Math Problem Statement

Solución del problema

Solution

The problem in the image states:

"Al dividir tanto 13ab13^{ab} como 175a17^{5a} entre 11, se obtiene 5 de residuo; ¿Cuál es el residuo de dividir 135b13^{5b} entre a+ba + b?"

Solution:

To solve this, we need to understand and utilize the given information effectively.

  1. Given Information:

    • 13ab5(mod11)13^{ab} \equiv 5 \pmod{11}
    • 175a5(mod11)17^{5a} \equiv 5 \pmod{11}
  2. Objective:

    • Find the residue when 135b13^{5b} is divided by a+ba + b.

Step 1: Simplify using the modulo property

Since 13ab5(mod11)13^{ab} \equiv 5 \pmod{11}, we know that: 13ab13k5(mod11)13^{ab} \equiv 13^{k} \equiv 5 \pmod{11} Where k=abk = ab.

Step 2: Check the modular equivalences

We need to check the equivalence 13kmod1113^{k} \mod 11. We can use Fermat's Little Theorem to simplify the calculations, where 132(mod11)13 \equiv 2 \pmod{11}. Thus, the problem becomes: 2ab5(mod11)2^{ab} \equiv 5 \pmod{11}

Let's calculate the powers of 22 modulo 1111: 21=22^1 = 2 22=42^2 = 4 23=82^3 = 8 24=165(mod11)2^4 = 16 \equiv 5 \pmod{11} 25=3210(mod11)2^5 = 32 \equiv 10 \pmod{11} 26=649(mod11)2^6 = 64 \equiv 9 \pmod{11} 27=1287(mod11)2^7 = 128 \equiv 7 \pmod{11} 28=2563(mod11)2^8 = 256 \equiv 3 \pmod{11} 29=5126(mod11)2^9 = 512 \equiv 6 \pmod{11} 210=10241(mod11)2^{10} = 1024 \equiv 1 \pmod{11}

So, 2ab5(mod11)2^{ab} \equiv 5 \pmod{11} implies abab must be congruent to 4(mod10)4 \pmod{10}.

Step 3: Apply similar logic to the second congruence

Given that 175a5(mod11)17^{5a} \equiv 5 \pmod{11}, note that 176(mod11)17 \equiv 6 \pmod{11}. Therefore: 65a5(mod11)6^{5a} \equiv 5 \pmod{11}

Similarly, calculating powers of 66 modulo 1111: 61=66^1 = 6 62=363(mod11)6^2 = 36 \equiv 3 \pmod{11} 63=187(mod11)6^3 = 18 \equiv 7 \pmod{11} 64=429(mod11)6^4 = 42 \equiv 9 \pmod{11} 65=5410(mod11)6^5 = 54 \equiv 10 \pmod{11} 66=605(mod11)6^6 = 60 \equiv 5 \pmod{11}

This implies 5a5a must be congruent to 5(mod10)5 \pmod{10}. Therefore, aa is congruent to 1(mod2)1 \pmod{2}.

Step 4: Determine the residue of 135bmod(a+b)13^{5b} \mod (a + b)

Given aa and bb values:

  • Since ab4(mod10)ab \equiv 4 \pmod{10} and a1(mod2)a \equiv 1 \pmod{2}, bb must satisfy these equations.
  • 135b25b(mod11)13^{5b} \equiv 2^{5b} \pmod{11}.

Use similar modular calculations to determine the residue when divided by a+ba + b.

Final Step: Calculation and Conclusion

Without the exact value of a+ba + b, further simplification isn't possible without assuming specific aa and bb values satisfying the modular conditions.

Would you like to proceed with assuming specific values for aa and bb, or do you have further details to specify these values?

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Math Problem Analysis

Mathematical Concepts

Modular Arithmetic
Fermat's Little Theorem

Formulas

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Theorems

Fermat's Little Theorem

Suitable Grade Level

Advanced High School