Determining the Last / Second Last / Last two Digit(s) - Ravi Handa


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    Finding Last Digit

    Last digit of the number depends only on the last digit of the numbers in the expression and not on the other digits

    What is the last digit of 1273^122!?

    Last digit of 1273^122!
    = Last digit of 3^122!
    = Last digit of 3^4n (Last digits of powers of 3 move in cycle of 4. The cycle is 3, 7, 9, and 1. 122! is a multiple of 4, so it can be written as 4n)
    = 1

    What would be the last digit of K = 1! +2! +3! ...+19! ?

    K = 1! + 2! + 3! .... 19!
    Last digit of K = Last digit (1!) + Last digit (2!) + Last digit (3!) ... Last digit (19!)
    Last digit (1!) = 1
    Last digit (2!) = 2
    Last digit (3!) = 6
    Last digit (4!) = 4
    Last digit (5!) = 0
    Last digit (6!) = 0
    .
    .
    .
    Last digit (19!) = 0
    Please note that Last digit (n!) such that n > 4 will be 0
    Last digit of K = 1 + 2 + 6 + 4 + 0 + 0 + 0 ... 0 = 3

    What is the remainder when N= (1! + 2! + 3! + 4! +...1000!)^40 is divided by 10?

    We have to find out remainder of (1! + 2! + 3! ... 1000!)^40 from 10
    => We have to find out last digit of (1! + 2! + 3! ... 1000!)^40
    Last digit of n! where n > 4 will be 0 and will have no impact on the answer.
    => We have to find out last digit of (1! + 2! + 3! + 4!)^40
    Last digit of (1! + 2! + 3! + 4!)^40
    = Last digit of (1 + 2 + 6 + 24)^40
    = Last digit of 33^40
    = Last digit of 3^40
    = Last digit of 81^10
    = Last digit of 1^10
    = 1

    Finding Second Last Digit & Last Two Digits

    Nike caused controversy with its advertising campaign during the 1996 Olympics by using the slogan, "You Don't Win Silver — You Lose Gold." Nike's use of this slogan drew harsh criticism from many former Olympic Silver medallists. In a way, it did undermine the importance of the second position but in Math things are often very different. Figuring out the second last digit is often tougher than figuring out the last digit. It is unlikely but definitely not impossible that in CAT you get a straightforward question that asks you to find out the second last digit of a number (abc^pqr). It did happen in CAT 2008. In few cases, you will be able to do it by forming a cycle and observing the pattern. Those will be the easier cases. Read on if you wish to do the same for the not so easy cases.

    The question becomes really simple if the last digit in abc^pqr is 0 or 5 because if it 0, second last digit will be 0 and if it is 5, second last digit will be 2 or 7 (which can be easily figured out by observing the cyclicity). All the other questions can be divided in two broad categories:
    a) Last digit is odd
    b) Last digit is even

    I recommend that before using any of the concepts given below, you should try and see if a pattern exists.

    Let us consider our number is abc^pqr where a,b,c,p,q and r are digits and c is not 0 or 5.

    Concept 1: What to do when the last digit is odd?

    The second last digit always depends on the last two digits of the number so anything before that can be easily neglected.
    We first convert the number in such a way that the last digit of the base becomes 1.
    Second last digit of the number will simply be:
    Last digit of (Second last digit of base) X (Last digit of power)

    Let us look at few examples
    Eg 1a: Second last digit of 3791^768
    = Last digit of 9×8
    = 2

    Eg 1b: Second last digit of 1739^768
    = Second last digit of 39^768
    = Second last digit of Second Last digit of 1521^384
    = Last digit of 2 × 4
    = 8

    Eg 1c: Second last digit of 9317^768
    = Second last digit of 17^768
    = Second last digit of (17^4)^192
    = Second last digit of (…21) 192
    = Last digit of 2 x 2
    = 4

    Concept 2: What to do when the last digit is even?

    The second last digit always depends on the last two digits of the number so anything before that can be easily neglected.
    We need to remember the following ideas:

    1. **2 raised to power 10 will always end in 24.
    2. 24 raised to an even power will always end in 76 and to an odd power will always end in 24.
    3. 76 raised to any power will always end in 76.

