Divisibility and Divisibility Rules

A whole number is divisible by 2 if its last digit is 0, 2, 4, 6 or 8.

124 ends in 4, so it is even. Checking: $124÷2=62$ with no remainder, so 2 goes into 124 exactly.

So the answer is Yes.

A whole number is divisible by 5 only if its last digit is 0 or 5.

40 ends in 0, and $40÷5=8$ exactly, so 40 is the number that is divisible by 5.

The others end in 3, 4 and 7: $23÷5=4$ remainder 3, $54÷5=10$ remainder 4 and $57÷5=11$ remainder 2.

A number is divisible by 3 when the sum of all its digits is a multiple of 3.

For 542: $5+4+2=11$. The multiples of 3 either side of 11 are 9 and 12, so 11 is not a multiple of 3 and neither is 542.

Check: $542=3\times 180+2$, so there is a remainder of 2.

Looking at only part of the number (for example, spotting that $42÷3=14$) is not enough. The test uses every digit.

A number is divisible by 9 when the sum of its digits is a multiple of 9.

For 675: $6+7+5=18$, and $18=9\times 2$, so 675 is divisible by 9.

Check: $675÷9=75$ exactly.

The digit sum only has to be a multiple of 9. It does not have to be 9 itself. Being an odd number does not stop 675 from being divisible by 9 either.

A whole number is divisible by 8 if its last three digits form a number that is divisible by 8.

432 has only three digits, so check it directly: $432÷8=54$ exactly, so 432 is divisible by 8.

A number is divisible by 6 when it is divisible by both 2 and 3.

528 is even, so it is divisible by 2. Its digits add to $5+2+8=15$, and $15=3\times 5$ is a multiple of 3, so 528 is divisible by 3 as well.

Both tests pass, so 528 is divisible by 6. Check: $528÷6=88$ exactly.

Note that the digit sum only has to be a multiple of 3. It does not need to be a multiple of 6.