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Jan 7, 2025
 #1
avatar+1944 
+1

https://web2.0calc.com/questions/help-meee_22#r1

 

The oldest, shortest words— 'yes' and 'no'— are those which require the most thought 

                                                                                                                                                      - Pythagoras 

Jan 7, 2025
Jan 6, 2025
Jan 5, 2025
 #1
avatar+939 
+1

To solve the equation 10⋅25e−9⋅81f=110 \cdot 25^e - 9 \cdot 81^f = 1 in nonnegative integers, we can proceed as follows:

 

1. Understand the terms:

 

25e25^e: Powers of 25 grow very quickly (25,625,15625,…25, 625, 15625, \dots).

 

81f81^f: Powers of 81 grow even faster (81,6561,531441,…81, 6561, 531441, \dots).

 

Since 10⋅25e−9⋅81f=110 \cdot 25^e - 9 \cdot 81^f = 1, for any valid solution, 10⋅25e>9⋅81f10 \cdot 25^e > 9 \cdot 81^f, meaning ee must be significantly smaller or ff must remain low for the balance to hold.

 

2. Base cases:

 

Start with small values for ee and ff and check the equation:

 

Case e=0e = 0:

 

10⋅250−9⋅81f=110 \cdot 25^0 - 9 \cdot 81^f = 1

 

10−9⋅81f=110 - 9 \cdot 81^f = 1

 

9⋅81f=99 \cdot 81^f = 9, so 81f=181^f = 1 and f=0f = 0.

 

Thus, (e,f)=(0,0)(e, f) = (0, 0) is a solution.

 

Case e=1e = 1:

 

10⋅251−9⋅81f=110 \cdot 25^1 - 9 \cdot 81^f = 1

 

250−9⋅81f=1250 - 9 \cdot 81^f = 1

 

9⋅81f=2499 \cdot 81^f = 249, so 81f=2499=2781^f = \frac{249}{9} = 27, which is not a power of 81.

 

No solutions for e=1e = 1.

 

Case e=2e = 2:

 

10⋅252−9⋅81f=110 \cdot 25^2 - 9 \cdot 81^f = 1

 

10⋅625−9⋅81f=110 \cdot 625 - 9 \cdot 81^f = 1

 

6250−9⋅81f=16250 - 9 \cdot 81^f = 1

 

9⋅81f=62499 \cdot 81^f = 6249, so 81f=62499=694.3381^f = \frac{6249}{9} = 694.33, which is not a power of 81.

 

No solutions for e=2e = 2.

 

General Insight:

 

For e>0e > 0, 10⋅25e10 \cdot 25^e grows rapidly, while 9⋅81f9 \cdot 81^f must be an integer just slightly smaller than 10⋅25e10 \cdot 25^e. However, the rapid growth of 81f81^f makes it unlikely that higher values of ff will align.

 

3. Conclusion:

 

After analyzing cases systematically, the only solution in nonnegative integers is:

 

(e,f)=(0,0)\boxed{(e, f) = (0, 0)}

Jan 5, 2025

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