+8 Let \(M=\frac{{9}^{40,000}}{{9}^{200}-2}\). If M is rounded to the nearest integer and then divided by 100, what is the remainder? Any thoughts would be very appreciated! :)
+666 We can rewrite the expression to make simplification easier:
M = 9^40000 / (9^200 - 2) = (9^200)^200 / (9^200 - 2)
Since 9^200 is a very large number, when we divide by (9^200 - 2), the remainder will be very close to the remainder of dividing 1 by (9^200 - 2).
Now, let's analyze (9^200 - 2):
The units digit of 9^n repeats in a cycle of 1, 9, 8, 7, 6, 5, 4, 3, 2 (powers of 9 ending in 1, 9, 8, etc.). Since 200 is a multiple of 4, the units digit of 9^200 will be 1.
Subtracting 2 from 1 results in a units digit of -1, which is equivalent to 9 (since remainders are non-negative).
Therefore, dividing 1 by (9^200 - 2) will have a remainder of 1.
When we take M, a very large number divided by (9^200 - 2) with a remainder of 1, rounding it to the nearest integer won't change the remainder when divided by 100.
So, the remainder of M divided by 100 will also be 9.
