R134 · Smallest number from three remainders

To find the smallest positive integer that satisfies the given conditions, we can use the Chinese Remainder Theorem or simply test numbers that fit the first two conditions and check if they satisfy the third.

Let's denote the number as \( n \).

1. \( n \equiv 1 \pmod{3} \)
2. \( n \equiv 2 \pmod{5} \)
3. \( n \equiv 3 \pmod{7} \)

First, we find numbers that satisfy \( n \equiv 1 \pmod{3} \) and \( n \equiv 2 \pmod{5} \).

- Numbers that leave a remainder of 1 when divided by 3 are: 1, 4, 7, 10, 13, 16, 19, 22, ...
- Among these, the numbers that also leave a remainder of 2 when divided by 5 are: 7, 22, ...

Now we check these candidates against \( n \equiv 3 \pmod{7} \):

- For \( n = 7 \):
  - \( 7 \mod 7 = 0 \) (does not satisfy \( n \equiv 3 \pmod{7} \))

- For \( n = 22 \):
  - \( 22 \mod 7 = 1 \) (does not satisfy \( n \equiv 3 \pmod{7} \))

We continue this process until we find a number that satisfies all three conditions. The next candidate in the sequence is \( n = 46 \):

- For \( n = 46 \):
  - \( 46 \mod 3 = 1 \) (satisfies \( n \equiv 1 \pmod{3} \))
  - \( 46 \mod 5 = 1 \rightarrow (46 + 2) 48 \mod 5 = 47 \) (satisfies \( n \equiv 2 \pmod{5} \))
  - \( 46 \mod 7 = 3 \) (satisfies \( n \equiv 3 \pmod{7} \))

Thus, the smallest number that satisfies all three conditions is \( n = 46.

This problem requires solving a system of linear congruences using the Chinese Remainder Theorem (CRT). Let $N$ be the positive integer. We are given:
1. $N \equiv 1 \pmod{3}$
2. $N \equiv 2 \pmod{5}$
3. $N \equiv 3 \pmod{7}$

**Step 1: Solve using the first two congruences.**
From (1), $N = 3k + 1$. Substituting this into (2):
$3k + 1 \equiv 2 \pmod{5}$
$3k \equiv 1 \pmod{5}$
Multiplying both sides by 2 (since $3 \times 2 = 6 \equiv 1 \pmod{5}$):
$6k \equiv 2 \pmod{5}$
$k \equiv 2 \pmod{5}$
So, $k = 5j + 2$. Substituting this back into the expression for $N$:
$N = 3(5j + 2) + 1 = 15j + 7$.
Thus, $N \equiv 7 \pmod{15}$.

**Step 2: Incorporate the third congruence.**
We use the result from Step 1 and substitute it into (3):
$15j + 7 \equiv 3 \pmod{7}$
Since $15 \equiv 1 \pmod{7}$ and $7 \equiv 0 \pmod{7}$, the congruence simplifies:
$1j + 0 \equiv 3 \pmod{7}$
$j \equiv 3 \pmod{7}$
So, $j = 7m + 3$.

**Step 3: Find the smallest N.**
Substitute this expression for $j$ back into the equation for $N$:
$N = 15(7m + 3) + 7$
$N = 105m + 45 + 7$
$N = 105m + 52$

To find the smallest positive integer, we set $m=0$.
$N = 52$.

**Verification:**
* $52 \div 3 = 17$ R $1$ (Correct)
* $52 \div 5 = 10$ R $2$ (Correct)
* $52 \div 7 = 7$ R $3$ (Correct)

The smallest such number is **52**.

## Step 1: Let's denote the smallest positive integer that satisfies all given conditions as N.
We need to find N.

## Step 2: Since N leaves a remainder of 1 when divided by 3, it can be expressed as 3k + 1 for some integer k.
This is because any number that leaves a remainder of 1 when divided by another number is one more than a multiple of that number.

