I asked 4o about these congruences in separate conversations, but decided to make them just one post. They are: that (39x) mod 5 = 13, which is true; that (95x) mod 5 = 13, which is false; that (x²) mod 6 = 3, which is true; that (5x²) mod 8 = 12, which is true; that (4x³ + 2x) mod 5 = 7, which is false; and that (5^x + 3) mod 100 = 5, which is false.
39x ≡ 13 (mod 5)
Is this true: ∃x, 39x ≡ 13 (mod 5)
We are given the congruence:
$$ \exists x,\ 39x \equiv 13 \pmod{5} $$
Step 1: Simplify the coefficients modulo 5
$$ 39 \equiv 4 \pmod{5} \quad \text{and} \quad 13 \equiv 3 \pmod{5} $$
So the congruence becomes:
$$ 4x \equiv 3 \pmod{5} $$
Step 2: Solve the congruence
We are looking for an integer $x$ such that:
$$ 4x \equiv 3 \pmod{5} $$
Try small values of $x$ modulo 5:
- $x = 1 \Rightarrow 4(1) = 4 \not\equiv 3$
- $x = 2 \Rightarrow 4(2) = 8 \equiv 3 \pmod{5}$
✅ Success! $x = 2$ is a solution.
✅ Conclusion:
True. There exists an integer $x$ such that $39x \equiv 13 \pmod{5}$.
95x ≡ 13 (mod 5)
Is this true: ∃x, 95x ≡ 13 (mod 5)
We are asked whether the following statement is true:
∃x, 95x ≡ 13 (mod 5)
Step 1: Reduce mod 5
We simplify both coefficients modulo 5:
- 95 mod 5 = 0 (since 95 is divisible by 5)
- So: 95x ≡ 0 mod 5
But the right-hand side is:
- 13 mod 5 = 3
So the equation becomes:
0 ≡ 3 mod 5
This is clearly false.
Step 2: Conclusion
There is no value of $x$ that satisfies $95x \equiv 13 \pmod{5}$, because the left-hand side is always 0 mod 5.
✅ Final Answer:
False.
x^2 ≡ 3 (mod 6)
Is this true: ∃x, x^2 ≡ 3 (mod 6)
We are asked whether the following statement is true:
∃x, x² ≡ 3 (mod 6) That is, does there exist an integer x such that the square of x is congruent to 3 modulo 6?
Step 1: Understand modulo 6
Modulo 6 means we only need to consider the residues modulo 6: 0, 1, 2, 3, 4, 5
We will compute $x^2 \mod 6$ for each of these values of $x \mod 6$:
| x mod 6 | x² mod 6 |
|---|---|
| 0 | 0 |
| 1 | 1 |
| 2 | 4 |
| 3 | 3 |
| 4 | 4 |
| 5 | 1 |
So from this table, we see that:
- $3^2 = 9 \equiv 3 \mod 6$
So yes, there does exist an integer $x$ such that $x^2 \equiv 3 \mod 6$.
✅ Final Answer:
Yes, the statement is true. There exists an $x \in \mathbb{Z}$ (e.g., $x = 3$) such that $x^2 \equiv 3 \mod 6$.
5x² ≡ 12 (mod 8)
Is this true: ∃x, 5x² ≡ 12 (mod 8)
We are asked whether the following congruence has a solution:
$$ \exists x, \quad 5x^2 \equiv 12 \pmod{8} $$
Step 1: Work modulo 8
We want to find an integer $x$ such that:
$$ 5x^2 \equiv 12 \pmod{8} $$
First, reduce the RHS and coefficients modulo 8:
- $12 \equiv 4 \pmod{8}$
So the congruence becomes:
$$ 5x^2 \equiv 4 \pmod{8} $$
Step 2: Try all values of $x \mod 8$
Since we're working modulo 8, try all residues $x = 0, 1, 2, ..., 7$:
Compute $5x^2 \mod 8$:
| $x$ | $x^2$ | $5x^2 \mod 8$ |
|---|---|---|
| 0 | 0 | 0 |
| 1 | 1 | 5 |
| 2 | 4 | 20 ≡ 4 |
| 3 | 9 ≡ 1 | 5 |
| 4 | 16 ≡ 0 | 0 |
| 5 | 25 ≡ 1 | 5 |
| 6 | 36 ≡ 4 | 20 ≡ 4 |
| 7 | 49 ≡ 1 | 5 |
We see that:
- $x = 2$ and $x = 6$ yield $5x^2 \equiv 4 \mod 8$
✅ Conclusion:
Yes, the statement is true. There exists $x$ such that $5x^2 \equiv 12 \mod 8$. For example, $x = 2$ or $x = 6$ are solutions.
