Simple and Interesting Math

Mathematics is a fascinating subject that encompasses a wide range of captivating calculations. Throughout my high school studies, I stumbled upon some specific calculations that I find both simple and intriguing. Consequently, I made the decision to compile these captivating mathematical findings and incorporate them into my personal collection. This way, I will be equipped to teach them to my children in the future 🙂

(A) Quickly determine square of an integer, using formula a²– b² = (a+b)(a-b)

For example, we can find 99×99 = 9801 quickly without actual multiplication.

Firstly, let’s rearrange the formula as

a² = (a+b)(a-b) + b²

Then come to the actual calculation. For example, to calculate 99*99 , we can express it as (99+1)*(99-1)+1² =100*98 + 1 = 9801. The key point is to modify one of the numbers to 10, 100, or 1000, etc., enabling us to easily determine the product.

Let’s try some calculations:

92*92 = ? Add 8 to 92 become 100. Then the other number is 92-8 =84. The square of 8 is 64. Thus, the answer is 100*84+64 = 8464.
988*988 = ? Add 12 to 988 become 1000. Then the other number is 988-12 = 976. The square of 12 = 144. Thus, the answer is 1000*976+144 = 976144.
213*213 = ? Subtract 13 from 213 become 200. Then the other number is 213+13 = 226. The square of 13 =169. Thus, the answer is 226*200+169 = 45369.
499*499 =? Add 1 to 499 become 500. Then the other number is 499-1 = 498. Calculate 500*498 using 498000 / 2 and can tell 249000 at once. Thus, the answer is 249000+1 = 249001.

(B) Quickly determine square of an integer ending in 5

For instance, we can effortlessly determine that 125 multiplied by 125 equals 15625 without actually performing the multiplication.

To begin, any integer ending in 5 can be represented as a*10 + 5, where ‘a’ is a positive integer. In the case of 125, ‘a’ is 12. The square of a*10 + 5 can be expressed as:

(a*10+5)(a*10+5) = a*a*10*10 + 2*5*a*10 + 25 = 100*a*(a+1) + 25

By observing the above expression, we notice that 25 fills the last two digits while a*(a+1) occupies the remaining digits. In the case of 125 * 125, a*(a+1) = 12 * 13 = 156. So you can give the answer 15625 at once without any calculation.

Let’s explore a couple of calculations:

65*65 = ? Given ‘a’ as 6, ‘a*(a+1)’ equals 6*7 = 42. Thus, the answer is 4225.
205*205 = ? With ‘a’ being 20, ‘a*(a+1)’ equals 20*21 = 420. Therefore, the answer is 42025.

(C) Quickly determine the product of two integers when only their unit digits differ and the sum of unit digits equal 10

Based on the information discussed in part B above, we can discover additional methods for performing rapid mental calculations. For example, we can easily determine that the result of multiplying 56 by 54 is 3024 without actually performing the multiplication. These two numbers only differ in unit digit, 6 and 4. The sum of unit digits 6 + 4 is 10.

To demonstrate, let’s consider two digits represented by (a*10 + b) and (a*10 + c), where the sum of b and c is 10. In the case of 56 and 54, ‘a’ is 5, ‘b’ is 6 and ‘c’ is 4.

(a*10+b) (a*10+c) = 100*a² +10*a*(b+c)+b*c = 100*a² +100*a+b*c = 100*a*(a+1) + b*c

Upon observing the above expression, we can find that b*c fills the last two digits and a*(a+1) occupies the digits other than the last two. In the case of 56 * 54, a*(a+1) = 5 * 6 = 30 and (b*c) = 6 * 4 = 24. So you can tell the answer 3024 very fast. No calculation at all.

Let’s try a few more calculations:

21*29 = ? Given ‘a’ as 2, ‘a*(a+1)’ equals 2*3 = 6. The product of the unit digits is 1*9 = 9. Thus, the answer is 609.
112*118 = ? Given ‘a’ as 11, ‘a*(a+1)’ equals 11*12 = 132. The product of the unit digits is 2*8 = 16. Thus, the answer is 13216.

(D) Quickly determine the product of any two integers just smaller than 100, 1000, 10000…

For example, we can find 88 * 95 = 8360 quickly without actual multiplication.

