) = x & sup2; + MX + N, find the values of M and N respectively
- 16. It is known that (m-2) x ^ | M-1 | - (n + 3) y ^ n & sup2-8 = 1 is a univariate linear equation with respect to x, y, and m, n satisfies {Ma + Nb = 5,2ma NB = 7
- 17. Given that the equation (m-2) x | m | - 1 + (n + 30) y n & amp; sup2; - 8 = 6 is a quadratic equation of two variables, find the value of M.N. if x = 1 / 2, find the corresponding value of Y The known equation (m-2) x ^| m | - 1 + (n + 30) y ^ n & # 178; - 8 = 6
- 18. Given that m and N are positive integers and N & sup2 = M & sup2 + 168, find the value of M and n
- 19. Given that a and B are natural numbers, and the quadratic power of a - the quadratic power of B = 45, find the value of a and B
- 20. According to the given conditions, the values of natural numbers x and y are obtained: (1) x, y satisfy X & sup2; + xy = 35; (2) x, y satisfy X & sup2; - Y & sup2; = 45 That's what the exercise book says