How to Multiply and Divide Integers

2024 Author: Samantha Chapman | [email protected]. Last modified: 2024-01-17 17:31

Integers are positive or negative numbers with no fractions or decimals. Multiplying and dividing 2 or more whole numbers isn't much different than the same operations on positive-only numbers. The substantial difference is represented by the minus sign, which must always be taken into account. Taking into account the sign, you can proceed to multiplication normally.

Steps

General informations

Step 1. Learn to recognize integers

An integer is a round number that can be represented without fractions or decimals. Integers can be positive, negative, or null (0). For example, these numbers are integers: 1, 99, -217 and 0. While these are not: -10.4, 6 ¾, 2.1².

• Absolute values can be integers, but they don't necessarily have to. An absolute value of any number is the “size” or “quantity” of the number, regardless of the sign. Another way to render this is that the absolute value of a number is its distance from 0. Therefore, the absolute value of an integer is always an integer. For example, the absolute value of -12 is 12. The absolute value of 3 is 3. Of 0 is 0.

  Absolute values of non-integers, however, will never be integers. For example, the absolute value of 1/11 is 1/11 - a fraction, so not an integer

Step 2. Learn the basic times tables

The process of multiplying and dividing integers, whether large or small, is much simpler and faster after memorizing the products of each pair of numbers between 1 and 10. This information is usually taught in school as "times tables". As a reminder, the 10x10 times table is shown below. The numbers in the first row and in the first column range from 1 to 10. To find the product of a pair of numbers, locate the intersection between the column and the row of numbers in question:

Times tables from 1 to 10

	1	2	3	4	5	6	7	8	9	10
Step 1.	1	2	3	4	5	6	7	8	9	10
Step 2.	2	4	6	8	10	12	14	16	18	20
Step 3.	3	6	9	12	15	18	21	24	27	30
Step 4.	4	8	12	16	20	24	28	32	36	40
Step 5.	5	10	15	20	25	30	35	40	45	50
Step 6.	6	12	18	24	30	36	42	48	54	60
Step 7.	7	14	21	28	35	42	49	56	63	70
Step 8.	8	16	24	32	40	48	56	64	72	80
Step 9.	9	18	27	36	45	54	63	72	81	90
Step 10.	10	20	30	40	50	60	70	80	90	100

Method 1 of 2: Multiply the whole numbers

Step 1. Count the minus signs within the multiplication problem

A common problem between two or more positive numbers will always give a positive result. However, each negative sign added to a multiplication transforms the final sign from positive to negative or vice versa. To start an integer multiplication problem, count the negative signs.

Let's use the example -10 × 5 × -11 × -20. In this problem, we can clearly see three less. We will use this data in the next point.

Step 2. Determine the sign of your answer based on the number of negative signs in the problem

As noted earlier, the response to a multiplication with only positive signs will be positive. For each minus in the problem, reverse the sign of the answer. In other words, if the problem has only one negative sign, the answer will be negative; if it has two, it will be positive and so on. A good rule of thumb is that odd numbers of negative signs give negative results and even numbers of negative signs give positive results.

In our example, we have three negative signs. Three is odd, so we know the answer will be negative. We can put a minus in the answer space, like this: -10 × 5 × -11 × -20 = - _

Step 3. Multiply the numbers from 1 to 10 using the multiplication tables

The product of two numbers less than or equal to 10 is included in the basic times tables (see above). For these simple cases, just write the answer. Remember that, in problems with multiplication only, you can move the integers as you like to multiply the simple numbers together.

• In our example, 10 × 5 is included in the multiplication tables. We do not have to take into account the minus sign on 10 because we have already found the sign of the answer. 10 × 5 = 50. We can insert this result into the problem like this: (50) × -11 × -20 = - _

  If you're having trouble visualizing basic multiplication problems, think of them as addition. For example, 5 × 10 is like saying "10 times 5". In other words, 5 × 10 = 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5

Step 4. If necessary, break larger numbers into simpler pieces

If your multiplication involves numbers larger than 10, you don't have to use long multiplication. First, see if you can break one or more numbers into more manageable chunks. Since, with multiplication tables, you can solve simple multiplication problems almost immediately, reducing a difficult problem into many easy problems is usually simpler than solving the single complex problem.

Let's move on to the second part of the example, -11 × -20. We can omit the signs because we have already obtained the sign of the answer. 11 × 20 seems complicated, but rewriting the problem as 10 × 20 + 1 × 20, it is suddenly much more manageable. 10 × 20 is only 2 times 10 × 10, or 200. 1 × 20 is only 20. Adding the results, we get 200 + 20 = 220. We can put it back into the problem like this: (50) × (220) = - _

Step 5. For more complex numbers, use long multiplication

If your problem includes two or more numbers greater than 10 and you can't find the answer by breaking the problem down into more feasible parts, you can still solve by long multiplication. In this type of multiplication, you line up your answers as you would in addition and multiply each digit in the bottom number with each digit of the top one. If the lower number has more than one digit, you need to account for the digits in the tens, hundreds, and so on by adding zeros to the right of your answer. Finally, to get the final answer, add up all the partial answers.