## Step 3: Similarly, since N leaves a remainder of 2 when divided by 5, it can also be expressed as 5m + 2 for some integer m.
This follows the same logic as in step 2 but for division by 5.

## Step 4: And since N leaves a remainder of 3 when divided by 7, it can be expressed as 7n + 3 for some integer n.
Again, following the pattern established in previous steps for division by 7.

## Step 5: We need to find the smallest number that satisfies all three expressions (3k+1), (5m+2), and (7n+3).
Since we're looking for the smallest such N, we can start by listing multiples of each expression until they align or find a pattern.

## Step 6: Starting with the first form (3k + 1), some numbers that fit this pattern are 4, 7, 10, 13, 16, 19, 22, etc.
These come from incrementing k by 1 and multiplying by 3, then adding 1.

## Step 7: For the second form (5m + 2), some numbers that fit this pattern are 2, 7, 12, 17, 22, 27, etc.
Increment m similarly to find these values.

## Step 8: And for the third form (7n + 3), some numbers that fit this pattern are 3, 10, 17, 24, 31, etc.
Again, increment n to generate these values.

## Step 9: The first number that appears in all three sequences is 17, which means it satisfies all the given conditions and is thus the smallest such positive integer N.
This is found by inspection of our lists generated from steps 6-8.

The final answer is: $\boxed{17}$

Let $N$ be the positive integer. We are given the system of congruences:
$$\begin{cases} N \equiv 1 \pmod{3} \\ N \equiv 2 \pmod{5} \\ N \equiv 3 \pmod{7} \end{cases}$$

**Step 1: Solve the first two congruences.**

From $N \equiv 1 \pmod{3}$, we have $N = 3k + 1$ for some integer $k$.
Substitute this into the second congruence:
$$3k + 1 \equiv 2 \pmod{5}$$
$$3k \equiv 1 \pmod{5}$$

To solve for $k$, we find the multiplicative inverse of 3 modulo 5. Since $3 \times 2 = 6 \equiv 1 \pmod{5}$, the inverse is 2.
Multiply both sides by 2:
$$2(3k) \equiv 2(1) \pmod{5}$$
$$6k \equiv 2 \pmod{5}$$
$$k \equiv 2 \pmod{5}$$

Therefore, $k = 5j + 2$ for some integer $j$. Substitute this back into the expression for $N$:
$$N = 3(5j + 2) + 1$$
$$N = 15j + 7$$

**Step 2: Incorporate the third congruence.**

Substitute $N = 15j + 7$ into the third congruence ($N \equiv 3 \pmod{7}$):
$$15j + 7 \equiv 3 \pmod{7}$$

We simplify the coefficients modulo 7:
$15 \equiv 1 \pmod{7}$
$7 \equiv 0 \pmod{7}$

The congruence becomes:
$$1j + 0 \equiv 3 \pmod{7}$$
$$j \equiv 3 \pmod{7}$$

**Step 3: Find the smallest number.**

Since we are looking for the smallest positive integer $N$, we choose the smallest non-negative value for $j$, which is $j=3$.

Substitute $j=3$ back into the expression for $N$:
$$N = 15(3) + 7$$
$$N = 45 + 7$$
$$N = 52$$

The smallest such number is $\mathbf{52}$.

To find the smallest positive integer $n$ that satisfies the given conditions, we can set up a system of linear congruences:
1.  $n \equiv 1 \pmod 3$
2.  $n \equiv 2 \pmod 5$
3.  $n \equiv 3 \pmod 7$

**Step 1: Solve the first two congruences.**
From the first congruence, we know $n$ can be written as $n = 3k + 1$. 
Substitute this into the second congruence:
$3k + 1 \equiv 2 \pmod 5$
$3k \equiv 1 \pmod 5$
To solve for $k$, we find the modular inverse of $3 \pmod 5$. Since $3 \times 2 = 6 \equiv 1 \pmod 5$, the inverse is $2$.
$k \equiv 2 \pmod 5$
This means $k = 5m + 2$. 

