Question:
A divisor of
\[
3^{6n}+56n-1,\qquad n\in\mathbb{N}
\]
is
A divisor of
\[
3^{6n}+56n-1,\qquad n\in\mathbb{N}
\]
is
Show Hint
For expressions of the form
\[
a^{2m}-1,
\]
always use
\[
a^{2m}-1=(a^m-1)(a^m+1).
\]
Then check divisibility of consecutive even factors or use modular arithmetic to identify constant divisors.
Updated On: Jul 9, 2026
- \(729\)
- \(625\)
- \(676\)
- \(784\) \bigskip
Show Solution
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The Correct Option is D
Solution and Explanation
Concept:
To find a constant divisor of an expression involving \(n\), try to rewrite the expression so that a standard factorization can be applied.
Useful identity:
\[
a^2-b^2=(a-b)(a+b).
\]
Step 1: Rewrite the given expression. Observe that \[ 3^{6n} = (3^{3n})^2. \] Hence, \[ 3^{6n}+56n-1 = 3^{6n}-1+56n. \] Now, \[ 3^{6n}-1 = (3^{3n}-1)(3^{3n}+1). \] Therefore, \[ 3^{6n}+56n-1 = (3^{3n}-1)(3^{3n}+1)+56n. \]
Step 2: Show that \(28\) divides \(3^{3n}-1\). Since \[ 3^3-1=27-1=26, \] consider modulo \(28\): \[ 3^3=27\equiv -1 \pmod{28}. \] Thus, \[ 3^{6}=729\equiv 1 \pmod{28}. \] Hence, \[ 3^{6n}\equiv 1 \pmod{28}. \] Therefore, \[ 3^{3n}-1 \] or \[ 3^{3n}+1 \] is always divisible by \(28\). Thus, \[ (3^{3n}-1)(3^{3n}+1) \] is divisible by \(28^2\). \[ 28^2=784. \]
Step 3: Check the remaining term. Also, \[ 56n=2\cdot 28\,n. \] Since the first term contributes a factor \(28^2\), the whole expression remains divisible by \[ 28^2=784. \] Therefore, \[ 784 \mid \left(3^{6n}+56n-1\right). \]
Step 4: Verify for a small value of \(n\). For \(n=1\), \[ 3^6+56-1 = 729+55 = 784, \] which is clearly divisible by \[ 784. \]
Step 5: Write the final answer. \[ \boxed{784} \]
Step 1: Rewrite the given expression. Observe that \[ 3^{6n} = (3^{3n})^2. \] Hence, \[ 3^{6n}+56n-1 = 3^{6n}-1+56n. \] Now, \[ 3^{6n}-1 = (3^{3n}-1)(3^{3n}+1). \] Therefore, \[ 3^{6n}+56n-1 = (3^{3n}-1)(3^{3n}+1)+56n. \]
Step 2: Show that \(28\) divides \(3^{3n}-1\). Since \[ 3^3-1=27-1=26, \] consider modulo \(28\): \[ 3^3=27\equiv -1 \pmod{28}. \] Thus, \[ 3^{6}=729\equiv 1 \pmod{28}. \] Hence, \[ 3^{6n}\equiv 1 \pmod{28}. \] Therefore, \[ 3^{3n}-1 \] or \[ 3^{3n}+1 \] is always divisible by \(28\). Thus, \[ (3^{3n}-1)(3^{3n}+1) \] is divisible by \(28^2\). \[ 28^2=784. \]
Step 3: Check the remaining term. Also, \[ 56n=2\cdot 28\,n. \] Since the first term contributes a factor \(28^2\), the whole expression remains divisible by \[ 28^2=784. \] Therefore, \[ 784 \mid \left(3^{6n}+56n-1\right). \]
Step 4: Verify for a small value of \(n\). For \(n=1\), \[ 3^6+56-1 = 729+55 = 784, \] which is clearly divisible by \[ 784. \]
Step 5: Write the final answer. \[ \boxed{784} \]
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