Articles
BY OUR STUDENT CONTRIBUTORS
Malhar Rajpal Most math competitions have stringent time limits and a prohibition of calculator use. Because of this, it is extremely important to learn basic strategies to perform quick mental calculations to solve more problems and thus be more successful in such competitions. Furthermore, performing mental math in front of a live audience is extremely impressive: imagine squaring huge numbers in your head quicker than any calculator can do it! In this article, you will learn the strategy of quickly squaring two digit numbers and the reasoning behind the method.
Squaring a two digit number in your head: Imagine seeing 56x56 on your exam paper, as you’re scramming for time. It would be a huge pain to calculate this using table multiplication and you would save at least half a minute by doing this in your head. How would we do this? The method: If we take any two digit number, say in this case we take 56, we first round it to the nearest 10. So if our number is 56, we round to 60. If our number is 42, we would round to 40. Then you find the difference between the rounded value and the initial value. So 60-56 = 4 or 42-40 = 2 and you perform the “reverse calculation” by that same value: If you initially chose 56, you have to add 4 to get 60, and hence the reverse calculation would be to subtract 4, and this number would be 56-4 = 52. If you had to square 42, you subtract 2 to get 40, and hence the reverse calculation would be to add 2 to get 42+2 = 44. After you get the two numbers, the rounded number and the number you obtained after performing the reverse calculation, you multiply them together. So for squaring 56, you would multiply 60 and 52, which is just 6x10x52 = (6x52)x10 = 3120 and for squaring 42, you would multiple 40 and 44, which is 4x10x44 = (4x44)x(10) = 1760. After you have this number, you take the difference between the rounded number and the initial value that you found before and square it. So for 56, 60 is the number rounded to the nearest 10 and 60-56 = 4. We take 4 and square it to get 16. For 42, 42-40 = 2, so we take 2 and square it to get 4. After we get this squared value, we add it to the product that we calculated(the product after multiplying the rounded number and the reverse calculation number). So for 56, we have 3120+16 = 3136, and for 42, we have 1760+4 = 1764, which are our final answers! Amazing how simple it is, isn’t it! It’s simpler than trying to do it manually because you are essentially multiplying a two digit number with a multiple of 10(which is the same as multiplying a two digit number by a one digit number and then multiplying the product by 10), and then adding a small number to that result to get the final answer. Let’s try one more example: 73. 73 rounded to the nearest 10 is 70. 73-3 = 70, hence the reverse calculation is to add 3 to 73: 73+3 = 76. We then multiply 76 with 70 which is the same as 76x7x10 = 532x10 = 5320. The difference between the rounded number, 70, and initial number to square, 73 is 3, and 3 squared = 9. We, thus, add 9 to our existing product to get: 73x73 = 5320 + 9 = 5329, our final answer! The mathematics and reasoning behind the method: The mathematics behind this method is actually very simple. Let’s say we have a number, a. We want to find a^2. By the method, we are adding or subtracting some value b from a to get the rounded value. We then perform the reverse operation subtracting or adding b from a. Finally we multiply these two numbers together and this can be represented as: (a+b)(a-b) where the first number is the rounded up value, and the second number is the number after performing the reverse operation on a. We could also have (a-b)(a+b)if the rounded number is rounded down instead of up, however, since xy =yx, (a+b)(a-b)=(a-b)(a+b) hence both expressions are equivalent. After we evaluate the product (a-b)(a+b), we add the value of b^2 to the product to obtain our final answer. We can thus hypothesise that a^2=(a-b)(a+b) + b^2 and now we have to prove this. The proof is direct and just requires an expansion of the right side: a^2=a^2-b^2+b2 and the b^2s cancel out leaving us with a^2=a^2, an equivalent statement so we know that a^2=(a-b)(a+b)+b^2 is true and our methodology therefore works for any real a and b!
0 Comments
Leave a Reply. |
Our AuthorsWe are high school and college students from around the world who are passionate about maths, and want to share that passion with others. Categories
All
|