Sohel Sahoo : Hello Guys,
In this article, we will study the digitsum method. This method is not used for quick calculation but for quick checking of answers. It will help us to verify the answer that we have obtained to a particular question. This technique has wonderful different applications for students giving
competitive exams as they are already provided with four alternatives(options) to every answer.
# For students giving competitive and other exams, this technique has a lot of utility. Many times they can check the digitsum of each of the alternatives with the digitsum of the question and try to arrive at the correct answer. This will eliminate the need for going through the whole calculation.
Note: (The digit sum will always be a single digit. You have to keep on adding the numbers until you get a singledigit answer.)
In digit sum method,while we are trying to make single digit value in the process of adition all the nines and all the digits that add up to nine are completely divisible by 9,hence leaves remainder 0.
#Therefore,While calculating the digitsum of a number, you can eliminate all the nines and all the digits that add up to nine. (Shortcut Method)
When you eliminate all the nines and all the digits that add up to nine you will be able to calculate
the digitsum of any number much faster. The elimination will have no effect on the final result.
Note: 1. If the digitsum of a number is 9, then we can eliminate the 9 straight away and the digit
sum becomes ‘zero.’ (Remember that 9 is synonymous with 0 and can be used interchangeably as 9 is also divisible by 9 leaves remainder 0.[9=0,In digit sum])
However, if the digitsum of the answer does not match with the digit sum of the question then you can be 100% sure that the answer is wrong.
In this article, we will study the digitsum method. This method is not used for quick calculation but for quick checking of answers. It will help us to verify the answer that we have obtained to a particular question. This technique has wonderful different applications for students giving
competitive exams as they are already provided with four alternatives(options) to every answer.
# For students giving competitive and other exams, this technique has a lot of utility. Many times they can check the digitsum of each of the alternatives with the digitsum of the question and try to arrive at the correct answer. This will eliminate the need for going through the whole calculation.
METHOD:
The digitsum method (also known as the addup method) can be used to check answers involving different arithmetic operations like multiplication, division, addition, subtraction, squares, square roots, cube roots, etc. The procedure involved is very simple. We have to convert any given number into a single digit by repetitively adding up all the digits of that number.Note: (The digit sum will always be a single digit. You have to keep on adding the numbers until you get a singledigit answer.)
Rule behind this !
We know that any no. is divisible by 9 iff the sum of the digits is divisible by 9.
So, in digit sum method we have the single digit value of the total no. ;which represents the remainder obtained on dividing that no. by 9.And any no. divisible by 9 leaves remainder 0.In digit sum method,while we are trying to make single digit value in the process of adition all the nines and all the digits that add up to nine are completely divisible by 9,hence leaves remainder 0.
#Therefore,While calculating the digitsum of a number, you can eliminate all the nines and all the digits that add up to nine. (Shortcut Method)
When you eliminate all the nines and all the digits that add up to nine you will be able to calculate
the digitsum of any number much faster. The elimination will have no effect on the final result.
Note: 1. If the digitsum of a number is 9, then we can eliminate the 9 straight away and the digit
sum becomes ‘zero.’ (Remember that 9 is synonymous with 0 and can be used interchangeably as 9 is also divisible by 9 leaves remainder 0.[9=0,In digit sum])
2. While solving problems if the digit sum value comes negative then just add it to 9 & result will be the actual digit sum value . (As remainder is always +ve ,so digit sum is +ve. For details I will explain in remainder theorem article)
Example: Find the digitsum of 2467539.
Ans: The number is 2467539. We add all the digits of that number. 2 + 4 + 6 + 7 + 5 + 3 + 9 = 36
Now, we take the number 36 and add its digits 3 + 6 to get the answer 9. Thus, we have converted the number 2467539 into its digitsum 9/0.(same meaning)
Alter: (shortcut method)
(Q). Find the digitsum of 637281995.
Ans: The digit sum of 6372819923 is: 6+3+7+2+8+1+9+9+2+3 = 50 and again 5+0 is 5.
Alter: (shortcut method)
Now, we will eliminate the numbers that add up to 9 (6 and 3, 7 and 2, 8 and 1 and also eliminate the two 9’s) We are left with the digits 2 and 3 which also add up to 5.
Example: Find the digitsum of 2467539.
Ans: The number is 2467539. We add all the digits of that number. 2 + 4 + 6 + 7 + 5 + 3 + 9 = 36
Now, we take the number 36 and add its digits 3 + 6 to get the answer 9. Thus, we have converted the number 2467539 into its digitsum 9/0.(same meaning)
Alter: (shortcut method)
Now, we will eliminate the numbers that add up to 9 (2 and 7, 4 and 5, 6 and 3 and also eliminate the 9) We obtain 0 or 9 ;same as above.
Ans: The digit sum of 6372819923 is: 6+3+7+2+8+1+9+9+2+3 = 50 and again 5+0 is 5.
Alter: (shortcut method)
Now, we will eliminate the numbers that add up to 9 (6 and 3, 7 and 2, 8 and 1 and also eliminate the two 9’s) We are left with the digits 2 and 3 which also add up to 5.
Number