• Let's go back to our example. Now, we need to multiply 50 by 220. It will be difficult to break down into easier pieces, so let's use long multiplication. Long multiplication problems are easier to handle if the smallest number is at the bottom, so let's write the problem with 220 above and 50 below.

  • First multiply the digit in the lower units by each digit of the upper number. Since 50 is below, 0 is the digit in units. 0 × 0 is 0, 0 × 2 is 0, and 0 × 2 is zero. In other words, 0 × 220 is zero. Write it under the long multiplication in units. This is our first partial answer.
  • Then, we will multiply the digit in the tens of the lower number by each digit of the higher number. 5 is the tens digit in 50. Since this 5 is in the tens instead of the units, we write a 0 below our first partial answer in the units before moving on. Then, we multiply. 5 × 0 is 0. 5 × 2 to 10, so write 0 and add 1 to the product of 5 and the next digit. 5 × 2 is 10. Usually, we would write 0 and report 1, but in this case we also add 1 from the previous problem, obtaining 11. Write "1". Returning the 1 from the tens of 11, we see that we have no more digits, so we simply write it on the left of our partial answer. By recording all this, we have 11,000 left.
  • Now, let's just add up. 0 + 11000 is 10000. Since we know that the answer to our original problem is negative, we can safely establish that -10 × 5 × -11 × -20 = - 11000.

  Method 2 of 2: Divide the whole numbers

  

  Step 1. As before, determine the sign of your answer based on the number of minus signs in the problem

  Introducing division into a mathematical problem does not change the rules regarding negative signs. If there is an odd number of negative signs, the answer is negative, if it is even (or null) the answer will be positive.

  Let's use an example involving both multiplication and division. In the problem -15 × 4 ÷ 2 × -9 ÷ -10, there are three minus signs, so the answer will be negative. As before, we can put a minus sign in place of our answer, like this: -15 × 4 ÷ 2 × -9 ÷ -10 = - _

  

  Step 2. Make simple divisions using your knowledge of multiplication

  Division can be thought of as a backward multiplication. When you divide one number by another, you are wondering "how many times is the second number included in the second?" or, in other words, “what do I have to multiply the second number by to get the first?”. See the basic 10x10 times tables for reference - if you are asked to divide one of the answers in the times tables by any number from 1 to 10, you know the answer is simply the other number from 1 to 10 that you need to multiply n to get it.

  • Let's take our example. In -15 × 4 ÷ 2 × -9 ÷ -10, we find 4 ÷ 2. 4 is an answer in the multiplication tables - both 4 × 1 and 2 × 2 give 4 as the answer. Since we are asked to divide 4 by 2, we know that we are basically solving the problem 2 × _ = 4. In space, of course, we will write 2, so that 4 ÷ 2 =

    Step 2.. We rewrite our problem as -15 × (2) × -9 ÷ -10.

  

  Step 3. Use long parting where needed

  As with multiplication, when you come across a division that is too difficult to solve mentally or with the multiplication tables, you have the opportunity to solve it with a long approach. In a long division, write the two numbers in a special L-shaped bracket, then divide digit by digit, shifting the partial answers to the right as you go to account for the decreasing value of the digits you're dividing - hundreds, then tens., then units and so on.

  • We use the long division in our example. We can simplify -15 × (2) × -9 ÷ -10 into 270 ÷ -10. We will ignore the signs as usual because we know the final sign. Write 10 on the left and place 270 below it.

    • Let's start by dividing the first digit of the number below the parenthesis by the number on the side. The first digit is 2 and the number on the side is 10. Since 10 is not included in the 2, we will use the first two digits instead. The 10 goes into the 27 - twice. Write "2" above the 7 below the parenthesis. 2 is the first digit in your answer.
    • Now, multiply the number to the left of the parenthesis by the newly discovered digit. 2 × 10 is 20. Write it under the first two digits of the number under the parenthesis - in this case, 2 and 7.
    • Subtract the numbers you just wrote. 27 minus 20 is 7. Write it under the problem.
    • Move to the next digit of the number below the parenthesis. The next digit in 270 is 0. Return it to the side of 7 to get 70.

    • Divide the new number. Then divide 10 by 70. 10 is included exactly 7 times in 70, so write it above next to 2. This is the second digit of the answer. The final answer is

      Step 27..

    • Note that in the event that 10 was not perfectly divisible into the final number, we would have had to take into account the advanced 10 odds - the remainder. For example, if our last task were to divide 71, instead of 70, by 10, we would notice that 10 is not perfectly included in 71. It fits 7 times, but one unit is left over (1). In other words, we can include seven 10s and a 1 in 71. We would then write our answer as "27 with remainder of 1" or "27 r1".

    Advice

    • In multiplication, the order of the factors can be varied, and they can be grouped. So a problem like 15x3x6x2 can be rewritten as 15x2x3x6 or (30) x (18).
    • Remember that a problem like 15x2x0x3x6 will equal 0. You don't have to calculate anything.
    • Pay attention to the order of operations. These rules apply to any group of multiplications and / or divisions, but not to subtraction or addition.