Find the largest three-digit number that can be written in the form 3^m + 2^n, where m and n are positive integers.
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I agree with Ms Sue
To show us that you are not merely sponging off us,
for the question above, I would be interested in your thinking to solve the problem
So you want whole numbers m and n so that
3^m + 2^n ≤ 999
To find the largest three-digit number that can be written in the form 3^m + 2^n, you need to find the largest values of m and n that satisfy the condition.
First, let's find the largest value for m. Since m is a positive integer, we start with m = 9. Calculating 3^9, we get 19683. This is already larger than any three-digit number, so we can conclude that the largest possible value for m is 8.
Next, let's find the largest value for n. As n is also a positive integer, we start with n = 9. Calculating 2^9, we get 512. Since 512 is greater than any three-digit number, we need to try a smaller value for n. Let's recursively decrease n until we find a value that works.
When n = 8, 2^8 = 256. Adding 256 to 3^8 gives us 2592, which is a four-digit number. So, n cannot be 8.
When n = 7, 2^7 = 128. Adding 128 to 3^8 gives us 1707, which is still larger than any three-digit number.
When n = 6, 2^6 = 64. Adding 64 to 3^8 gives us 1152, which is a three-digit number. Therefore, the largest value for n is 6.
Now, we know that the largest three-digit number that can be written in the form 3^m + 2^n is obtained when m = 8 and n = 6.
Calculating 3^8 + 2^6, we get 656, which is indeed the largest three-digit number that satisfies the given condition.
So, the largest three-digit number that can be written in the form 3^m + 2^n is 656.