    Now we can use these to find out the second last digit. We reduce the number in such a way that the last two digits of the base become 76.

    Eg 2a: Second last digit of 1372^482
    = Second last digit of 72^482
    = Second last digit of 72^480 x 72^2
    = Second last digit of (72^10)^48 x (**84)
    = Second last digit of 24^48 x (**84)
    = Second last digit of 76 x 84
    = Second last digit of 6384 = 8

    Eg 2b: Second last digit of 48^307
    = (48^3)^102 x 48
    = (****92)^102 x 48
    = Second last digit of 92^100 x 92^2 x 48
    = 76 x (**64) x 48
    Second last digit of (****72) = 7

    Eg 2c: Second last digit of 154^84
    = Second last digit of (54)^84
    = Second last digit of (54^5)^16 x 54^4
    = (***24)^16 x (54^2)^2
    = Second last digit of 76 x (2916)^2
    = Second last digit of 76 x 56
    = Second last digit of 4256 = 5

    Practice Questions :

    How do you find the remainder when 7^26 is divided by 100?

    Finding out the remainder from 100, is the same as finding out the last two digits of a number
    Last two digits of 7^1 are 07
    Last two digits of 7^2 are 49
    Last two digits of 7^3 are 43
    Last two digits of 7^4 are 01
    After this, the same pattern will keep on repeating.
    So, 7^(4n+1) will end in 07, 7^(4n+2) will end in 49, 7^(4n + 3) will end in 43, and 7^4n will end in 01
    7^26 = 7^(4n+2) will end in 49
    => Rem [7^26/100] = 49

    What is the remainder when 787^777 is divided by 100?

    We have to find out the remainder of 787^777 divided by 100
    This is the same as finding out the last two digits of 787^777

    Last two digits of the answer depend on the last two digits of the base.
    => We need to find out last two digits of 87^777

    The key in questions like these is to reduce the number to something ending in 1.
    87^777
    = 87 * 87^776
    = 87 * (..69)^388 {Just looking at last two digits of 87^2}
    = 87 * (...61)^194 {Just looking at last two digits of 69^2}
    = 87 * (...41) {a number of the format ..a1^..b will end in (a * b)1}
    = 67

    What are the last two digits of 2^1997 ?

    For finding out the last two digits of an even number raised to a power, we should first try and reduce the base to a number ending in 24.
    After that, we can use the property
    Last two digits of 24^Odd = 24
    Last two digits of 24^Even = 76

    Last two digits of 2^1997
    = Last two digits of 2^7 * (2^1990)
    = Last two digits of 128 * (1024^199)
    = Last two digits of 28 * 24
    = 72

    What are the last two digits of 2^2012 ?

    For finding out the last two digits of an even number raised to a power, we should first try and reduce the base to a number ending in 24.
    After that, we can use the property
    Last two digits of 24^Odd = 24
    Last two digits of 24^Even = 76

    Last two digits of 2^2012
    = Last two digits of 2^2 * (2^2010)
    = Last two digits of 4 * (1024^201)
    = Last two digits of 4*24
    = 96

    How do I find the last 2 digits of (123)^123!?

    For finding out the last two digits of an odd number raised to a power, we should first try and reduce the base to a number ending in 1.
    After that, we can use the property
    Last two digits of (...a1)^(...b) will be [Last digit of a*b]1

    Let us try and apply this concept in the given question
    Last two digits of 123^123!
    = Last two digits of 23^123!
    = Last two digits of (23^4)^(123!/4)
    = Last two digits of (529^2)^(123!/4)
    = Last two digits of (...41)^(a large number ending in a lot of zeroes)
    = Last two digits of (...01) {Here I have used the concept mentioned above}
    = 01

    Find the last two digits of 2025^2052 + 1392^1329?

    Let us break the problem into two parts.