Substituting $k$ back into the expression for $n$:
$n = 3(5m + 2) + 1$
$n = 15m + 7$
So, $n \equiv 7 \pmod{15}$.

**Step 2: Solve for the third congruence.**
Now we use the condition $n \equiv 3 \pmod 7$ with our new expression $n = 15m + 7$:
$15m + 7 \equiv 3 \pmod 7$
Since $15 \equiv 1 \pmod 7$ and $7 \equiv 0 \pmod 7$, the equation simplifies to:
$1m + 0 \equiv 3 \pmod 7$
$m \equiv 3 \pmod 7$

This means $m$ can be written as $m = 7p + 3$.

**Step 3: Find the final value of $n$.**
Substitute $m$ back into the expression for $n$:
$n = 15(7p + 3) + 7$
$n = 105p + 45 + 7$
$n = 105p + 52$

To find the smallest positive integer, we set $p = 0$:
$n = 52$

**Verification:**
*   $52 \div 3 = 17$ remainder $1$
*   $52 \div 5 = 10$ remainder $2$
*   $52 \div 7 = 7$ remainder $3$

The smallest such number is **52**.

This problem requires finding the smallest positive integer $N$ that satisfies a system of congruences, which can be solved using techniques related to the Chinese Remainder Theorem.

We are given:
1. $N \equiv 1 \pmod{3}$
2. $N \equiv 2 \pmod{5}$
3. $N \equiv 3 \pmod{7}$

**Step 1: Combine the first two congruences.**
From (1), $N = 3k + 1$. Substituting this into (2):
$3k + 1 \equiv 2 \pmod{5}$
$3k \equiv 1 \pmod{5}$

To solve for $k$, we multiply by the inverse of 3 modulo 5. Since $3 \times 2 = 6 \equiv 1 \pmod{5}$, we use 2:
$2(3k) \equiv 2(1) \pmod{5}$
$6k \equiv 2 \pmod{5}$
$k \equiv 2 \pmod{5}$

Thus, $k = 5j + 2$. Substituting this back into the expression for $N$:
$N = 3(5j + 2) + 1$
$N = 15j + 7$
This means $N \equiv 7 \pmod{15}$.

**Step 2: Incorporate the third congruence.**
We use the result $N = 15j + 7$, and apply condition (3):
$15j + 7 \equiv 3 \pmod{7}$

Simplify modulo 7. Since $15 \equiv 1 \pmod{7}$ and $7 \equiv 0 \pmod{7}$:
$(1)j + 0 \equiv 3 \pmod{7}$
$j \equiv 3 \pmod{7}$

Thus, the smallest non-negative value for $j$ is 3. We set $j = 7m + 3$.

**Step 3: Find N.**
Substitute $j=3$ (setting $m=0$) back into the expression for $N$:
$N = 15(3) + 7$
$N = 45 + 7$
$N = 52$

The smallest such number is **52**.

*(Verification: $52 \div 3 = 17$ R 1; $52 \div 5 = 10$ R 2; $52 \div 7 = 7$ R 3.)*

To solve the problem, we are given the following conditions:

- $ N \equiv 1 \pmod{3} $
- $ N \equiv 2 \pmod{5} $
- $ N \equiv 3 \pmod{7} $

We are asked to find the **smallest positive integer** $ N $ that satisfies all three congruences.

---

### Step 1: Combine the First Two Congruences

Let’s first solve the system:

$$
\begin{cases}
N \equiv 1 \pmod{3} \\
N \equiv 2 \pmod{5}
\end{cases}
$$

Assume $ N = 3k + 1 $. Substitute into the second equation:

$$
3k + 1 \equiv 2 \pmod{5} \Rightarrow 3k \equiv 1 \pmod{5}
$$

To solve this, multiply both sides by the modular inverse of 3 modulo 5. Since $ 3 \cdot 2 = 6 \equiv 1 \pmod{5} $, the inverse is 2. So:

$$
k \equiv 2 \pmod{5} \Rightarrow k = 5m + 2
$$

Substitute back into $ N = 3k + 1 $:

