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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! :)

 Jun 26, 2024
 #3
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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​.

 Jun 28, 2024

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