The Daring Book for Girls - Andrea J. Buchanan [94]
Let’s try another example: 152
We know the last part of our answer will be 25. Following the “by one more than the one before” rule, we need to multiply the first numeral in “15” by one more than itself. So that’s 1 (our first numeral) multiplied by 2 (one more than our first numeral, 1), which equals 2. So our answer is 225.
Another example: 1052
We know the last part of our answer will be 25. Following the “by one more than the one before” rule, we need to multiply 10 by 11 (one more than 10), which equals 110. So our answer is 11025.
“All from 9 and the last from 10”
This is an easy rule for subtracting numbers from 100, 1000, 10000, etc.
In the equation “10,000 - 6347,” you can figure the answer by using “all from the 9 and the last from 10”: subtracting each of the digits in 6347 from 9, except the last digit, which you subtract from 10. So that’s 9 minus 6 (which is 3), 9 minus 3 (which is 6), 9 minus 4 (which is 5), and 10 minus 7 (which is 3), which gives you the answer 3653. This rule works when you have one zero for each digit you’re subtracting—no more, no less. Here are some examples in action:
“Vertically and crosswise”
This can be used for multiplying numbers, and also adding and subtracting fractions. Let’s tackle fractions first, adding 6/7/6/7 and 5/3/5/3. The way we have traditionally been taught to compute this can get a bit complicated. But using “vertically and crosswise,” we can do this in our heads.
To get the “top” part of our answer, we multiply 6 by 3 and 7 by 5. That gives us 18 and 35. Add those together to get our final top number, 53. For the bottom number we multiply the two bottom numbers of our equation, 7 and 3. That gives 21, and so our answer is 53/21/53/21.
Let’s try another example: 3/2/3/2 + 5/6/5/6
To get the top number of our answer, multiply 3 × 6 (that gives us 18) and 2 × 5 (that gives us 10), then add those together (28). To get the bottom number, multiply the two bottom numbers of the equation, 2 and 6. That gives us 12. So our answer is 28/12/28/12.
This works the same with subtracting fractions. Let’s use our second example, subtracting instead of adding this time: 3/2/3/2 - 5/6/5/6
To get the top number of our answer, multiply 3 × 6 (that makes 18) and 2 × 5 (that’s 10), then subtract instead of add: 18 - 10 = 8. That’s our top number. Multiply the bottom two numbers of the equation, 2 × 6, and that gives us our bottom number, 12. Our answer is 8/12/8/12 (which can be further reduced to 2/3/2/3).
“Vertically and crosswise” also works with multiplying numbers. If you’ve memorized your timestables, you might know some basic multiplication by rote. But Vedic math offers a creative way to arrive at answers to long multiplication problems that makes multiplying even more fun.
Multiplying 21 × 23 the usual way will get us an answer of 483, but using Vedic math will help us get there faster. Imagine 23 sitting just below 21 and multiply vertically and crosswise, using the following three steps, to arrive at the answer:
Multiply vertically on the right to get the final digit of the answer: In this case, that’s 1 × 3, which equals 3.
Multiply crosswise and then add to get the middle digit of the answer: In this case, that’s 2 × 3 added to 1 × 2, which gives us 8. (If multiplying crosswise and adding gives you 10 or over, you’ll have to carry over the first digit of the number and add that to the answer in step 3.)
Multiply vertically on the left (and then add any carried-over number, if necessary) to get the first digit of the answer: In this case, that’s 2 × 2, which equals 4.
Here’s another example: 61 × 31
Multiply vertically on the right (1 x 1) to get the final digit of the answer (1); multiply crosswise (6 × 1 and 1 × 3) and then add to get the middle digit (9); and multiply vertically on the left (6 × 3) to get the first digit of your answer (18). The result is 1891.
With two-digit numbers that are close to 100, you can use “vertically and crosswise” as follows. Let’s try 88 x 97. Write out the equation, and then subtract both 88