4x³ + 2x ≡ 7 (mod 5)
Is this true: ∃x, 4x³ + 2x ≡ 7 (mod 5)
We are asked whether the congruence
$$ \exists x, \quad 4x^3 + 2x \equiv 7 \pmod{5} $$
has a solution modulo 5.
Step 1: Work modulo 5
Modulo 5, we simplify the congruence:
$$ 4x^3 + 2x \equiv 7 \pmod{5} $$
But $7 \equiv 2 \pmod{5}$, so we are solving:
$$ 4x^3 + 2x \equiv 2 \pmod{5} $$
Step 2: Try all values of $x \mod 5$
Since there are only 5 residue classes modulo 5, we can just try each:
- $x = 0$: $4(0)^3 + 2(0) = 0 \mod 5$
- $x = 1$: $4(1)^3 + 2(1) = 4 + 2 = 6 \equiv 1 \mod 5$
- $x = 2$: $4(8) + 2(2) = 32 + 4 = 36 \equiv 1 \mod 5$
- $x = 3$: $4(27) + 2(3) = 108 + 6 = 114 \equiv 4 \mod 5$
- $x = 4$: $4(64) + 2(4) = 256 + 8 = 264 \equiv 4 \mod 5$
Modulo 5 equivalents:
- $x = 0$: 0
- $x = 1$: 1
- $x = 2$: 1
- $x = 3$: 4
- $x = 4$: 4
None of these are congruent to 2 mod 5.
✅ Final Answer:
No, the statement is false:
$$ \not\exists x, \quad 4x^3 + 2x \equiv 7 \pmod{5} $$
There is no integer $x \mod 5$ that satisfies the congruence.
5^x +3 ≡ 5 (mod 100)
Is this true: ∃x∈ℕ, 5^x +3 ≡ 5 (mod 100)
We are asked whether the following congruence has a solution:
$$ \exists x \in \mathbb{N}, \quad 5^x + 3 \equiv 5 \pmod{100} $$
Step 1: Simplify the congruence
Subtract 3 from both sides:
$$ 5^x \equiv 2 \pmod{100} $$
So the question becomes: Is there a natural number $x$ such that $5^x \equiv 2 \mod 100$?
Step 2: Compute powers of 5 modulo 100
Let's compute a few values of $5^x \mod 100$:
- $5^1 = 5 \mod 100 = 5$
- $5^2 = 25$
- $5^3 = 125 \mod 100 = 25$
- $5^4 = 625 \mod 100 = 25$
- In fact, for $x \geq 2$, $5^x \equiv 25 \mod 100$
That’s because:
- $5^2 = 25$
- $5^3 = 5 \cdot 25 = 125 \equiv 25 \mod 100$
- $5^4 = 5 \cdot 125 = 625 \equiv 25 \mod 100$
So by induction:
- For all $x \geq 2$, $5^x \equiv 25 \mod 100$
Step 3: Check if any value of $x$ gives $5^x \equiv 2 \mod 100$
From above:
- $5^1 = 5$
- $5^2 = 25$
- For $x \geq 2$, result is stuck at 25
So the powers of 5 modulo 100 cycle among 5 and 25, never hitting 2.
✅ Final Answer:
No, the statement is false. There is no $x \in \mathbb{N}$ such that:
$$ 5^x + 3 \equiv 5 \pmod{100} $$
equivalently,
$$ 5^x \equiv 2 \pmod{100} $$
This congruence has no solution.
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