A math class has fewer than $40$ students. When the students try to sit in rows of $7$, $3$ students sit in the last row. When the students try to sit in rows of $5$, $1$ student is left in the last row. How many students are in this class?
Let $n$ be the number of students in the class. When we divide $n$ by $7$, we get a remainder of $3$. Therefore, we need to add multiples of $7$ to $3$ until we get a number which, when we divide by $5$, has a remainder of $1$. When we add $28$ to $3$, we get $31$, a number that gives a remainder of $1$ when dividing by $5$. By the Chinese remainder theorem, the other integers which leave a remainder of 3 when divided by 7 and a remainder of 1 when divided by 5 differ from 31 by a multiple of $7\cdot 5=35$. Therefore, the only integer between 0 and 40 satisfying these conditions is 31, and there are $\boxed{31}$ students in this math class.
31
Level 1
Number Theory
Problem: A math class has fewer than $40$ students. When the students try to sit in rows of $7$, $3$ students sit in the last row. When the students try to sit in rows of $5$, $1$ student is left in the last row. How many students are in this class?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>Let the number of students be denoted as \( n \). The problem states that \( n < 40 \). 

When the students sit in rows of \( 7 \), there are \( 3 \) students in the last row. This can be expressed mathematically as:
\[
n \equiv 3 \mod 7
\]
This means that \( n \) can be written in the form:
\[
n = 7k + 3
\]
for some integer \( k \).

When the students sit in rows of \( 5 \), there is \( 1 \) student left in the last row. This can be expressed as:
\[
n \equiv 1 \mod 5
\]
This means that \( n \) can also be written in the form:
\[
n = 5m + 1
\]
for some integer \( m \).

Now, we have two equations:
1. \( n = 7k + 3 \)
2. \( n = 5m + 1 \)

Setting these two expressions for \( n \) equal gives:
\[
7k + 3 = 5m + 1
\]
Rearranging this equation leads to:
\[
7k - 5m = -2
\]

Next, we need to find integer solutions for \( k \) and \( m \). Rearranging gives:
\[
7k = 5m - 2
\]
This implies that \( 5m - 2 \) must be divisible by \( 7 \). We can express this condition as:
\[
5m - 2 \equiv 0 \mod 7
\]
or equivalently:
\[
5m \equiv 2 \mod 7
\]

To solve for \( m \), we can find the multiplicative inverse of \( 5 \) modulo \( 7 \). Testing values, we find:
\[
5 \times 3 = 15 \equiv 1 \mod 7
\]
Thus, the inverse is \( 3 \). Multiplying both sides of \( 5m \equiv 2 \) by \( 3 \) gives:
\[
m \equiv 6 \mod 7
\]
This means:
\[
m = 7j + 6
\]
for some integer \( j \).

Substituting \( m \) back into the equation for \( n \):
\[
n = 5(7j + 6) + 1 = 35j + 30 + 1 = 35j + 31
\]

Now, since \( n < 40 \), we can set up the inequality:
\[
35j + 31 < 40
\]
This simplifies to:
\[
35j < 9 \quad \Rightarrow \quad j < \frac{9}{35}
\]
The only integer value for \( j \) that satisfies this is \( j = 0 \).

Substituting \( j = 0 \) back into the equation for \( n \):
\[
n = 35(0) + 31 = 31
\]

Thus, the number of students in the class is \( 31 \).<|END_OF_LATENT|> The final answer is $31$.