Digit Sum & Last digit(s)  Kamal Lohia

Digital sum of a number is the single digit obtained by adding all the digits of a number continuously. Important point to observe is that ‘Digital Sum’ of a number is nothing but its remainder when divided by 9 :)
For example: DS (12345) = DS (1+2+3+4+5) = DS (15) = DS (1+5) = DS (6) = 6.
Let’s take one more: What is the sum of sum of sum of digits of 25! Where n! is product of first n positive integers starting from 1?
As 25! < 10^{25 }
So sum of digits of 25! Is certainly less than or equal to the sum of digits of greatest number less than or equal to 10^{25 }which is for 9999… 25 digits= 9 + 9 + … 25 digits= 225.
With similar logic sum of sum of digits of 25! Is certainly less than or equal to the sum of digits of greatest number less than or equal to 225 which is for 199 = 1 + 9 + 9 = 19.
Same way sum of sum of sum of digits of 25! Is certainly less than or equal to the sum of digits of greatest number less than or equal to 19 which is for 19 = 1 + 9 = 10.
That means required answer cannot be more than 10. But, also, we know that 25! Is a multiple of 9, so sum of sum of sum of digits of the number will always be multiple of 9.
So answer here is 9.
Confusing??? Let’s take easier one this time :)
Find the digital sum of 7^{100}
We just need to find the remainder of the number with 9. Answer is 7 and explanation, I am leaving for the reader.
Concept of digital sum finds its use in mainly checking calculations or in, a better manner, eliminating options in an exam like CAT.
Some properties of Digital Sum:
If two numbers are multiplied to give an answer, then we can check the calculation using concept of digital sum. Digital sum of product of digital sums of the two numbers will be same as the digital sum of the answer.
E.g. 25 × 35 = 875
DS (25) = 7
DS (35) = 8
DS (7 × 8 ) = DS (56) = DS (5 + 6) = DS (11) = 2
Also DS (875) = DS (8 + 7 + 5) = DS (20) = 2.
Remember, if digital sum of the product is not matching, then there IS some calculation error certainly. But vice versa is not true. That is, matching of digital sum doesn’t ensure whether product is correct or not.
Digital sum of perfect squares is 1, 4, 7 or 9. (Again that doesn’t say that every number with a digital sum of 1, 4, 7 or 9 is a perfect square).
Digital sum of perfect cubes is always 1, 8 or 9.
Last digit of a numberWe know that unit digit of the number 2012 is 2.
what is the unit digit of 2012 × 2012?
Certainly it is 4 as it depends only on the unit digits of the numbers being multiplied 2012 × 2012 i.e. 2 × 2 = 4.
what is the unit digit of 2012 × 2013 × 2014?
With similar logic, it also depends on the unit digit of numbers being multiplied 2012 × 2013 × 2014 i.e. 2 × 3 × 4 i.e. unit digit of 24 i.e. 4.
Next a better problem: What is the unit digit of 2012^{2012}?
With the logic discussed above, unit digit of 2012^{2012 }depends on unit digit of 2^{2012}.
Now we observe a pattern in the unit digits of all the digits as follows:
So we find that after every fourth power, unit digit of powers of 2 starts repeating. Strange and important point is that, it is true for all digits. Some repeat at every power (e.g. 0, 1, 5, 6), some after every two powers (e.g. 4, 9) and remaining after every fourth power (e.g. 2, 3, 7, 8 ).
The numbers which give same last digit as itself, when raised to any positive integral power are known as singledigit Automorphic numbers and they are 0, 1, 5, 6. Similarly 2digit automorphic numbers are 00, 01, 25 and 76. We’ll use them later. :)
This is the table of the unit digits of powers of all the digits (i.e. numbers of the form a^{b})
You don’t need to mug up the table. But this is just for your reference. You can always count mentally the unit digits of any digit up to its fourth power. Important Point to observe is that fourth power of all even digit (except 0) ends in 6 and that of all odd digits (except 5) ends in 1.
HOW TO FIND LAST TWO DIGITS OF A NUMBERAs unit digit depends on product of unit digits of the numbers being multiplied, same way last two digits of a number depends on the product of last two digits of the numbers being multiplied.
That means Last two digits of 2012 × 2013 = TD(12 × 13) = TD(156) = 56.
A question for you now: What are the last two digits of 2012 × 2013 × … × 2031.
It’s simple. Answer is ZERO. Because the product contains enough 5’s and 2’s (e.g. 2025 and 2012).
Now comes the important point for which this topic was written. Please read every word carefully. I am not giving the reason behind every step. So please be cautious that it is to be used Asitis. No malfunctioning should be done without expertise.
LAST TWO DIGITS OF A NUMBER ENDING IN 1
71^{83} : Unit digit (UD) of this number is always going to be 1 (because 1 is a single digit Automorphic number, as told above). And ten’s digit is obtained by unit digit of product of ten’s digit of the number (7 here) with unit digit of exponent (3 here).
That means TD(71^{83}) = UD(3 × 7)1 = UD(21)1 = 11
I hope this is clear.
Let’s take one more example:
Find the last two digits of 561^{165}.
Now TD (561^{165}) = TD(61^{165}) = UD(5 × 6)1 = UD(30)1 = 01.
One more: Find the last two digits of 79^{78}.
Caution – You cannot directly use the method here as the unit digit of the base number is not 1. But remember, as we talked earlier, every odd digit (except 5) can be raised to a positive integral power such that it’s unit digit becomes 1.
So we first need to convert the base number’s unit digit to be 1 by using some part of exponent.
TD(79^{78}) = TD(79^{2})^{39 }= TD(41^{39}) = UD(9 × 4)1 = UD(36)1 = 61.
Important points to remember:
Last two digits of (50n + x)^2 = Last two digits of (50n  x)^2 = Last two digits of x^2 (x and n are integers)
Last two digits of a number are same as the remainder obtained when the number is divided by 100.
LAST TWO DIGITS OF EVEN NUMBERS
Every even number contains some odd part and some even.
For example 62^{31 }can be written as 2^{31 }× 31^{31}.So to find the last two digits of an even number, we just need to know how to find last two digits of powers of 2, in addition to previously acquired knowledge.
Now, again, automorphic number comes to rescue. I told you above that 76 is two digit automorphic number i.e. which when raised to any positive integral power, always ends in 76 as last two digits.
i.e. TD(76^{n})= 76.
For your good fortune, 20 is the smallest power of 2 which ends in 76. That means TD(2^{20k}) = 76 where k is a positive integer.
Also one more property is associated with 76.
i.e. TD(76 × 2^{n}) = 2^{n }where n is a positive integer greater than 1.
Let’s take an example: Find the last two digits of 2^{2012}.
TD(2^{2012}) = TD(2^{20})^{100 }× TD(2^{12}) = TD(76)^{100 }× TD(2^{12}) = TD(76 × 2^{12}) = TD(2^{12}) = 96.
I hope this is clear.