The least common multiple of 4 and 6 is 12, so any number divisible by both must also be divisible by 12.

Any number divisible by 10 is also a multiple of 10.

The least common multiple of 6 and 8 is 24, so any number divisible by both must also be divisible by 24.

The least common multiple of 8 and 12 is 24, so the smallest number divisible by both is 24.

The least common multiple of 8 and 12 is 24, so any number divisible by both must also be divisible by 24.

The least common multiple of 8 and 12 is 24, so the number must be divisible by 24.

The least common multiple of 9 and 12 is 36, and since 36 is divisible by 18, the number must also be divisible by 18.

The term 5k is divisible by 5, so the remainder is solely from 3, which is 3.

The remainder is 3, as 5k is divisible by 5.

Let the original number be x. Then, x + 0.15x = 115, or 1.15x = 115. Thus, x = 115 / 1.15 = 100.

Let the original number be x. Then, x + 0.2x = 120, which gives 1.2x = 120, so x = 100.

Let the original number be x. Then, x + 0.2x = 120, which gives 1.2x = 120. Therefore, x = 120 / 1.2 = 100.

Let the number be 100. After a 20% increase, it becomes 120. After a 20% decrease, it becomes 96. Net change = (96-100)/100 * 100% = -4%.

Net change = (1.2x * 0.8) - x = 0.96x, which is a decrease of 4%.

Any number multiplied by 0 is always 0.

Let the number be x. Then, 3x - 4 = 11. Solving gives x = 5.

Let the number be x. Then, 3x - 5 = 10. Solving gives x = 5.

Let the number be x. Then 3x - 9 = 6. Solving gives 3x = 15, so x = 5.

Let the number be x. Then, 9x - 27 = 0, so 9x = 27, and x = 3.

The smallest number that satisfies both conditions is 11 (11 % 3 = 2 and 11 % 4 = 3).

The smallest number satisfying both conditions is 9, as 9 mod 4 = 1 and 9 mod 5 = 4.

The maximum number satisfying both conditions is 19 (19 % 4 = 3 and 19 % 5 = 4).

The smallest number that satisfies both conditions is 9.

The smallest number that satisfies both conditions is 7 (7 % 4 = 3 and 7 % 5 = 2).

The smallest number satisfying both conditions is 12, as 12 mod 5 = 2 and 12 mod 7 = 5.

The number can be expressed as 5k + 2, which when divided by 10 will leave a remainder of 2.

17 divided by 5 gives a remainder of 2 (17-15=2).

The smallest number that satisfies both conditions is 14 (14 % 6 = 2 and 14 % 5 = 4).

The smallest number satisfying both conditions is 23.

8 leaves a remainder of 3 when divided by 5 (5*1 + 3 = 8).