Addition Is Fun
Addition is one of the most common mathematical operations, which even a layman uses frequently. Whether you go for shopping or purchasing your grocery, books, or other daily used commodities you need to calculate your expenditure and here the addition technique comes handy.
Have you ever visited a bank? I’m sure you have. Now my question is how does the cashier add the cashbook account so fast? The answer is very simple, he uses some tricks or adopts some methods which comes with practice and develops a method of his own to calculate fast.
You must have noticed the greengrocer mentally adding the cost of objects purchased. A more astonishing fact, which you will not believe is that some of them haven’t been to school but are equally efficient at calculations.
In my book, The Essentials of Vedic Mathematics, I have focused on the Vedic technique and discussed the various methods which are better than the conventional method taught in schools, but here I shall focus on some of the unconventional methods and tricks which will help you calculate faster than the method you use and many a times you will see that you can compete with a calculator.
Addition of 5 Consecutive Numbers
Addition of 5 consecutive numbers can be done by various
methods but the simplest one is to multiply the highest number
by 5 and subtract 10 from the result obtained.
Example:- 13 + 14+ 15 + 16 + 17 =7
Solution:- Here the highest number is 17 Multiply it by 5: 17 x 5 = 85
Subtract 10: 85 – 10 = 75
Example:- 66+ 67+ 68 + 69 + 70 =?
Solution:- Here the highest number is 70
Multiply it by 5: 70 x 5 = 350
Subtract 10: 350 – 10 = 340
Now the question you may ask: If there are more consecutive numbers then what strategy shall you use to find the total?
I am giving here some points and expect you to generalize
a rule to add consecutive numbers on your own.
If there are 4 consecutive numbers; multiply the highest by 4 and subtract 6 (11⁄2 × 4)
If there are 5 consecutive numbers; multiply the highest by 5 and subtract 10 (2 × 5)
If there are 6 consecutive numbers; multiply the highest by 6 and subtract 15 (21⁄2 x 6)
If there are 7 consecutive numbers; multiply the highest by 7 and subtract 21 (3 × 7)
If there are 8 consecutive numbers; multiply the highest by 8 and subtract 28 (31⁄2 x 8)
You can add consecutive numbers more smartly if you understand
the two cases.
Case 1:- If there are odd numbers of consecutive terms. Rule:- Multiply the middle term by the number of terms.
Example:- 12 + 13 +14 + 15 + 16 + 17 + 18+ 19 + 20
Solution:- Since there are 9 terms so the middle term = 16
Multiply it by 9= 16 x 9
144
case 2:- If there are even numbers of consecutive terms. Rule:- Take the mean of the two middle terms and multiply it with the number of terms.
Example:- 23 + 24+ 25 + 26 +27 + 28 + 29 + 30 ?
Solution:- You can see that there are 8 consecutive numbers to be added. The two middle numbers are obviously 26 and 27 Take the mean of the two = 26.5
Multiply it with the number of terms = 26.5 x 8 = 212
I am hopeful that you would have generalized the rule to add the consecutive numbers. Leave this apart and learn some interesting methods to add fast.
Addition of Two Numbers in Case One of the Numbers Ends With Repeated 9s or 8s
You can add two numbers in your mind without doing much effort from your side if you see one of them has repeated 9s or 8s.
Example:- a) 98765 + 999 = ?
- b) 78976 + 9998 = ?
Rule:- Round off one of the number into multiple of 10s and do the balancing act.
Solution:-
= 98765 + 999
-1 | |+1
= 98764 1000
= 98764 + 1000 = 99764
Example:- 78976 + 9998 = ?
Solution:
78976 + 9998
-2 | | + 2
78974 10000
= 78974 + 10000 = 88974
Addition of Similar Digits Repeated
Addition of similar digits with digits repeated in ascending order sometimes creates problem as you have to arrange them according to their place value before adding. Let me simplify this problem to make such addition a child’s play.
Example:- 5 + 55 + 555 =?
