Find the least positive integer that leaves the remainder 3 when divided by 7, remainder 4 when divided by 9, and remainder 8 when divided by 11

You can try the CRT (Chinese remainder theorem).

Master sun proceeded as follows:

Given
N≡3 mod 7
N≡4 mod 9
N≡8 mod 11
We need to find the smallest positive N.

First we find the unit remainders,
99≡1 mod 7 (99 is product of 9*11_
154≡1 mod 9 (154 is multiple of 7*11)
441≡1 mod 11 (441 is multiple of 7*9)

N mod(7*9*11) is the required number such that
N=3*99+4*154+8*441 mod 693
=4441 mod 693
=283

Check:
283≡3 mod 7
283≡4 mod 9
283≡8 mod 11 Yeah!

Hint: Chinese Remainder Theorem

To find the least positive integer that satisfies these conditions, we can use the Chinese Remainder Theorem.

The Chinese Remainder Theorem states that for a system of congruences of the form:

1. x ≡ a(mod m)
2. x ≡ b(mod n)
3. x ≡ c(mod p)

where m, n, and p are pairwise coprime (i.e., have no common factors), there exists a unique solution for x (modulo m * n * p).

In this case, we have the following system of congruences:

1. x ≡ 3(mod 7)
2. x ≡ 4(mod 9)
3. x ≡ 8(mod 11)

To find the least positive integer satisfying these congruences, we'll solve the system step-by-step:

Step 1: Find the product of the modulo values: m * n * p
m = 7, n = 9, p = 11
m * n * p = 7 * 9 * 11 = 693

Step 2: Solve each congruence individually:

For the first congruence (x ≡ 3(mod 7)), notice that 693 is divisible by 7. Therefore, any multiple of 693 is congruent to 0(mod 7). So, we can start by finding the first multiple of 693 that is congruent to 3(mod 7).

The first multiple of 693 greater than or equal to 3 is 693 itself. Therefore, x = 693 satisfies the first congruence.

For the second congruence (x ≡ 4(mod 9)), notice that 693 is divisible by 9. Similarly to the previous step, we can find the first multiple of 693 that is congruent to 4(mod 9).

The first multiple of 693 greater than or equal to 4 is 693 itself. Therefore, x = 693 satisfies the second congruence.

For the third congruence (x ≡ 8(mod 11)), notice that 693 is not divisible by 11. To find the first multiple of 693 that is congruent to 8(mod 11), we can check multiples of 693 incrementally.

By checking, we find that the first multiple of 693 that leaves a remainder of 8 when divided by 11 is 693 * 2 = 1386. Therefore, x = 1386 satisfies the third congruence.

Step 3: Combine the solutions
Since x = 693 satisfies the first two congruences and x = 1386 satisfies the third congruence, we need to find a solution that satisfies all three congruences.

To do this, we can calculate the least common multiple (LCM) of the moduli for the first two congruences: LCM(7, 9) = 63.

Next, we need to find a common solution for the first two congruences (mod 63). Starting from x = 693, we can add multiples of 63 until we find a common solution.

x = 693 + k * 63, where k is an integer

By checking, we find that for k = 2, we have x = 693 + 2 * 63 = 819, which satisfies both congruences (x ≡ 3(mod 7) and x ≡ 4(mod 9)).

Finally, we need to find the common solution for all three congruences (mod 693). Starting from x = 819, we can add multiples of 693 until we find a common solution.

x = 819 + k * 693, where k is an integer

By checking, we find that for k = 2, we have x = 819 + 2 * 693 = 2205, which satisfies all three congruences (x ≡ 3(mod 7), x ≡ 4(mod 9), and x ≡ 8(mod 11)).

Therefore, the least positive integer that satisfies all the given congruences is x = 2205.

To find the least positive integer that satisfies these conditions, we can use the Chinese Remainder Theorem (CRT).

The Chinese Remainder Theorem states that given a set of congruences with pairwise coprime moduli, there exists a unique solution modulo the product of the moduli.

In this case, the moduli are 7, 9, and 11. Since these numbers are pairwise coprime (i.e., they have no common factors), we can find a unique solution.

Let's solve the congruences step by step:

1. The first congruence is x ≡ 3 (mod 7) or x mod 7 = 3. Start by finding the solutions for x that satisfy this congruence:
3, 10, 17, 24, 31, ...

2. The second congruence is x ≡ 4 (mod 9) or x mod 9 = 4. Find the solutions for x that satisfy this congruence:
4, 13, 22, 31, 40, ...

3. The third congruence is x ≡ 8 (mod 11) or x mod 11 = 8. Find the solutions for x that satisfy this congruence:
8, 19, 30, 41, 52, ...

Now, we need to find a solution that satisfies all three congruences. To do this, we need to find an integer that appears in all three lists.

We notice that the number 31 satisfies all three congruences. It leaves a remainder of 3 when divided by 7, a remainder of 4 when divided by 9, and a remainder of 8 when divided by 11.

Therefore, the least positive integer that satisfies these conditions is 31.