Finding the Greatest 3-Digit Number with a Specific Remainder

The problem asks us to find the largest 3-digit number that, when divided by 3, 4, 5, and 8, always leaves a remainder of 2.

A number that leaves the same remainder 'r' when divided by several numbers (divisors) is of the form $\text{k} \times \text{LCM(divisors)} + \text{r}$.

In this case, the divisors are 3, 4, 5, and 8, and the remainder is 2. So, the required number must be of the form $\text{k} \times \text{LCM(3, 4, 5, 8)} + 2$, where $\text{k}$ is a whole number.

Calculating the Least Common Multiple (LCM)

First, we need to find the LCM of the divisors 3, 4, 5, and 8. We can do this by finding the prime factorization of each number:

• 3 = 3
• 4 = $2^2$
• 5 = 5
• 8 = $2^3$

The LCM is found by taking the highest power of all prime factors that appear in the factorizations:

$\text{LCM(3, 4, 5, 8)} = 2^3 \times 3^1 \times 5^1 = 8 \times 3 \times 5 = 120$

Formulating the Number and Finding the Greatest 3-Digit Value

The required numbers are of the form $120\text{k} + 2$. We are looking for the greatest 3-digit number of this form.

The greatest 3-digit number is 999.

We need to find the largest integer value of $\text{k}$ such that $120\text{k} + 2 \le 999$.

Subtract 2 from both sides:

$120\text{k} \le 999 - 2$

$120\text{k} \le 997$

Now, divide by 120:

$\text{k} \le \frac{997}{120}$

$\text{k} \le 8.308...$

Since $\text{k}$ must be an integer, the largest possible integer value for $\text{k}$ is 8.

Calculating the Specific Number

Now, substitute the largest integer value of $\text{k}$ (which is 8) back into the formula $120\text{k} + 2$:

Number = $120 \times 8 + 2$

Number = $960 + 2$

Number = $962$

Verification

Let's check if 962 is a 3-digit number (yes) and if it leaves a remainder of 2 when divided by 3, 4, 5, and 8:

• $962 \div 3 = 320$ with remainder $962 - (320 \times 3) = 962 - 960 = 2$.
• $962 \div 4 = 240$ with remainder $962 - (240 \times 4) = 962 - 960 = 2$.
• $962 \div 5 = 192$ with remainder $962 - (192 \times 5) = 962 - 960 = 2$.
• $962 \div 8 = 120$ with remainder $962 - (120 \times 8) = 962 - 960 = 2$.

The number 962 satisfies all the conditions and is the greatest 3-digit number of the form $120k + 2$.

Comparing with Options

Let's look at the given options:

Our calculated number, 962, matches Option 1.

Revision Table: LCM and Remainders

Additional Information: Finding Numbers with Specific Remainders

Problems involving finding numbers that satisfy multiple division conditions often rely on the concept of the LCM. If a number leaves a remainder 'r' when divided by 'a', 'b', and 'c', it means the number can be written as $a\text{k}_1 + \text{r}$, $b\text{k}_2 + \text{r}$, and $c\text{k}_3 + \text{r}$ for some integers $\text{k}_1, \text{k}_2, \text{k}_3$. This implies that the number minus the remainder ($N - \text{r}$) is divisible by a, b, and c. Therefore, $N - \text{r}$ must be a multiple of $\text{LCM(a, b, c)}$.

So, the number $N$ is of the form $\text{m} \times \text{LCM(a, b, c)} + \text{r}$ for some integer m. To find the smallest such number, we typically use m=1 (if the result is positive and meets any digit requirements). To find the greatest number within a range (like greatest 3-digit number), we find the largest multiple of the LCM (plus the remainder) that fits within that range.

For example, to find the smallest number that leaves remainder 1 when divided by 6 and 9:

1. Find $\text{LCM(6, 9)}$. $6 = 2 \times 3$, $9 = 3^2$. $\text{LCM(6, 9)} = 2 \times 3^2 = 18$.
2. The numbers are of the form $18\text{k} + 1$.
3. For the smallest number, take $\text{k}=1$. Number = $18 \times 1 + 1 = 19$. (Note: For smallest *positive* number, if $18 \times 0 + 1 = 1$ is considered, check if it meets criteria). $1 \div 6$ gives remainder 1, $1 \div 9$ gives remainder 1. So 1 is the smallest. If *positive* numbers are implied, then 19 is the smallest positive number greater than the remainder. The context usually clarifies this.