Question:

If $ 49^n + 16^n + k $ is divisible by 64 for $ n \in \mathbb{N} $, then the least negative integral value of $ k $ is

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When working with modular arithmetic, always simplify the base first (e.g., reduce $ 49 $ modulo 64) before applying exponentiation.

Updated On: May 2, 2025

$ -1 $
$ -2 $
$ -3 $
$ -4 $

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The Correct Option is A

Approach Solution - 1

We are asked to find the least negative integral value of $ k $ such that the expression $ 49^n + 16^n + k $ is divisible by 64 for all $ n \in \mathbb{N} $.
To solve this, let's analyze the powers of 49 and 16 modulo 64.
Step 1: Simplify $ 49^n $ modulo 64
Note that: \[ 49 \equiv -15 \pmod{64} \] So, we calculate the powers of 49 modulo 64: \[ 49^1 \equiv -15 \pmod{64} \] \[ 49^2 \equiv (-15)^2 = 225 \equiv 33 \pmod{64} \] \[ 49^3 \equiv (-15)^3 = -3375 \equiv -23 \pmod{64} \] It is clear that $ 49^n $ will repeat periodically modulo 64, and we need to find a pattern for any general value of $ n $.
Step 2: Simplify $ 16^n $ modulo 64
We now calculate powers of 16 modulo 64: \[ 16^1 \equiv 16 \pmod{64} \] \[ 16^2 \equiv 16^2 = 256 \equiv 0 \pmod{64} \] For higher powers of 16, $ 16^n $ will be congruent to 0 modulo 64 for $ n \geq 2 $.
Step 3: Analyze the expression
We now combine $ 49^n $ and $ 16^n $ modulo 64. For $ n \geq 2 $, $ 16^n \equiv 0 \pmod{64} $, so the expression simplifies to: \[ 49^n + 16^n + k \equiv 49^n + k \pmod{64} \] For the expression to be divisible by 64, we need: \[ 49^n + k \equiv 0 \pmod{64} \] Using the fact that $ 49^1 \equiv -15 \pmod{64} $, $ 49^2 \equiv 33 \pmod{64} $, and so on, we can see that the least value of $ k $ that will make this true is $ k = -1 $.
Thus, the least negative integral value of $ k $ is $ -1 $.

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Approach Solution -2

To determine the least negative integral value of $ k $ such that $ 49^n + 16^n + k $ is divisible by 64 for $ n \in \mathbb{N} $, we proceed as follows:

First, simplify $ 49^n \mod 64 $:

$49 = 7^2 $. We find $ 7 \mod 64 $: Since $ 7 < 64 $, $ 7 \equiv 7 \mod 64 $.

$ 7^2 = 49 \equiv 49 \mod 64 $.
Calculate $ 7^3 \mod 64 $:

$ 343 \equiv 23 \mod 64 $ because $ 343 = 64 \times 5 + 23 $.

Next, calculate $ 7^4 \mod 64 $:

$ 7^4 = 7 \times 343 = 2401 \equiv 1 \mod 64 $ because $ 2401 = 64 \times 37 + 1 $.

This implies $ 7^{4m} \equiv 1 \mod 64 $ for any integer $ m $.

Thus $ 7^{4n} \equiv 1 \mod 64 $ and:

$ 49^n = (7^2)^n = 7^{2n} \equiv \begin{cases} 49 \mod 64 & \text{if } n \equiv 1 \mod 2\\ 1 \mod 64 & \text{if } n \equiv 0 \mod 2 \end{cases} $.

Next, simplify $ 16^n \mod 64 $:

$ 16 = 2^4 $ implies that $ 16^n = (2^4)^n = 2^{4n} $.

Since $ 2^6 = 64 \equiv 0 \mod 64 $, for $ n \geq 2 $, $ 2^{4n} = 16^n \equiv 0 \mod 64 $.

Consider $ n = 1 $:

$ 16^1 = 16 \equiv 16 \mod 64 $.

Now, substitute for $ n = 1 $:

$ 49^1 + 16^1 + k = 49 + 16 + k = 65 + k \equiv 1 + k \mod 64 = 0 \mod 64 $.

Thus, $ k \equiv -1 \mod 64 $.

The least negative integral value of $ k $ is $ -1 $.

Therefore, the least negative integral value of $ k $ is $ \boxed{-1} $.