= 9025 + 95 = 9120. The first step is an application of the distributive law, the second an application of a rule for the square of numbers ending in 5. To compute 95?95, first drop five to obtain 9. Multiply the latter by one more than the number itself: 9?10 = 90. Next, append 25.

As another example, what is 97?103? Quickly,

97?103= 100·100 - 3·3 = 10000 - 9 = 9991. Why? Because of the general formula (a - b)·(a + b) = a·a - b·b, which comes in handy if we observe that 97 = 100 - 3 and 103 = 100 + 3.

One of the simplest entertaining tricks is this. T

Mental mathematics - under various guises - is a skill of rapid calculations; arithmetic calculations being the most common variety. Those who possess an innate aptitude for mental mathematics are known as human calculators. They may not even know why their methods work, but with a few rare exceptions, those methods have a simple algebraic explanation. Practice and understanding of the basics help the regular folks who were not born the human calculators acquire the necessary skill. But is the skill of mental calculations indeed necessary? Candidly? Nothing is. However, mastering a skill may prove an invaluable asset.

Mental mathematics has many uses that mainly fall into three categories which I call defensive, offensive, and entertaining. The distinction between the three categories may at times be fuzzy, but in essence you may need to compute fast in order to verify a result obtained somehow else and to avoid a mistake, or to get a result fast in the first place, or just to show off and stun and please friends with your mastery of mysterious tricks.

In the way of example, without actually computing say 127?18 select the product from the list of numbers below:

• 2284,
• 22886,
• 2286,
• 2886.

Well, note that 18 is divisible by 9 (and 3) so is the product 127?18. The number is divisible by 3 if the sum of its digits is divisible by 3. This disqualifies the first number 2284 as the sum of its digits is 16 and is not divisible by 3. (Also, the last digit of the product is defined by the last digits if the multiplicands, which in this case are 7 and 8. So that last digit of the product is bound to be 6.) The second number 22886 is way too big. To see that, replace 127 and 18 with bigger numbers, e.g. 200 and 20. Easily, 200?20 = 4000 which is bigger than the product 127?18 but far off the oversized candidate 22886. The third number 2286 does not fail in an obvious way, but the fourth one, 2886, does. Similarly, to the second possible answer, this one is also too big, but not as much. If we replace 127 with 130 and 18 with 20, the product will grow. But 130?20 is 2600 still less than 2886 and so is the sought product. Thus the third number is the only plausible choice. Can you compute fast 95?96? Why, there are many ways to do that. For example,

95?96= 95?95 + 95 = 9025 + 95 = 9120. The first step is an application of the distributive law, the second an application of a rule for the square of numbers ending in 5. To compute 95?95, first drop five to obtain 9. Multiply the latter by one more than the number itself: 9?10 = 90. Next, append 25.

As another example, what is 97?103? Quickly,

97?103= 100·100 - 3·3 = 10000 - 9 = 9991. Why? Because of the general formula (a - b)·(a + b) = a·a - b·b, which comes in handy if we observe that 97 = 100 - 3 and 103 = 100 + 3.

One of the simplest entertaining tricks is this. Ta

kill may prove an invaluable asset.

Mental mathematics has many uses that mainly fall into three categories which I call defensive, offensive, and entertaining. The distinction between the three categories may at times be fuzzy, but in essence you may need to compute fast in order to verify a result obtained somehow else and to avoid a mistake, or to get a result fast in the first place, or just to show off and stun and please friends with your mastery of mysterious tricks.

In the way of example, without actually computing say 127?18 select the product from the list of numbers below:

• 2284,
• 22886,
• 2286,
• 2886.

Well, note that 18 is divisible by 9 (and 3) so is the product 127?18. The number is divisible by 3 if the sum of its digits is divisible by 3. This disqualifies the first number 2284 as the sum of its digits is 16 and is not divisible by 3. (Also, the last digit of the product is defined by the last digits if the multiplicands, which in this case are 7 and 8. So that last digit of the product is bound to be 6.) The second number 22886 is way too big. To see that, replace 127 and 18 with bigger numbers, e.g. 200 and 20. Easily, 200?20 = 4000 which is bigger than the product 127?18 but far off the oversized candidate 22886. The third number 2286 does not fail in an obvious way, but the fourth one, 2886, does. Similarly, to the second possible answer, this one is also too big, but not as much. If we replace 127 with 130 and 18 with 20, the product will grow. But 130?20 is 2600 still less than 2886 and so is the sought product. Thus the third number is the only plausible choice. Can you compute fast 95?96? Why, there are many ways to do that. For example,

95?96= 95?95 + 95 = 9025 + 95 = 9120. The first step is an application of the distributive law, the second an application of a rule for the square of numbers ending in 5. To compute 95?95, first drop five to obtain 9. Multiply the latter by one more than the number itself: 9?10 = 90. Next, append 25.

