A 4-digit number N is such that when divided by 3, 5, 6, 9 leaves a remainder 1, 3, 4, 7 respectively. What is the smallest value of N?

Correct Answer: 1078

To find the smallest 4-digit number $N$ that leaves remainders 1, 3, 4, and 7 when divided by 3, 5, 6, and 9 respectively, let's analyze the relationship between the divisors and their respective remainders:

Divisor 3, Remainder 1 ⇒ Difference = 3 - 1 = 2
Divisor 5, Remainder 3 ⇒ Difference = 5 - 3 = 2
Divisor 6, Remainder 4 ⇒ Difference = 6 - 4 = 2
Divisor 9, Remainder 7 ⇒ Difference = 9 - 7 = 2

Since the difference between each divisor and its corresponding remainder is constant (equal to 2), the number $N+2$ must be completely divisible by 3, 5, 6, and 9.

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

Let's calculate the Least Common Multiple (LCM) of 3, 5, 6, and 9:

Prime factorizations:
3 = $3$
5 = $5$
6 = $2\times 3$
9 = $3^2$

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

Any number satisfying this condition is of the form:

$N+2=90k$

$N=90k-2$ (where $k$ is a positive integer)

We need to find the smallest 4-digit number $N$. The smallest 4-digit number is 1000.

Let's find the required value of $k$ by dividing 1000 by 90:

$1000÷90=11.11\dots$

Taking the smallest integer $k=12$ to ensure $N$ has 4 digits:

$N=90\times 12-2$

$N=1080-2=1078$

Thus, the smallest 4-digit number satisfying the given condition is 1078.