Is there a way to tell if a number is divisible by 4 if the figure is of 2 digits

Question

I'm trying to help my daughter learn maths. She is struggling with factors, which is to work out what numbers go into a larger number (division).

I've already learned that by summing numbers, if they make 3, it can divide by 3. I also know the rules for 2, 5, 6, 9 and 10.

I'm trying to see if there is a rule for 4. I'm thinking not.

https://www.quora.com/Why-does-the-divisibility-rule-for-the-number-4-work shows the following

The divisibility rule for 4 is in any large number, if the digits in tens and units places is divisible by then the whole number is divisible by 4.

This doesn't make sense. 56 divides by 4. However, the 2 numbers add to 11, and so can't be divided by 4.

It may very well get a "no" answer, but is there any pattern/method I can use for determining if a number can be divided by 4 if it is less than 100 (and greater than 4)

Answer 1

How to make sense of that rule for divisibility by $4$: it's not saying to add the last two digits; it's merely saying to look at the last two digits. Because $4$ divides $100$, a number is divisible by $4$ if and only if its last two digits (ten's place and one's place) are divisible by $4$. Robert Israel's answer gives a method for determining whether a two-digit number is divisible by $4$, and the rule is saying that's essentially all you need.

For example, if you want to know whether $2389080349$ is divisbile by $4$, you merely have to determine whether $49$ is divisible by $4$. (It's not.)

Answer 2

Tens place even and units $0$, $4$ or $8$ (i.e. divisible by $4$), or tens place odd and units $2$ or $6$ (even but not divisible by $4$).

Answer 3

The test for divisibility by $4$ is given any integer $n$ consider the last two digits; if that two-digit number is divisible by $4$ then so is $n$.

Example.

Consider 96. Since $96$ is divisible by $4$, so is $196.$

Reason: $196 = 100 + 96$. The number on the left (which will always be the case even if it is $0$) is divisible by $4$; hence, it suffices to consider only the number represented by the last two digits of the integer $n$.

Finally, in regards to your last question, say you had the number 8. Describing $8$ as $08$, the test applies to single digit numbers as well.

Answer 4

The key is that 100 is divisible by 4. So, we have have:

$$12345678956 = (123456789)(100) + 56 = (123456789)(25)(4) + (14)(4) = ((123456789)(25)+14)(4)$$

Therefore, if the first two digits are divisible by four, the whole thing is divisible by four. In fact, the remainder when divided by four is equal to that when you just divide the last two digits, because the 3 digits onwards have remainder zero.

Answer 5

There is a rule of divisibility for the number $4$. Here it is:

To figure out if a number is divisible by four, you first need to look at the last two digits, and if they're divisible by four together, you can assume that the whole number is divisible by $4$.

Why does this work? Well, $100$ is divisible by four, any number value in the place values greater than the hundreds place is a multiple of $100$. For example, in the number $2,375$, the $2$ in the thousands place stands for $2,000$, and $100\times 20=2,000$, so $2,000$ is a multiple of $100$. If we then add the two digits below the hundreds place, we can say that if all the digits above the ones place, and the tens place are divisible by four, if the two digits left are also divisible by four, it doesn't change anything!

Hopefully this helped you with your question.