Normal Digit Sum

No. left after
applying Shortcut Method

Digit Sum After
Elimination

199999

1

1

1

63727

7

7

7

45231

6

231

6

8001573

6

573

6

56789

8

5678

8

Application:
In any mathematical operation involving calculation you can simply use the digitsum instead of the actual no.s given & continue the expression as given in the question.Finally, check the digit sum of the question obtained to the options.If it matches, then that is the most likely correct option.
Let me illustrate this method by solving previous year competitive questions as explained below.
1. 58 x 96 x 62 =?
(a) 345246
(b) 345226
(c) 345216
(d) 345236
Sol: By putting digit sum value of no.s in question we have,
4 x 6 x 8=24 x 8=6 x 8=48=3 (digit sum)
From options,(a)345246= 6
(b)345226= 4
(c)345216= 3
(d)345236= 5
Clearly option (c) is the correct answer as both digit sum matches. Isn't it easy?
2. 2387123+980=?145+945
(a)2244 (b)2434 (c)2444 (d)2544
Sol: As above, 26+8=?1+0
=>4=?1
=>?=4+1=5 (digit sum)
Digit sum of options are; (a)3 (b)4 (c) 5 (d)6
So, option (c) is correct answer.
(Q). Find digit sum of ? in 2134200 =?
Ans: Apply digit sum 12=?
=> ?= 1
= 1+9 = 8 (Remember it !)
Mathematical Procedure involved in
Division:
(Q). Find digit sum of ? in 2134200 =?
Ans: Apply digit sum 12=?
=> ?= 1
= 1+9 = 8 (Remember it !)
Mathematical Procedure involved in
Division:
Understand the logic carefully while dealing with division
as illustrated below.
In division the Numerator(N^{r}) doesn’t put impact
. So, it’s upto the Denominator(D^{r}) used in the expression.
We can face the denominators in digit sum division process
as written below:
N^{r}/1 ; N^{r}/2 ; N^{r}/3 ;N^{r}/4
; N^{r}/5 ; N^{r}/6 ; N^{r}/7 ; N^{r}/8 ; N^{r}/ 9or 0 .
As, N^{r}/1 is simply equals to the Numerator(N^{r})
. Therefore, our moto should be to make the Denominator(D^{r}) digit
sum “1” .
·
N^{r}/1 is equal to N^{r}.
·
N^{r}/2 = N^{r }x 5/2 x 5 = N^{r} x 5/10 = N^{r}
x 5/1 = N^{r} x 5
·
N^{r}/3 ; we can’t make the D^{r}
digit sum 1.
·
N^{r}/4 = N^{r }x 7/4 x 7= N^{r }x 7/28 = N^{r }x 7/10 = N^{r} x 7
·
N^{r}/5 = N^{r }x 2/5 x 2 = N^{r} x 2/10 = N^{r}
x 2/1 = N^{r} x 2
·
N^{r}/6 ; we can’t make the D^{r}
digit sum 1.
·
N^{r}/7 = N^{r }x 4/7 x 4= N^{r }x 4/28 = N^{r }x 4/10 = N^{r} x 4
·
N^{r}/8 = N^{r }x 8/8 x 8 = N^{r
}x 8/64 = N^{r }x 8/10 = N^{r
}x 8/1 = N^{r }x 8
·
N^{r}/9 ; we can’t make the D^{r}
digit sum 1.
Notes:
1.
When the Denominator(D^{r}) is 2 ;
simply multiply the Numerator(N^{r}) by 5.
2.
When the Denominator(D^{r}) is 4 ;
simply multiply the Numerator(N^{r}) by 7.
3.
When the Denominator(D^{r}) is 5 ;
simply multiply the Numerator(N^{r}) by 2.
4.
When the Denominator(D^{r}) is 7 ;
simply multiply the Numerator(N^{r}) by 4.
5.
When the Denominator(D^{r}) is 8 ;
simply multiply the Numerator(N^{r}) by 8.
6.
When the Denominator(D^{r}) is either 3
or 6 or 9/0 then if it goes divide by them respectively then it is alright
&we can have our desired result. If not then the value is nearly equal & digit sum directly can’t applied but can be solved .we will see it later in problems.
3. 5016 x 1001333 x 77 +22=? x 11
(a)435570=6 (b)454127=5 (c) 572240 =2 (d)366520=4
Sol: From question, 3 x 20 x 77+4=?x 2
=> 60+4=?x 2
=>1=?x 2
=>?=1/2=1x5/2x5=5/10=5/1=5 (Or simply multiply 5 to 1)
so,option(b)is correct answer.
# You can neglect decimal point (.) & percentage (%) sign in question. Simply proceed as there was no such signs.
So, option (b) is correct.
# You can neglect decimal point (.) & percentage (%) sign in question. Simply proceed as there was no such signs.
( This is because in decimal no. & percentage the Denominator is powers of 10 or 100 respectively . And in both case digit sum will be 1.)
t 4. 12.8% of 8800  16.4 of 5550 = 20% of ?
(a)965 =2 (b)996 =6 (c)1004=5 (d)1081=1
Sol:Now replacing digit sum values in question we have,
2 of 7  2 of 6 = 2 of ?
=> 2 x 7 2 x6 =2 x ? [ 'of ' means multiplication (x)]
=> 2=2 x ?
=>?=2/2 =1 [Or you have 2 x 5=10=1;as written division rule]
So, option (d) is correct.
5. 4160.715 ÷ 3369 =?
(a)
1.315 = 1 (b) 1.235 = 2 (c)1.035 = 0 (d)1.225 = 1
Sol: 4160.715/3369 =?
=>6/3=? (Replacing digit sum value &ignoring decimal point)
=>?=2
The digitsum method can only tell us whether an answer is wrong or not. It cannot tell us with complete accuracy whether an answer is correct or not.complete accuracy whether an answer is correct or not.
However, if the digitsum of the answer does not match with the digit sum of the question then you can be 100% sure that the answer is wrong.
Thank you very much.Crisp and clear
ReplyDelete55% of 900 + 70% of 1050=?% of 3000
ReplyDeleteI know the answer to this question is 41 whose digit sum is 5
But using the digit sum method i am getting answer as 2
1×0 + 7×6=3x
0+6=3x
2=x
Please help
7×6/3= 7×2 (cancel 6/3)=14=5
DeleteIf you have any doubt, please let me know.