Rule:-
- a) Take the digit repeated and replace all the repeated digits with 1
- b) Count number of digits in each number.
- c) Multiply the result 123… by the repeated digit.
Solution:- a) Take 5 common and replace all 5 with 1
5 (1 + 11+ 111)
- b) Count number of digits in each number. Here the digit 1 has been repeated 1, 2 and 3 times.
-
- a) 45 +46 +47 +48 +49 + 50 = ?
- b) 78 + 79 +80+81 + 82 = ?
- c) 2.2 + 2.22 + 2.222 + 2.2222 + 2.22222 = ?
- d) 0.6 + 0.66 +0.666 +0.6666 = ?c) Multiply 123 by 5 9 (1 + 11 + 111 + 1111 + 11111 + 111111) b) Count number of digits in each number. Here the digit 1 has been repeated 1, 2, 3, 4, 5 and 6 times. c) Multiply 123456 by 9Hence, 9 + 99 + 999 + 9999 + 99999 + 999999 = 9 X 123456 = 1111104
Addition of Repeated Digits with Decimals
Addition of repeated digits after the decimal can be done easily in the same manner as discussed above with the only difference that here the counting of numbers will be in the reverse order. Example:- 0.9 +0.99 +0.999 + 0.9999 =?
Solution:- Here the repeated digits are in ascending order. The number of 9s after decimal parts are 1, 2, 3 and 4. Reverse the order of repetition and multiply it by the repeated digit 9 Hence, 0.9 +0.99 + 0.999 + 0.9999 = 9 x 0.4321 = 3.8889
Example:- 0.7 +0.77 +0.777 +0.7777 + 0.77777 +0.777777
=?
Solution:- Here the repeated digits are in ascending order. The number of 7s after decimal parts are 1, 2, 3, 4, 5 and 6. Reverse the order of repetition and multiply it by the repeated digit 7
Hence, 0.7 + 0.77 +0.777 +0.7777 +0.77777 +0.777777 = 7 × 0.654321 = = 4.580247
Example:-0.4 +0.44 + 0.444 + 0.4444 + 0.44444 + 0.444444 +0.4444444 + 0.44444444
?
Solution:- Here the repeated digits are in ascending order. The number of 4s after decimal parts are 1, 2, 3, 4, 5, 6, 7 and 8. Reverse the order of repetition and multiply it by the repeated digit 4
Hence,
0.4 + 0.44 + 0.444 + 0.4444 + 0.44444 + 0.444444
+0.4444444 + 0.44444444 = 4 x 0.87654321 =
3.50617284
Addition of Repeated Digits in the Whole and Decimal Parts
Together
So far we have seen two cases. In the first one we only dealt with whole numbers and in the second case we dealt with decimal fractions. In the third case we will deal with both the cases
together. Let’s learn the rule.
Rule:-
- a) Calculate the decimal part and do as done in case
1
- b) Calculate the whole part.
- c) Add both the results and you will get the answer.
Example:- 3.3 + 3.33 +3.333 + 3.3333 = ?
Solution:- a) In the second case we learned how to deal with numbers with only decimal parts.
0.3 + 0.33 + 0.333 +0. 3333 = 3 x 0.4321
= 1.2963
- b) Here the number 3 in the whole part is repeated
4 times; i.e. 3 x 4 = 12
- c) Add both the parts
Hence, 3.3 +3.33 +3.333 +3.3333 = 12 + 1.2963 = 13.2963. I hope you have enjoyed the magic of addition a lot. Let’s practise this before I move on to the next method for addition
addition.
- a) 45 +46 +47 +48 +49 + 50 = ?
- b) 78 + 79 +80+81 + 82 = ?
- c) 2.2 + 2.22 + 2.222 + 2.2222 + 2.22222 = ?
- d) 0.6 + 0.66 +0.666 +0.6666 = ?
Faster Addition in the Group of 10
Have you ever tried to discover the reason behind your mistake in adding large numbers? If I am not wrong then it is the carry- over problem. Adding larger numbers in a length takes a lot of time as we are not accustomed to adding larger numbers.
