Finding the Smallest 4-Digit Number with Specific Remainders

The problem asks us to find the smallest 4-digit number, let's call it N, which satisfies several conditions related to remainders upon division. These conditions are:

• When N is divided by 3, the remainder is 1.
• When N is divided by 5, the remainder is 3.
• When N is divided by 6, the remainder is 4.
• When N is divided by 9, the remainder is 7.

Analyzing the Remainder Conditions

Let's look at the relationship between the divisors and their respective remainders for the number N:

• For divisor 3, remainder is 1. Difference = $3 - 1 = 2$.
• For divisor 5, remainder is 3. Difference = $5 - 3 = 2$.
• For divisor 6, remainder is 4. Difference = $6 - 4 = 2$.
• For divisor 9, remainder is 7. Difference = $9 - 7 = 2$.

We observe a consistent difference of 2 between the divisor and the remainder in each case. This is a crucial observation. If a number N leaves a remainder 'r' when divided by 'd', it means N can be written as $N = dq + r$, where q is the quotient. In our case, $N \equiv r \pmod{d}$, which means $N - r$ is divisible by $d$. Alternatively, adding the difference $(d-r)$ to N makes it divisible by $d$. Since the difference $d-r$ is 2 for all given conditions, adding 2 to N will make it perfectly divisible by 3, 5, 6, and 9.

So, the number $N + 2$ must be a common multiple of 3, 5, 6, and 9.

Finding the Least Common Multiple (LCM)

To find the smallest such number $N+2$, we need to find the least common multiple (LCM) of the divisors 3, 5, 6, and 9.

Let's find the prime factorization of each divisor:

• $3 = 3^1$
• $5 = 5^1$
• $6 = 2 \times 3 = 2^1 \times 3^1$
• $9 = 3 \times 3 = 3^2$

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

Prime factors are 2, 3, and 5.

• Highest power of 2 is $2^1$ (from 6).
• Highest power of 3 is $3^2$ (from 9).
• Highest power of 5 is $5^1$ (from 5).

LCM(3, 5, 6, 9) = $2^1 \times 3^2 \times 5^1 = 2 \times 9 \times 5 = 90$.

This means $N + 2$ must be a multiple of 90. We can write this as $N + 2 = 90k$, where k is an integer.

Therefore, $N = 90k - 2$.

Finding the Smallest 4-Digit Value of N

We are looking for the smallest 4-digit number N. A 4-digit number is between 1000 and 9999, inclusive. So, we need to find the smallest integer value of k such that $N = 90k - 2$ is greater than or equal to 1000.

$90k - 2 \ge 1000$

$90k \ge 1000 + 2$

$90k \ge 1002$

$k \ge \frac{1002}{90}$

$k \ge \frac{100.2}{9} \approx 11.133...$

Since k must be an integer, the smallest integer value for k that satisfies this inequality is 12.

Calculating the Smallest Number N

Substitute the smallest valid value of k (which is 12) back into the equation for N:

$N = 90k - 2$

$N = 90 \times 12 - 2$

$N = 1080 - 2$

$N = 1078$

Verification

Let's check if 1078 satisfies all the given conditions:

• $1078 \div 3$: $1078 = 3 \times 359 + 1$ (Remainder 1) - Correct.
• $1078 \div 5$: $1078 = 5 \times 215 + 3$ (Remainder 3) - Correct.
• $1078 \div 6$: $1078 = 6 \times 179 + 4$ (Remainder 4) - Correct.
• $1078 \div 9$: $1078 = 9 \times 119 + 7$ (Remainder 7) - Correct.

Also, 1078 is indeed a 4-digit number (between 1000 and 9999). Since we found the smallest integer k that makes N a 4-digit number, 1078 is the smallest such number satisfying all the given remainder conditions.

The smallest value of N is 1078.

Revision Table: Key Concepts

Additional Information on Remainder Problems

Problems involving remainders and divisibility often utilize concepts from modular arithmetic and number theory. When a number leaves different remainders with different divisors, we might use the Chinese Remainder Theorem for more complex cases. However, if there's a constant difference between the divisor and remainder (as in this problem) or a constant remainder, the problem simplifies significantly, and the LCM concept is directly applicable.

For a number N such that $N \equiv r_1 \pmod{d_1}$, $N \equiv r_2 \pmod{d_2}$, ..., $N \equiv r_k \pmod{d_k}$:

• If $r_1 = r_2 = ... = r_k = r$ (constant remainder), then $N - r$ is divisible by $d_1, d_2, ..., d_k$. So $N = \text{LCM}(d_1, ..., d_k) \times q + r$.
• If $d_1 - r_1 = d_2 - r_2 = ... = d_k - r_k = c$ (constant difference), then $N + c$ is divisible by $d_1, d_2, ..., d_k$. So $N = \text{LCM}(d_1, ..., d_k) \times q - c$. This is the case in our problem where $c=2$.

Understanding the relationship between the number, the divisor, and the remainder is key to solving such problems. The smallest positive number satisfying the conditions will be of the form $k \times \text{LCM} \pm \text{constant}$, and we find the smallest k that fits the required range (e.g., 4-digit number).