$$
N = 3(5m + 2) + 1 = 15m + 7
$$

So all $ N $ that satisfy the first two congruences are of the form:

$$
N = 15m + 7
$$

---

### Step 2: Incorporate the Third Congruence

Now, substitute $ N = 15m + 7 $ into the third congruence:

$$
15m + 7 \equiv 3 \pmod{7}
$$

Simplify:

$$
15m \equiv -4 \pmod{7} \Rightarrow 15m \equiv 3 \pmod{7}
$$

Note that $ 15 \equiv 1 \pmod{7} $, so:

$$
m \equiv 3 \pmod{7} \Rightarrow m = 7n + 3
$$

Substitute back into $ N = 15m + 7 $:

$$
N = 15(7n + 3) + 7 = 105n + 45 + 7 = 105n + 52
$$

---

### Step 3: Find the Smallest Positive $ N $

The smallest positive $ N $ occurs when $ n = 0 $:

$$
N = 105(0) + 52 = 52
$$

---

### Step 4: Verification

Check that $ N = 52 $ satisfies all three original conditions:

- $ 52 \div 3 = 17 $, remainder 1 ✅
- $ 52 \div 5 = 10 $, remainder 2 ✅
- $ 52 \div 7 = 7 $, remainder 3 ✅

All conditions are satisfied.

---

### Final Answer

$$
\boxed{52}
$$

To find the smallest positive integer that satisfies the following conditions:

\[
\begin{cases}
n \equiv 1 \pmod{3} \\
n \equiv 2 \pmod{5} \\
n \equiv 3 \pmod{7}
\end{cases}
\]

**Step-by-Step Solution:**

1. **List Numbers Satisfying Each Condition Individually:**
   - **Condition 1:** \( n \equiv 1 \pmod{3} \)
     Possible values: 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, ...
   
   - **Condition 2:** \( n \equiv 2 \pmod{5} \)
     Possible values: 2, 7, 12, 17, 22, 27, 32, 37, ...

   - **Condition 3:** \( n \equiv 3 \pmod{7} \)
     Possible values: 3, 10, 17, 24, 31, 38, 45, ...

2. **Find Common Values in Conditions 1 and 2:**
   - Compare the lists from Condition 1 and Condition 2.
   - The common value is **7**.

3. **Check if This Value Satisfies Condition 3:**
   - \( 7 \div 7 = 1 \) with a remainder of **0**, which does not satisfy \( n \equiv 3 \pmod{7} \).

4. **Continue Finding Common Values in Conditions 1 and 2 Beyond the First Match:**
   - Next common value is **12**.
     - \( 12 \div 7 = 1 \) with a remainder of **5**, which also doesn't satisfy Condition 3.

   - Next common value is **17**.
     - \( 17 \div 7 = 2 \) with a remainder of **3\), which satisfies all three conditions.

Therefore, the smallest positive integer that meets all the given conditions is:

\[
\boxed{17}
\]

The smallest such number is $43$. 

We can find this through a systematic process of elimination using the Chinese Remainder Theorem:

1. It must be greater than $20$ since it leaves remainder $2$ when divided by $5$, which means it's at least $5 \times 4 + 2 = 22$. 

2. It also must be less than or equal to $36$ as it should leave a remainder of $1$ when divided by $3$, and the largest multiple of $3$ in the range we're looking for is $3\times12=36$.

3. So, we've narrowed our search between $20 < N \leq 36$. The number must also have a remainder of $3$ when divided by $7$, so it can't be any multiple of $7$ in that range (like $28$ or $35$) as they leave remainders of $0$ and not $3$. That leaves us with the options: 
- $21, 24, 26, 30, 33$, which have remainders of $2, 1, 1, 0,$ and $5$ when divided by $7$ respectively. 

4. Among these numbers, only one gives the required remainder of $1$ when divided by $3$: that number is $N = 24$.

5. Finally, we check if this value satisfies our initial condition: it does! When you divide $24$ by $5$, you get a quotient of $4$ and a remainder of $2$.