    For the first part :
    Last two digits of 2025^2052
    = Last two digits of 25^2052
    = 25 I{Any power of 25 will have the last two digits as 25}

    For the second part:
    For finding out the last two digits of an odd number raised to a power, we should first try and reduce the base to a number ending in 1.
    After that, we can use the property
    Last two digits of (...a1)^(...b) will be [Last digit of a*b]1

    For finding out the last two digits of an even number raised to a power, we should first try and reduce the base to a number ending in 24.
    After that, we can use the property
    Last two digits of 24^Odd = 24
    Last two digits of 24^Even = 76

    Let us try and apply these concepts in the given question
    1392^1329
    = 92^1329
    = 4^1329 * 23^1329
    = 2^2658 * 23 * 23^1328
    = 2^8 * 2^2650 * 23 * (23^4)^332
    = 256 * (1024^265) * 23 * (...41)^332
    = 56 * 24 * 23 * 81
    = 72

    So, our overall answer will be the sum of the two parts
    = 25 + 72 = 97

    What are the last two digits of (86789)^41?

    For finding out the last two digits of an odd number raised to a power, we should first try and reduce the base to a number ending in 1.
    After that, we can use the property
    Last two digits of (...a1)^(...b) will be [Last digit of a*b]1

    Let us try and apply this concept in the given question
    (86789)^41
    = 89^41
    = 89 * 89^40
    = 89 * (..21)^40
    = 89 * 01 {Here I have used the concept mentioned above}
    = 89

    What will be the last two digits of 57^69?

    For finding out the last two digits of an odd number raised to a power, we should first try and reduce the base to a number ending in 1.
    After that, we can use the property
    Last two digits of (...a1)^(...b) will be [Last digit of a * b]1

    Let us try and apply this concept in the given question
    Last two digits of 57^69
    = Last two digits of 57 * (57^2)^34
    = Last two digits of 57 * (..49)^34
    = Last two digits of 57 * (..01)^17
    = Last two digits of 57 * (..01)
    = 57

    How do we find the last 2 digits of 19^39?

    For finding out the last two digits of an odd number raised to a power, we should first try and reduce the base to a number ending in 1.
    After that, we can use the property : Last two digits of (...a1)^(...b) will be [Last digit of a*b]1

    Let us try and apply this concept in the given question
    19^39
    = 19 * 19^38
    = 19 * (361)^19
    = 19 * (...41) {Here I have applied the concept given above}
    = (...79)

    What is the remainder when 767^1009 is divided by 25?

    Remainder of a number from 25 will be the same as the remainder of the last two digits of the number from 25.

    For finding out the last two digits of an odd number raised to a power, we should first try and reduce the base to a number ending in 1.
    After that, we can use the property : Last two digits of (…a1)^(…b) will be [Last digit of a*b]1

    Let us try and apply this concept in the given question
    Last two digits of 767^1009
    = Last two digits of 67^1009
    = Last two digits of 67 * 67^1008
    = Last two digits of 67 * (67^2)^504
    = Last two digits of 67 * (..89)^504
    = Last two digits of 67 * ((..89)^2)^252
    = Last two digits of 67 * (...21)^252
    = Last two digts of 67 * (...41) {Here I have used the property mentioned above}
    = 47

    Rem [767^1009 / 25]
    = Rem [47/25]
    = 22

    What are the remainders when 2^222 and 11^100 are divided by 25?

    We need to solve two questions here. In both, we need to find out the remainder from 25. Solving both of them would be easier if we just find out the last two digits.

    Remainder of the last two digits of a number from 25 will be the same as the remainder of the number from 25.

    Rem [2^222 / 25]
    Last two digits of 2^222
    = Last two digits of 4 * 2^220
    = Last two digits of 4 * 1024^22
    = Last two digits of 4 * (...76) {24^Even will always end in 76}
    = Last two digits of (...04)
    = 04

    So, Rem [04 / 25] = 4
    => Rem [2^222 / 25] = 4

    Rem [11^100 / 25]
    Last two digits of 11^100 = 01 {Last two digits of (…a1)^(…b) will be [Last digit of a*b]1}
    => Rem [11^100 / 25] = 1

    I hope that after reading this post you will be at ease in figuring out the second last digit. I also hope that you will not mind winning silver medals either


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