Let’s us define two integers as s and w. Then s’ = 100-s and w’ = 100-w. In the case of 88 * 95, s=88, s’ =12, w=95, w’=5

s*w=(100-s')*(100-w')=(100-s')*100-100*w'+s'*w'=s*100-w'*100+s'*w'=100*(s-w') + s'*w'

By observing the above expression, we notice that s’*w’ fills the last two digits while (s-w’) occupies the remaining digits. In the case of 88 * 95, s’*w’ =12 * 5 = 60 and (s-w’) = 88-5=83. So the answer is 8360.

Let’s try some calculations:

95 * 98 =? Given s=95, s’=5, w’=2, s’*w’ = 5 * 2 =10, (s-w’) = 95 – 2 = 93. Thus the answer is 9310.
888 * 995 = ? Given s= 888, s’=112, w’=5, s’*w’ = 112 * 5 =560, (s-w’) = 888 – 5 = 883. Thus the answer is 883560.

(E) Quickly determine the product of any two integers just larger than 100, 1000, 10000…

For example, we can find 107 * 111 = 11877 quickly without actual multiplication.

It uses the same theory as part (D). Let’s us define two integers as s and w. Then s’ = 100+s and w’ = 100+w. In the case of 107 *111, s=107, s’ =7, w=111, w’=11

s*w=(100+s')*(100+w')=(100+s')*100+100*w'+s'*w'=s*100+w'*100+s'*w'=100*(s+w') + s'*w'

By observing the above expression, we notice that s’*w’ fills the last two digits while (s+w’) occupies the remaining digits. In the case of 107 *111, s’*w’ =7 * 11 = 77 and (s+w’) = 107+11=118. So the answer is 11877.

let’s try one more example:

112 *108 = ? Given s=112, s’=12, w’=8, s’*w’ = 12* 8 =96, (s+w’) = 112 + 8 = 120. Thus the answer is 12096.

(F) Quickly determine the product of any two integers close to 100,1000, 10000,

When multiplying two integers, where one is less than 100 (denoted as s) and the other is more than 100 (denoted as w), we obtain

s*w=(100-s')*(100+w')=(100-s')*100+100*w'-s'*w'=s*100+w'*100-s'*w'=100*(s+w') - s'*w'

Conversely, if s exceeds 100 and w falls below 100, we encounter

s*w=(100+s')*(100-w')=(100+s')*100-100*w'-s'*w'=s*100-w'*100-s'*w'=100*(s-w') - s'*w'

By combining the formulas from Part D and Part E, we can deduce the following equation:

s*w = 100*(s + w') + s'*w' when w is larger than 100 
s*w = 100*(s - w') + s'*w' when w is smaller than 100, 
   where the value of s'*w' is positive when both integers are greater or smaller than 100, and 
   it is negative when one integer is greater than 100 and the other is smaller than 100.

Let’s try some calculations using the ultimate formula

95*98 = 100 * (95-2) +(5*2) = 9310
112 *108 = 100 * (112+8) + (12*8) = 12096
112*93 = 100 * (112-7) – (12 * 7) = 10416
93*108 = 100 * (93 + 8) – (7 * 8) = 10044 or flip the order of two numbers to 108 * 93
108*93 = 100 * (108 – 7) – (7 * 8) = 10044

Try one example from part A

988*988 = 1000 * (988-12) + (12 * 12) = 976144

The calculation follows a similar approach as Part A, although the methodology behind it differs slightly.

Try one example with integer ending in 5

95*95 = 100 * (95-5) + (5*5) = 9025

Of course you can use the method in Part B to calculate. Given ‘a’ as 9, ‘a*(a+1)’ equals 9*10 = 90. Thus, the answer is 9025.

(G) Quickly determine the product of any two integers that are approximately in the same tens, same hundreds, or similar magnitude.

Applying the same approach as in part (F), we can derive the following equation through further deduction:

s*w = i*10*(s + w') + s'*w' when w is larger than i*10 
s*w = i*10*(s - w') + s'*w' when w is smaller than i*10, 
   where the value of s'*w' is positive when both integers are greater or smaller than i*10, and 
   it is negative when one integer is greater than 100 and the other is smaller than i*10.

Let’s attempt a few questions. With practice, you’ll discover that mental calculations are quicker than relying on a calculator.

49 *45 = ? Let me do that in this way: s’ = 1, w’ = 5, 49 *45 = (49-5) * 50 +1*5 = 4400/2 + 5 = 2205
198 *192 = ? Let me do that in this way: s’ = 2, w’ =8, 198 *192 = (198-8) * 200 + 2*8 = 38016
198 * 202 = ? Let me do that in this way: s’ = 2, w’ =2, 198 *192 = (198+2) * 200 – 2*2 = 39996