To find the greatest number of 4 digits that when divided by any of the numbers 6, 9, 12, 17 leaves a remainder of 1, we need to use the Chinese Remainder Theorem method.


The Chinese Remainder Theorem states that if we have a system of linear congruences, then there exists a unique solution modulo the product of all the moduli. In this case, the moduli are 6, 9, 12, and 17.

We can write the system of linear congruences as follows:

x ≡ 1 (mod 6)
x ≡ 1 (mod 9)
x ≡ 1 (mod 12)
x ≡ 1 (mod 17)

We can solve this system of congruences using the Chinese Remainder Theorem method as follows:

First, we find the product of the moduli:
M = 6 × 9 × 12 × 17 = 9,612

Next, we find the values of Mi, which are the products of the remaining moduli divided by the current modulus:
M1 = 9 × 12 × 17 = 1,836
M2 = 6 × 12 × 17 = 1,428
M3 = 6 × 9 × 17 = 918
M4 = 6 × 9 × 12 = 648

Then, we find the inverse of each Mi modulo its corresponding modulus:
a1 = 6⁻¹ ≡ 3 (mod 9)
a2 = 9⁻¹ ≡ 6 (mod 6)
a3 = 12⁻¹ ≡ 15 (mod 17)
a4 = 17⁻¹ ≡ 4 (mod 12)

Finally, we can find the solution x using the formula:
x = Σ(ai × Mi × xi) (mod M)

where xi is the solution to the corresponding linear congruence:
x1 = 1
x2 = 1
x3 = 1
x4 = 1

Substituting the values, we get:
x = (3 × 1,836 × 1) + (6 × 1,428 × 1) + (15 × 918 × 1) + (4 × 648 × 1) (mod 9,612)
x = 5,237

Therefore, the greatest number of 4 digits that when divided by any of the numbers 6, 9, 12, 17 leaves a remainder of 1 is 5,237.


The Chinese Remainder Theorem method is used to solve systems of linear congruences. In this case, we had a system of four linear congruences, and we used the Chinese Remainder Theorem method to find the solution to the system. The solution we obtained, 5,237, is the greatest number of 4 digits that when divided by any of the numbers 6, 9, 12, 17 leaves a remainder of 1.

First we have to find the number which is divided by all the given numbers i.e 6,9,12,17

For that we have to find the LCM of those numbers.

The LCM of 6,9,12,17 is 612.

Now we shall find the largest 4-digit number which is divisible by 6,9,12,17 .

For that we have to find the largest 4-digit multiple of 612.

If we consider (612*17)= 10404 (since it is a 5-digit number we should reduce the multiple )

i.e (612*16)

=>9792

so it is the largest possible 4-digit number which is divisible by all the given numbers

But the question is to find the number which leaves a remainder of 1 when it is divided by all those numbers

So for that we have to add 1 to the obtained number…

so the answer is (9792 +1)

=> 9793