Prove that the 10th power of 99 can be divided by 1000. Who will

Prove that the 10th power of 99 can be divided by 1000. Who will

The last digit of the positive integer power of 99 is 9 or 1, which cannot be 0,
So the tenth power of 99 cannot be divided by 1000

How many numbers from 10 to 99 can be divided by 3 or 7 Please talk about the mathematical knowledge points of solving this kind of problems. I often encounter such a headache. I either ask what the number you can be divided by, or ask how much is left. I feel dizzy‘ Well, don't be wordy. Thank you very much The answer on the second floor is a positive solution. I have some questions, but the answer 39 can't divide 7. Isn't it the meaning of division? What is your algorithm? Hehe, I don't understand it,

three × 3＝9 < 10 < 3 × 4＝12,3 × There are 33-3 = 30 7 numbers that can be divided by 3 in 33 = 99 ≠ 10 to 99 × 1＝7 < 10 < 7 × 2＝14,7 × 14＝98 < 99 < 7 × There are 14-1 = 13 21 numbers that can be divided by 7 in 15 = 105 ≠ 10 to 99 × 0＝0 < 10 < 21 × 1＝21,21 × 4＝84 < 99 < 2...

Given that n is a positive integer and the 2nd power of (the nth power of x) = 9, find the value of the 2nd power of (the 3N power of 1 / 3x) - the 2n power of 3 (the 2nd power of x) [urgent! I'm a math idiot... Everyone... Please, OTZ

(n-th power of x) ²= 9, i.e. 2n power of x = 9 (3N power of 1 / 3 x) ²- 3(X ²) 2n power of = (1 / 3) ² (2n power of x) ³- 3 (2n power of x) ²= (1/9)x9 ³- 3x9 ²= nine ²- 3x9 ²=- 2x9 ²=- one hundred and sixty-two

Given that n is a positive integer and the 2n power of x = 4, find the 2n power of 9 (3N power of x) 2-13 (x2) I'll finish it in a minute

x^2n=4
9(x^3n)^2-13(x^2)^2n
=9*(x^2n)^3-13*(x^2n)^2
=9*4^3-13*4^2
=9*64-13*16
=368

If n is a positive integer and X2N = 2, try to find (- 3x3n) 2-4 (- x2) 2n

Original formula = 9x6n-4x4n = 9 (X2N) 3-4 (X2N) 2,
When X2N = 2, the original formula = 9 × 23-16=56．

Binomial theorem proof and division problem Verify that 2 ^ (6n-3) + 3 ^ (2n-1) can be divided by 11~

Binomial theorem proof 2 ^ (6n-3) + 3 ^ (2n-1) = (11-3) ^ (2n-1) + 3 ^ (2n-1) = 11 ^ (2n-1) + (2n-1) 11 ^ (2n-2) (- 3) ^ 2 +... + C (2n-1,2n-2) * 11 * (- 3) ^ (2n-2) + (- 3) ^ (2n-1) + 3 ^ (2n-1) = 11q (q is an integer), so 11 divides 2 ^ (6n-3) +