As another example, what is 97?103? Quickly,

97?103= 100·100 - 3·3 = 10000 - 9 = 9991. Why? Because of the general formula (a - b)·(a + b) = a·a - b·b, which comes in handy if we observe that 97 = 100 - 3 and 103 = 100 + 3.

One of the simplest entertaining tricks is this. T

om the list of numbers below:

• 2284,
• 22886,
• 2286,
• 2886.

Well, note that 18 is divisible by 9 (and 3) so is the product 127?18. The number is divisible by 3 if the sum of its digits is divisible by 3. This disqualifies the first number 2284 as the sum of its digits is 16 and is not divisible by 3. (Also, the last digit of the product is defined by the last digits if the multiplicands, which in this case are 7 and 8. So that last digit of the product is bound to be 6.) The second number 22886 is way too big. To see that, replace 127 and 18 with bigger numbers, e.g. 200 and 20. Easily, 200?20 = 4000 which is bigger than the product 127?18 but far off the oversized candidate 22886. The third number 2286 does not fail in an obvious way, but the fourth one, 2886, does. Similarly, to the second possible answer, this one is also too big, but not as much. If we replace 127 with 130 and 18 with 20, the product will grow. But 130?20 is 2600 still less than 2886 and so is the sought product. Thus the third number is the only plausible choice. Can you compute fast 95?96? Why, there are many ways to do that. For example,

95?96= 95?95 + 95 = 9025 + 95 = 9120. The first step is an application of the distributive law, the second an application of a rule for the square of numbers ending in 5. To compute 95?95, first drop five to obtain 9. Multiply the latter by one more than the number itself: 9?10 = 90. Next, append 25.

As another example, what is 97?103? Quickly,

97?103= 100·100 - 3·3 = 10000 - 9 = 9991. Why? Because of the general formula (a - b)·(a + b) = a·a - b·b, which comes in handy if we observe that 97 = 100 - 3 and 103 = 100 + 3.

One of the simplest entertaining tricks is this. T

bigger numbers, e.g. 200 and 20. Easily, 200?20 = 4000 which is bigger than the product 127?18 but far off the oversized candidate 22886. The third number 2286 does not fail in an obvious way, but the fourth one, 2886, does. Similarly, to the second possible answer, this one is also too big, but not as much. If we replace 127 with 130 and 18 with 20, the product will grow. But 130?20 is 2600 still less than 2886 and so is the sought product. Thus the third number is the only plausible choice. Can you compute fast 95?96? Why, there are many ways to do that. For example,

95?96= 95?95 + 95 = 9025 + 95 = 9120. The first step is an application of the distributive law, the second an application of a rule for the square of numbers ending in 5. To compute 95?95, first drop five to obtain 9. Multiply the latter by one more than the number itself: 9?10 = 90. Next, append 25.

As another example, what is 97?103? Quickly,

97?103= 100·100 - 3·3 = 10000 - 9 = 9991. Why? Because of the general formula (a - b)·(a + b) = a·a - b·b, which comes in handy if we observe that 97 = 100 - 3 and 103 = 100 + 3.

One of the simplest entertaining tricks is this. T

= 9025 + 95 = 9120. The first step is an application of the distributive law, the second an application of a rule for the square of numbers ending in 5. To compute 95?95, first drop five to obtain 9. Multiply the latter by one more than the number itself: 9?10 = 90. Next, append 25.

As another example, what is 97?103? Quickly,

97?103= 100·100 - 3·3 = 10000 - 9 = 9991. Why? Because of the general formula (a - b)·(a + b) = a·a - b·b, which comes in handy if we observe that 97 = 100 - 3 and 103 = 100 + 3.

One of the simplest entertaining tricks is this. Take a three digits number and subtract from it the sum of its digits. (You may need to sum up the digits one more time until you get a 1-digit number.) Tell me any two digits of the result. I'll respond with the remaining digit. For example, assume you chose number 573 whose digits ad up to 15. This is a two digit number. So we replace it with the sum of the digits, which is 6. Subtract now 6 from 573. 573 - 6 = 567. Note that the three digits of the result (5, 6, 7) add up to 18, a multiple of 9. This is always the case. Now, if you announce the first two digits of the result, 5 and 6, I compute their sum, which is 11, and find the first multiple of 9 exceeding that number. The first multiple of 9 greater than 11 is 18. Now the difference 18 - 11 = 7 gives me the last digit of your result.

Try this trick with your friends. And if you like to learn more, have a look at my page Fast Arithmetic Tips.