1.4.4 Finding Remainders of Other Integers

Another popular question on number sense tests include finding the remainder when dividing by 6 or 12 or some combination of the tricks mentioned above. When dividing seems trivial, sometimes it is best to just long divide to get the remainder (for example $1225 \div 6 = 1$ from obvious division), however, when this seems tedious, you can use a combination of the two of the tricks mentioned above (depending on the factors of the number you are dividing). Let’s look at an example:

556677 ÷ 6 has what remainder?

• Dividing by 2: Remainder 1 (odd number)
• Dividing by 3: $(5 + 5 + 6 + 6 + 7 + 7) = 36 \div 3$ Remainder 0

So now the task is to find an appropriate remainder (less than 6) such that it is odd (has a remainder of 1 when dividing by 2) and is divisible by 3 (has a remainder of 0 when dividing by 3). From this information, you get r = 3.

Let’s look at another example to solidify this procedure:

54259 ÷ 12 has what remainder?

• Dividing by 4: $59 \div 4$ Remainder 3
• Dividing by 3: $(5 + 4 + 2 + 5 + 9) = 25 \div 3$ Remainder 1

So for this instance, we want an appropriate remainder (less than 12) that has a remainder of 3 when dividing by 4, and a remainder of 1 when dividing by 3. Running through the integers of interest (0 - 11), you get the answer r = 7.

The best way of getting faster with this trick is through practice and familiarization of the basic principles.

Problem Set 1.4.4