Finding the Least Number with Specific Remainders and Divisibility

The problem asks us to find the sum of the digits of the smallest number that satisfies two conditions:

1. When divided by 12, 16, and 20, it leaves a remainder of 6 in each case.
2. It is divisible by 9.

Using the Remainder Condition

A number that leaves the same remainder 'r' when divided by several numbers (say, $d_1, d_2, \ldots, d_n$) can be expressed in the form $\text{k} \cdot \text{lcm}(d_1, d_2, \ldots, d_n) + r$, where $\text{k}$ is a non-negative integer.

In this problem, the divisors are 12, 16, and 20, and the remainder is 6. So, the number $N$ can be written as:

$N = \text{k} \cdot \text{lcm}(12, 16, 20) + 6$

Let's calculate the Least Common Multiple (LCM) of 12, 16, and 20.

• Prime factorization of 12: $12 = 2^2 \times 3^1$
• Prime factorization of 16: $16 = 2^4$
• Prime factorization of 20: $20 = 2^2 \times 5^1$

The LCM is found by taking the highest power of all prime factors involved:

$\text{lcm}(12, 16, 20) = 2^{\max(2, 4, 2)} \times 3^{\max(1, 0, 0)} \times 5^{\max(0, 0, 1)}$

$\text{lcm}(12, 16, 20) = 2^4 \times 3^1 \times 5^1 = 16 \times 3 \times 5 = 48 \times 5 = 240$

So, the number $N$ must be of the form:

$N = 240\text{k} + 6$, where $\text{k}$ is a non-negative integer ($\text{k} = 0, 1, 2, \ldots$).

Applying the Divisibility by 9 Condition

The second condition states that the number $N$ must be divisible by 9. This means $N \equiv 0 \pmod{9}$.

Substituting the form of $N$ we found:

$240\text{k} + 6 \equiv 0 \pmod{9}$

Let's find the remainder of 240 when divided by 9:

$240 \div 9$

$240 = 9 \times 26 + 6$

So, $240 \equiv 6 \pmod{9}$.

The congruence becomes:

$6\text{k} + 6 \equiv 0 \pmod{9}$

We can factor out 6:

$6(\text{k} + 1) \equiv 0 \pmod{9}$

This means $6(\text{k} + 1)$ must be a multiple of 9. Let's analyze the factors. $6 = 2 \times 3$ and $9 = 3 \times 3$. For $6(\text{k}+1)$ to be divisible by 9, $(\text{k}+1)$ must contribute the remaining factor of 3 that $6$ doesn't provide for divisibility by $9$. Thus, $(\text{k}+1)$ must be a multiple of 3.

So, $\text{k} + 1 = 3\text{m}$ for some integer $\text{m}$.

$\text{k} = 3\text{m} - 1$

We are looking for the least number, which corresponds to the smallest possible non-negative integer value of $\text{k}$.

• If $\text{m} = 1$, $\text{k} = 3(1) - 1 = 2$.
• If $\text{m} = 2$, $\text{k} = 3(2) - 1 = 5$.
• And so on...

The smallest non-negative value for $\text{k}$ occurs when $\text{m}=1$, which gives $\text{k}=2$. Let's check $k=0$ and $k=1$.

• If $\text{k}=0$, $6(0+1) = 6$, which is not divisible by 9.
• If $\text{k}=1$, $6(1+1) = 12$, which is not divisible by 9.
• If $\text{k}=2$, $6(2+1) = 18$, which is divisible by 9. This is the smallest non-negative $\text{k}$ that satisfies the condition.

Finding the Least Number

Using the smallest valid value $\text{k}=2$, we find the least number $N$:

$N = 240\text{k} + 6$

$N = 240(2) + 6$

$N = 480 + 6$

$N = 486$

Let's quickly verify this number:

• $486 \div 12 = 40$ remainder 6. (Correct)
• $486 \div 16 = 30$ remainder 6. (Correct)
• $486 \div 20 = 24$ remainder 6. (Correct)
• $486 \div 9 = 54$. Divisible by 9. (Correct)

So, the least number satisfying both conditions is 486.

Calculating the Sum of Digits

The problem asks for the sum of the digits of this least number, which is 486.

Sum of digits = $4 + 8 + 6 = 18$

The sum of the digits of the least number is 18.

Revision Table: Number Properties Summary

Additional Information: Exploring Related Concepts

Understanding remainders and LCM is fundamental in number theory. The structure of numbers leaving a specific remainder when divided by multiple numbers is always based on the LCM of the divisors.

For example, if a number leaves remainder 3 when divided by 4 and 6, it must be of the form $\text{lcm}(4, 6)\text{k} + 3$. $\text{lcm}(4, 6) = 12$, so the numbers are $12\text{k} + 3$ (e.g., 3, 15, 27, ...).

Combining remainder conditions with divisibility conditions (like divisibility by 9 in this problem) often leads to linear congruences, which can be solved to find the possible values for the multiplier $\text{k}$. The "least number" requirement directs us to the smallest valid non-negative integer value for $\text{k}$.

The sum of digits is a useful property for checking divisibility by 3 and 9, as demonstrated in the solution where we used the divisibility rule of 9.