# Short Tricks for Multiplication – Part 1

Mathematics possesses not only truth but supreme beauty—a beauty cold and austere, like that of a sculpture, and capable of stern perfection, such as only great art can show.

Bertrand RUSSELL

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Multiplication tricks are vital to calculate long and difficult multiplication problems quickly. These tricks are best applied when we have large digit numbers, but they are equally fit for small digit numbers. In this blog post, we will discuss math tricks for fast multiplication.

Few Important Terms

The multiplicand is the number that is being multiplied. The multiplier is the number that is multiplying the first number.

For example, in 55 × 30, 55 is the multiplicand and 30 is the multiplier.

Bases are the numbers starting with 1 and followed by any number of 0’s. For example, 10, 100, 1000, 10000, …

A base number must start with 1 and should be followed only by 0’s. They are the first number for those many digits. Like, 10 for two-digit numbers, 100 for three-digit numbers, and so on.

The complement of a number is the number obtained from subtracting the number to its nearest base. For example,

• complement of 43 is 100 – 43 = 57;
• complement of 729 is 1000 – 729 = 271.

# Trick 1

While multiplying two numbers, let

■ the sum of the unit digits is 10;

■ the number formed by the remaining digits are same.

Popularly known as Last Totalling to Ten.

From the first condition, it follows that the pair of unit digits can be (1, 9), (2, 8), (3, 7), (4, 6), (5, 5) or vice versa. If we add 1 to a whole number, we get its successor.

Following are the steps:

Step 1. Multiply the unit digits of the numbers.

Step 2. Next, multiply the number obtained by the remaining digits with the next higher number (or the successor);

Step 3. Finally, append the result obtained in Step 1 to the right of the result obtained in Step 2.

Note 1. The number of digits in Step 1 must be double the number of 0’s in the base. If required, put extra 0’s to the left.

Solution. Here, sum of the unit digits is 10 (= 2 + 8). Ten’s digits are the same (= 7).

• Step 1.        Multiply the unit digits: 2 × 8 = 16.
• Step 2.        Next higher number (or the successor) of 7 is 8 and 7 × 8 = 56.
• Step 3:        Append 16 to the right side of 56. Hence, it becomes 5616.

Therefore, 72 × 78 is 5616.

Solution. Here, sum of the unit digits is 10 (= 9 + 1). Number obtained by the remaining digits (= 4) are same.

• Step 1.        Multiply the unit digits: 9 × 1 = 9 = 09. [see Note 1]
• Step 2.        Next higher number of 4 is 5 and 4 × 5 = 20.
• Step 3:        Append 09 to the right side of 20. Hence, it becomes 2009.

Therefore, 49 × 41 is 2009.

Example 3. Multiply 213 by 217.

Solution. Here, sum of the unit digits is 10 (= 3 + 7). Number obtained by the remaining digits (= 21) are same.

• Step 1.        Multiply the unit digits: 3 × 7 = 21.
• Step 2.        Next higher number of 21 is 22 and 21 × 22 = 462.
• Step 3:        Append 21 to the right side of 462. Hence, it becomes 46221.

Therefore, 213 × 217 is 46221.

Note 2. The method can also be extended in such cases where groups in Step 1 in both multiplicand and multiplier adds up to 100, 1000, . . . and the number obtained by the remaining digits is same.

Solution. Here, 6 and 4 on unit’s place add up to 10, but 19 and 10 are different. But, 96 + 04 = 100 and 1 on hundred’s place is same [see Note 2].

• Step 1.        Here, 96 × 04 = 384 = 0384 [see Note 1].
• Step 2.        Next higher number of 1 is 2 and 1 × 2 = 2.
• Step 3:        Append 0384 to the right side of 2. Hence, it becomes 20384.

Therefore, 196 × 104 is 20384.

Solution. Here, 200 + 800 = 1000 and 20 (the remaining number) is same in both numbers [see Note 2].

• Step 1.        Here, 200 × 800 = 160000 384 = 0384 [see Note 1].
• Step 2.        Next higher number of 20 is 21 and 20 × 21 = 420.
• Step 3:        Append 160000 to the right side of 420. Hence, it becomes 420160000.

Therefore, 196 × 104 is 420160000.

• 1. 25 × 35
• 2. 101 × 109
• 3. 2088 × 2012
• 4. 310040 × 310060

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