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Find the 434th power of 57 * the 617th power of 725 - the 24th power of 143. What's the number?
The power law of 7 is 7, 9, 3, 1, four numbers and one cycle
434/4=108+2/4
So the 434th power of 57 is 9
A place to the power of 5 must be 5
So the 434th power of 57 * the 617th power of 725 is 4
The power law of 3 is 3,9,7,1, which is also a period of four numbers
24/4=6
So the single digit of 143 to the 24th power is 1
So the number of digits is 4-1 = 3
What is the one digit number of 6 to the power of 2013
6¹=6
6²=36
6³=216
……
It can be seen that any power of 6 is equal to 6
So the number to the power of 6 equals 6
We know that the range of function f (x) = - x ^ 2 + ax + B (a, B belong to R) is (negative infinity, 0]. If the solution set of inequality f (x) > C-1 about X is (M-4, M + 1), we can find the value of C
The square of a + 4B = 0, the square of F (x) = - (x-a / 2), f (x) is greater than C-1, C-1 must be negative, so 2 radical (1-C) = (M + 1) - (M-4) = 5, C = - 21 / 4
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