Applying Divisibility Proofs in Problems

When a number is divided by 3 and leaves a remainder of 1, it can be expressed as 3k + 1 because it is 1 more than a multiple of 3.

To make 5x + 2 divisible by 3, substitute values: if x = 2, then 5(2) + 2 = 12, which divides evenly by 3.

For a number that’s divisible by 6 and leaves remainder 2 when divided by 5, 42 works because 42 ÷ 6 = 7 and 42 ÷ 5 leaves a remainder of 2.

If 2x + 3 is divisible by 5, substitute x = 1: 2(1) + 3 = 5, which divides evenly by 5.

A number that leaves remainder 4 when divided by both 5 and 7 is 39 because 39 ÷ 5 and 39 ÷ 7 each leave a remainder of 4.

If x is even and 3x + 1 is divisible by 5, try x = 8: 3(8) + 1 = 25, which divides by 5. 8 is even, so it satisfies both conditions.

Numbers divisible by both 2 and 3 but leaving remainder 1 when divided by 5 can be expressed using their LCM (30): 30k + 6 gives the needed pattern.

A number divisible by 6 but not by 9 is 24 because 24 ÷ 6 = 4, yet 24 is not a multiple of 9.

When a number is divided by 8 and leaves a remainder of 3, it can be written as 8k + 3.

A number divisible by 5 and leaving a remainder of 2 when divided by 3 is 20 because 20 ÷ 5 = 4 and 20 ÷ 3 leaves remainder 2.

A number divisible by both 5 and 9 must be a multiple of their LCM, 45, so it can be written as 45k.

A number divisible by 4 and leaving remainder 1 when divided by 3 fits the pattern 12k + 4 because 12 is the least common multiple of 4 and 3, and adding 4 creates the required remainder.

Solving 4x − 7 = 3x + 5 gives x = 12 after adding 7 and subtracting 3x.

The smallest number greater than 50 that is divisible by both 4 and 6 is 60 because the LCM of 4 and 6 is 12, and the next multiple of 12 after 50 is 60.

For 7x + 2 to be divisible by 3, substitute x = 1: 7(1) + 2 = 9, which is divisible by 3.