Question

A number is such that when divided by 27, 30 or 45 , the remainder is always 5 . Find the smallest value of x

Answer 1

Given that a number \( x \) leaves a remainder of 5 when divided by 27, 30, and 45, we can set up the following congruences:

\[
x \equiv 5 \mod{27}
\]
\[
x \equiv 5 \mod{30}
\]
\[
x \equiv 5 \mod{45}
\]

To simplify our problem, we can express \( x \) in terms of a variable \( k \):

\[
x = 27m + 5
\]
\[
x = 30n + 5
\]
\[
x = 45p + 5
\]

for some integers \( m, n, p \).

These equations imply that \( x - 5 \) is a multiple of 27, 30, and 45. Thus, we define:

\[
y = x - 5
\]

so we need to find \( y \) such that:

\[
y \equiv 0 \mod{27}
\]
\[
y \equiv 0 \mod{30}
\]
\[
y \equiv 0 \mod{45}
\]

This means \( y \) must be a multiple of the least common multiple (LCM) of the numbers 27, 30, and 45. We start by finding the LCM.

### Step 1: Find the prime factorizations
- \( 27 = 3^3 \)
- \( 30 = 2^1 \cdot 3^1 \cdot 5^1 \)
- \( 45 = 3^2 \cdot 5^1 \)

### Step 2: Determine the LCM
The LCM takes the highest power of each prime:
- For \( 2 \), the highest power is \( 2^1 \).
- For \( 3 \), the highest power is \( 3^3 \).
- For \( 5 \), the highest power is \( 5^1 \).

Thus, the LCM is:

\[
\text{lcm}(27, 30, 45) = 2^1 \cdot 3^3 \cdot 5^1 = 2 \cdot 27 \cdot 5
\]

Calculating this step by step:

\[
27 \cdot 5 = 135
\]
\[
2 \cdot 135 = 270
\]

Therefore, the LCM is 270.

### Step 3: Find \( y \)
Since \( y \) must be a multiple of 270, the smallest value for \( y \) is:

\[
y = 270
\]

### Step 4: Compute \( x \)
Now we substitute back to find \( x \):

\[
x = y + 5 = 270 + 5 = 275
\]

Thus, the smallest value of \( x \) that satisfies all the original conditions is:

\[
\boxed{275}
\]