Which is the least number, which, when divided by 12 and 16, leave the same remainder, but is exactly divisible by 11?*1 point9955132176

1. Break Down the Problem

To find the least number $x$ that:

• Leaves the same remainder when divided by 12 and 16.
• Is exactly divisible by 11.

2. Relevant Concepts

• If $x$ leaves the same remainder $r$ when divided by 12 and 16, we can express this as: $x \equiv r \mod 12$ $x \equiv r \mod 16$
• Additionally, since $x$ is exactly divisible by 11, we have: $x \equiv 0 \mod 11$

3. Analysis and Detail

• For $x \equiv r \mod 12$ and $x \equiv r \mod 16$, we can express $x$ as: $x = 12k + r \quad (1)$ $x = 16m + r \quad (2)$ where $k$ and $m$ are integers.

• Setting (1) equal to (2), we get: $12k + r = 16m + r$ Simplifying gives us: $12k = 16m$ or $\frac{k}{m} = \frac{4}{3}$ This implies that $k$ must be a multiple of 4 and $m$ must be a multiple of 3.

• Let $k = 4n$ and $m = 3n$ for some integer $n$. Substituting gives us: $x = 12(4n) + r = 48n + r$ The value of $r$ must satisfy $r \equiv 0 \mod \text{lcm}(12, 16)$.

• The least common multiple of 12 and 16 is: $\text{lcm}(12, 16) = 48$ Thus, we can set: $r = 48p \quad \text{for some integer } p.$

• Therefore, we rewrite $x$: $x = 48n + 48p = 48(n + p).$

4. Verify and Summarize

Now we need $x$ to be divisible by 11: $48(n + p) \equiv 0 \mod 11$ Calculating $48 \mod 11$: $48 = 4 \cdot 11 + 4 \quad \Rightarrow \quad 48 \equiv 4 \mod 11$ Thus, we require: $4(n + p) \equiv 0 \mod 11 \quad \Rightarrow \quad n + p \equiv 0 \mod \frac{11}{\gcd(4, 11)} = 11$ Let $n + p = 11k$ for the smallest $k = 1$: $n + p = 11$ So, we can have $n = 0, p = 11$ or other combinations. The smallest positive $x$ occurs at: $n = 11, p = 0 \Rightarrow x = 48 \cdot 11 = 528.$

Final Answer

The least number, which when divided by 12 and 16 leaves the same remainder and is exactly divisible by 11, is 528.