5+3+2 + 8 = 18 is right but 9 + 8 + 7+9+ 6+ 8 +8 +5 + 9 + 7 + 4 + 9 = ? is difficult to answer. So what I have done here is to break the sum in different parts for convenience. In simple words, whenever the sum of digits exceeds 10, we put a dot over the digit in the preceding column and take the unit digit to add to the next number and the process keeps continuing. It is important to understand here that the dot marks on any digit placed has the value 1 more than itself.
Example:- 329736
465728
0623999
+554321
1973784
Explanation: Addition for each column will be done as you generally do in the normal addition. For the convenience of readers the arrow has been shown moving downwards.
In the first column:
- 6+ 8 = 14 > 10 so place a dot over 2, the number preceding 8. Take the excess 4 (14-10 = 4) to the next number of 1st column.
4+9=13 > 10 so place a dot over 9, number preceding 9 of the first column and take the excess 3 (13-10 =
3) to the next number of the 1″ column.
3+1 = 4 since there is no more digits left in the first column so final sum 4 should be written in the bottom
line.
In the second column:
.
3 + 2 (=3) + 9 (=10)
= 16 > 10 so place a dot over 9 in the third column and take the excess 6 (16 – 10 = 6) to the next number of the second column.
6+2= 8. Since there is no more digits left in the
second column so the final sum of the second column is written below the line.
The sum of the rest of the columns will be done accordingly. Now you may ask a question:- What if the sum of digits is more than 10 and there is no column left?
The answer is very simple-put a zero behind the number in the preceding column.
In the sixth column 3 + 4 + 6 = 14 >10, so a dot needs to be put to the preceding number in the seventh column hence a zero (0) is placed in the seventh column.
Let us take another example to understand the whole process of addition
Example: 124 + 4234 + 8238+ 646 + 5321 + 3510 + 8989 =
Explanation:
In the first column:
- 4 + 4 + 8 = 16. It is more than 10, so a dot is placed next to 8 i.e. on 3. Here dot on 3 makes it one richer by itself.
3=3+1 = 4
Take the excess 6 (16-10 = 6) and move to the next digit. The sum of 6 + 6 = 12 > 10. So place a dot on 4, next to 6 in the preceding column.
Again 2(excess of the previous stage) + 1 + 0 + 9 = 12 so placing a dot on 8 write the final 2 at the end of first column.
In the second column:
2 + 3 + 3(= 4) + 4 (= 5) = 14 is more than 10, so dot is placed to the preceding number 6 in the third column. = 4 and add the Proceed with the excess 14-10 remaining digits in the second column.
4 +2 + 1 + 8 (=9)
= 16
Write the excess 6 (16-10) below the second column and move to the third column likewise.
This process will be continued till every column is taken into account. Once you practise the example shown in the first method, it will be easier for you to understand the second example and do some fast calculation.
Now this is the time for practising some problems.
Hope you have enjoyed this method too. Before I wind up this chapter let me show another interesting method to you. This method simply works on manipulation of numbers and breaking
them.
Let’s see the breaking up of numbers in terms of 10s to add conveniently.
Example:- 26 + 59 + 394 +66 +11 + 14 = ?
Solution:-
Step 1:- Here, the complete observation shows us that 26 + 14 are likely to yield a rounded result. The same would be the case if 59 and 11 were paired. Moreover, the pair 394 and 66 yield a rounded result.
26 + 59 + 394 +66 + 11+ 14
Step 2: – Rearrange the pair and add as per the pairing done above.
Addition by Double Column Method
In the double column method we break up the numbers to be added in columns, taking two numbers in each columns.
Suppose you have to add 2345 + 5768 + 8764 + 9127, we will break these numbers into two parts and keep adding two numbers by breaking in the multiples of ten so that we can do
addition conveniently.
Now you can invent your own rule to club the numbers according to your choice in any order by considering the fact that you have to make multiples of 10 to add smoothly. The above examples will help you invent your own rule to add numbers.
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