If a collection of pencils is placed in rows of 4, there are 2 pencils left; if placed in rows of 5, there are 3 left; and if placed in rows of 7, there are 5 left. What is the smallest possible number of pencils in the collection?

Answer 1

Let $n$ be the total number of pencils. Given that when $n$ is divided by 4, 5, and 7, there are remainders of 2, 3, and 5, respectively, we can set up the following system of congruences:
\begin{align*}
n &\equiv 2\pmod{4}, \\
n &\equiv 3\pmod{5}, \\
n &\equiv 5\pmod{7}.
\end{align*}We look for the smallest possible positive integer $n$ that satisfies these congruences. To begin, we can find one solution to each congruence.

From the first congruence, an integer $n$ that is 2 more than a multiple of 4 is 6, so in this case, $n = 6$.
From the second congruence, an integer $n$ that is 3 more than a multiple of 5 is 8, so in this case, $n = 8$.
From the third congruence, an integer $n$ that is 5 more than a multiple of 7 is 5, so in this case, $n = 5$.

Let $m_1 = \frac{n - 2}{4}$. Then
\[n = 2 + 4m_1 = 4m_1 + 2\]for some positive integer $m_1$.

Let $m_2 = \frac{n - 3}{5}$. Then
\[n = 3 + 5m_2 = 5m_2 + 3\]for some positive integer $m_2$.

Let $m_3 = \frac{n - 5}{7}$. Then
\[n = 5 + 7m_3\]for some positive integer $m_3$.

Substituting $5m_2 + 3$ for $n$ in $n = 4m_1 + 2$, we get
\[5m_2 + 3 = 4m_1 + 2,\]so $4m_1 = 5m_2 + 1$.

By the Euclidean algorithm, $\text{gcd}\,(4,5) = 1$, so $m_1$ must be one more than a multiple of 5.

Substituting $4m_1 + 2$ for $n$ in $n = 5 + 7m_3$, we get
\[4m_1 + 2 = 5 + 7m_3.\]Since $m_1$ is one more than a multiple of 5, $4m_1 + 2$ is 2 more than a multiple of 5, so $5 + 7m_3$ must be 2 more than a multiple of 5.

By the Euclidean algorithm, $\text{gcd}\,(5,7) = 1$, so $m_3$ must be two more than a multiple of 5.

Since $m_1$ is one more than a multiple of 5, and $m_3$ is two more than a multiple of 5, $m_1$ and $m_3$ must be 2 more than a multiple of 5.

Let $m_4 = \frac{m_1 - 1}{5}$, and let $m_5 = \frac{m_3 - 2}{5}$. We have that $m_4$ and $m_5$ are integers.

We can write $m_1 = 5m_4 + 1$ and $m_3 = 5m_5 + 2$.

Then $4m_1 = 20m_4 + 4$, and $5m_2 + 1 = 20m_4 + 4$. Since $5m_2 + 1$ is one more than a multiple of 5, $20m_4 + 4$ must be one more than a multiple of 5, so $20m_4$ must be one less than a multiple of 5, which means $m_4$ must be one less than a multiple of 5.

Since $m_4$ must be one less than a multiple of 5, $m_5$ must be 4 less than a multiple of 5.

Let $m_4 = 5m_6 - 1$, and let $m_5 = 5m_7 - 4$. Then
\begin{align*}
n &= 2 + 4(5m_4 + 1) \\
&= 4 + 20m_4 \\
&= 4 + 20(5m_6 - 1) \\
&= 20m_6 + 4.
\end{align*}We also have
\begin{align*}
n &= 3 + 5(5m_2 + 1) \\
&= 3 + 25m_2 + 5 \\
&= 8 + 25m_2.
\end{align*}Since $5m_2 + 3 = 8 + 25m_2,$ $23m_2 = 5,$ so
\[m_2 = \frac{5}{23}.\]This contradicts the fact that $m_2$ is an integer, so there are no positive integer solutions. Therefore, there are no pencils. Hence, the smallest possible number of pencils is $\boxed{0}$.

Answer 2

wrong. 138 is the smallest number of pencils.

Answer 3

We are given that $n \equiv 2 \pmod 4$, $n \equiv 3 \pmod 5$, and $n \equiv 5 \pmod 7$. From the condition $n \equiv 2 \pmod 4,$ we can write
\[n = 4a + 2\]for some integer $a$. Substituting into the second congruence, we get $4a + 2 \equiv 3 \pmod 5,$ so $4a \equiv 1 \pmod 5.$ Multiply both sides by 4, to get $16a \equiv 4 \pmod 5$. Since $16a \equiv a \pmod 5,$ this becomes $a \equiv 4 \pmod 5.$ Hence, $a = 5b + 4$ for some integer $b,$ so
\[n = 4(5b + 4) + 2 = 20b + 18 = 2(10b + 9).\]Since $n$ is a multiple of 2, $n = 2m$ for some integer $m$, so
\[m = 10b + 9.\]From this congruence, we see that $m \equiv 9 \pmod{10}.$

Substituting into the third congruence, we get $2m \equiv 5 \pmod 7,$ so $m \equiv 3 \pmod 7.$ Since $m \equiv 9 \pmod{10}$ and $m \equiv 3 \pmod 7,$ we can write
\[m = 10c + 9 = 7d + 3.\]Subtracting $m = 7d + 3$ from $m = 10c + 9,$ our equation becomes $3c = 7d + 6,$ so $3c \equiv 6 \pmod 7.$ Multiplying both sides by 5, we get $15c \equiv 30 \pmod 7,$ so $c \equiv 2 \pmod 7.$ Hence, $c = 7e + 2$ for some integer $e,$ so
\[m = 10(7e + 2) + 9 = 70e + 29.\]Since $n = 2m$, it follows that
\[n = 2(70e + 29) = 140e + 58.\]We want the smallest possible integer value of $n,$ so we set $e = 0,$ which gives $n = \boxed{58}.$

Answer 4

still wrong. 58 = 2(mod 7)

Answer 5

You are correct; I apologize for the error in my previous response.

To find the smallest possible number of pencils, we can use the Chinese Remainder Theorem. The given congruences are:

$n \equiv 2 \pmod 4$
$n \equiv 3 \pmod 5$
$n \equiv 5 \pmod 7$

From the first congruence, we can write $n = 4a + 2$ for some integer $a$. Substituting this into the second congruence, we have $4a + 2 \equiv 3 \pmod 5$, which simplifies to $4a \equiv 1 \pmod 5$. Solving for $a$ gives $a \equiv 4 \pmod 5$, or $a = 5b+4$ for some integer $b$.

Substituting this back into $n = 4a + 2$, we have $n = 4(5b+4) + 2 = 20b + 18$. From the third congruence, we then have $20b + 18 \equiv 5 \pmod 7$, which simplifies to $6b \equiv 5 \pmod 7$. Solving for $b$, we find $b \equiv 6 \pmod 7$, or $b = 7c + 6$ for some integer $c$.

Substituting this back into $n = 20b + 18$, we have $n = 20(7c+6) + 18 = 140c + 138$. Thus, the smallest possible value of $n$ that satisfies the given conditions is $\boxed{138}$.