Modern Math

Set Theory

Medium

If $A=\left\{6^{2 n}-35 n-1: n=1,2,3, \ldots\right\}$ and $B=\{35(n-1): n=1,2,3, \ldots\}$ then which of the following is true?

Every member of $B$ is in $A$

Every member of $A$ is in $B$ and at least one member of $B$ is not in $A$

Neither every member of $A$ is in $B$ nor every member of $B$ is in $A$

At least one member of $A$ is not in $B$

✅ Correct Option: 2

We need to determine the relationship between sets A and B.

Let's see what set B contains:

When n = 1: B contains 35(1-1) = 0

When n = 2: B contains 35(2-1) = 35

When n = 3: B contains 35(3-1) = 70

When n = 4: B contains 35(4-1) = 105

Set B = {0, 35, 70, 105, 140, ...} contains all multiples of 35 (including 0).

Now let's calculate the first few terms of A where A = { $6^{2n} - 35n - 1$ }:

When n = 1: $6^{2×1} - 35×1 - 1 = 36 - 35 - 1 = 0$

When n = 2: $6^{2×2} - 35×2 - 1 = 1296 - 70 - 1 = 1225$

When n = 3: $6^{2×3} - 35×3 - 1 = 46656 - 105 - 1 = 46550$

So A = {0, 1225, 46550, ...}

Let's check if all elements of A are multiples of 35:

0 ÷ 35 = 0

1225 ÷ 35 = 35

46550 ÷ 35 = 1330

To prove this is always true, we'll use modular arithmetic.

Since 35 = 5 × 7, we need to show that $6^{2n} - 35n - 1$ is divisible by both 5 and 7.

Divisibility by 5:

$6 ≡ 1 \pmod{5}$ [since 6 = 5 + 1]

So $6^{2n} ≡ 1^{2n} ≡ 1 \pmod{5}$

And $35n ≡ 0 \pmod{5}$ [since 35 is divisible by 5]

Therefore: $6^{2n} - 35n - 1 ≡ 1 - 0 - 1 ≡ 0 \pmod{5}$

Divisibility by 7:

$6 ≡ -1 \pmod{7}$ [since 6 = 7 - 1]

So $6^{2n} ≡ (-1)^{2n} ≡ 1 \pmod{7}$ [since 2n is always even]

And $35n ≡ 0 \pmod{7}$ [since 35 is divisible by 7]

Therefore: $6^{2n} - 35n - 1 ≡ 1 - 0 - 1 ≡ 0 \pmod{7}$

Since every element of A is divisible by both 5 and 7, every element of A is divisible by 35.

Every element of A is a multiple of 35, so every element of A belongs to set B.

However, not every element of B belongs to A. For example, 35 is in B, but there's no integer n such that $6^{2n} - 35n - 1 = 35$ .

A ⊂ B (A is a proper subset of B)

This means every member of A is in B, but at least one member of B is not in A.

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If $A=\left\{6^{2 n}-35 n-1: n=1,2,3, \ldots\right\}$ and $B=\{35(n-1): n=1,2,3, \ldots\}$ then which of the following is true?