Therefore, the smallest such number is $\boxed{24}$.

To find the smallest positive integer that satisfies the following conditions:

1. When divided by 3, the remainder is 1.
2. When divided by 5, the remainder is 2.
3. When divided by 7, the remainder is 3.

Let's denote the number as \( x \).

### Step 1: Translate the Conditions into Congruences
- **Condition 1:**  
  \( x \equiv 1 \pmod{3} \)

- **Condition 2:**  
  \( x \equiv 2 \pmod{5} \)

- **Condition 3:**  
  \( x \equiv 3 \pmod{7} \)

### Step 2: Solve the System of Congruences

We can use the Chinese Remainder Theorem to solve this system.

1. **Find a common solution for the first two congruences:**
   
   - From Condition 1:  
     \( x = 3k + 1 \), where \( k \) is an integer.
   
   - Substitute into Condition 2:
     \[
     3k + 1 \equiv 2 \pmod{5} \\
     3k \equiv 1 \pmod{5}
     \]
     Multiply both sides by the modular inverse of 3 modulo 5 (which is 2):
     \[
     k \equiv 2 \times 1 \pmod{5} \\
     k \equiv 2 \pmod{5}
     \]
   
   - Thus, \( k = 5m + 2 \), where \( m \) is an integer.
   
   - Substitute back into the expression for \( x \):
     \[
     x = 3(5m + 2) + 1 = 15m + 7
     \]

2. **Incorporate Condition 3:**
   
   - From above, \( x = 15m + 7 \).
   
   - Substitute into Condition 3:
     \[
     15m + 7 \equiv 3 \pmod{7} \\
     15m \equiv -4 \pmod{7} \\
     15m \equiv 3 \pmod{7}
     \]
     Since \( 15 \equiv 1 \pmod{7} \):
     \[
     m \equiv 3 \pmod{7}
     \]
   
   - Thus, \( m = 7n + 3 \), where \( n \) is an integer.
   
   - Substitute back into the expression for \( x \):
     \[
     x = 15(7n + 3) + 7 = 105n + 52
     \]

### Step 3: Find the Smallest Positive Integer

The smallest positive integer occurs when \( n = 0 \):
\[
x = 105 \times 0 + 52 = 52
\]

**Final Answer:**  
\(\boxed{52}\)

Given a positive integer \( x \) that satisfies the following conditions:

1. \( x \equiv 1 \pmod{3} \)
2. \( x \equiv 2 \pmod{5} \)
3. \( x \equiv 3 \pmod{7} \)

We start by expressing \( x \) in terms of the first congruence:
\[ x = 3k + 1 \]

Substituting into the second congruence:
\[ 3k + 1 \equiv 2 \pmod{5} \]
\[ 3k \equiv 1 \pmod{5} \]

The multiplicative inverse of 3 modulo 5 is 2 (since \( 3 \times 2 \equiv 6 \equiv 1 \pmod{5} \)). Thus:
\[ k \equiv 2 \times 1 \equiv 2 \pmod{5} \]
\[ k = 5j + 2 \]

Substituting back into \( x \):
\[ x = 3(5j + 2) + 1 = 15j + 7 \]

Now, substituting into the third congruence:
\[ 15j + 7 \equiv 3 \pmod{7} \]
Since \( 15 \equiv 1 \pmod{7} \) and \( 7 \equiv 0 \pmod{7} \):
\[ j \equiv 3 \pmod{7} \]
\[ j = 7k + 3 \]

Substituting back into \( x \):
\[ x = 15(7k + 3) + 7 = 105k + 52 \]

The smallest positive solution is when \( k = 0 \):
\[ x = 52 \]

Verification:
- \( 52 \div 3 \) gives remainder 1
- \( 52 \div 5 \) gives remainder 2
- \( 52 \div 7 \) gives remainder 3

Thus, the smallest such number is \(\boxed{52}\).