Gyan Room - Number Theory - Concepts & Shortcuts


  • QA/DILR Mentor | Be Legend


    How many ways can you express 360 as product of 2 co-primes

    Unordered : 2^(n-1)
    Ordered : 2^n

    How many factor pairs of 360 exist which are co-primes to each other

    Let me explain the concept of factor pair with a small example.
    let say you need co-prime factor pairs of 10. then what would be the answer?
    (1,1), (1,2) ,(1,5), (1,10) , (2,5) ..
    now lets decipher n generalize it

    Factor Pairs of a number x = 2^m * 3^n * .....
    Case - 1 (1,1 ) --> 1
    Case -2 = ; (1,2^m) ; (1,3^n) ; (As HCF should be 1)
    total arrangements = 2(m+n)
    case - 3 = (1,2^m*3^n) (As HCF should be 1)
    total arrangements = 2 * m * n
    Case - 4 (2^m , 3^n) (HCF = 1)
    total arrangements = 2 * m * n
    So total ordered cases for factor pairs = 1 + 2(m + n) + 4mn = (2m + 1) * (2n + 1)
    this is the general formula for factor pairs (ordered )

    in case of x = 360
    360 = 2^3 * 3^2 * 5
    Ordered factor pairs = (2 * 3 + 1) * (2 * 2 + 1) * (2 * 1+ 1 ) = 105


  • NMIMS, Mumbai (Marketing)


    Useful Tip :

    If N is a natural number, number of natural numbers in the range :
    Case 1 : A < = N < = B
    Number of natural numbers in the range : (B - A) + 1
    Case 2 : A < = N < B OR A < N < = B
    Number of natural numbers in the range : (B - A)
    Case 3 : A < N < B
    Number of natural numbers in the range : (B - A) - 1

    Practice examples :
    Q1) How many numbers,N, in the range :
    a) 100 < = N < = 200 b) 100 < = N < 200 c) 100 < N < 200
    Q2) How many even numbers,N, in the range :
    a) 100 < = N < = 250 b) 120 < = N < 360
    Q3) How many numbers,N, of the form 3k,where k is any positive integer in the range :
    a) 50 < N < 300 b) 100 < = N < 500 c) 100 < = N < = 1000

    Should help at a lot of places. So practice well if you are starting your preparations.


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    How to find the Number of digits in a^b ?

    Number of digits in a^b = [1 + log (a^b)]
    Here [ x ] denotes the Greatest Integer Function
    We will calculate the value of log(a^b) in base 10

    Example : What is the number of digits in 35^29
    Solution: [ 1+ log 10 (35^29) ]
    = [ 1+ 29 log 10 (35) ]
    = [ 1+ 29 * 1.54 ]
    = [ 1+ 44.776 ]
    = [ 45.776 ]
    = 45 digits


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    Perfect Squares :

    If the unit digit of a perfect square is 1, the “tens” digit must be even. Example : 1^2 = 01, 9^2 = 81, 11^2 = 121
    If the unit digit of the perfect square is 5, then the tens digit must be 2.
    If the Unit digit of a perfect square is 6, then the tens digit must be “odd”
    The number of zeroes present at the end of any perfect square cannot be “odd”. They must be in the form of 2k, where k =0,1,2,3…..
    A perfect Square of the form “aabb” is 7744 ( 88^2 = 7744 )


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    Factors :

    The number of divisors of a given number N (including 1 and the number itself) where N = a^m * b^n * c^p where a, b and c are prime numbers Is (1+m)(1+n)(1+p).

    Find the number of divisors of 120
    120 = 2^3 * 3 * 5
    number of divisors = 4 * 2 * 2 = 16

    In the above case, sum of divisors is ((a^(m+1) )-1)/(a – 1) * ((b^n+1)-1)/(b-1) * ((c^(p+1) -1 )/(c-1)

    Find the sum of divisors of 120
    120 = 2^3 * 3 * 5
    sum = 2^4 – 1 / 1 * 3^2-1/2 * 5^2 – 1 / 4 = 15 * 4 * 6 = 360

    The number of ways in which a number N can be expressed as the product of two factors which are relatively prime to each other is 2^(m-1) where m is the number of different prime factors of N.

    The number of ways 120 can be expressed as the product of two factors which are relatively prime to each other
    120 = 2^3 * 3 * 5
    Number of prime factors = 3
    number of ways = 2^2 = 4
    { (1,120),(3,40),(5,24),(8,15)}

    Product of factors of N = N^(f/2) where f is number of factors .

    Find the product of factors of 24
    24 = 2^3 * 3
    Number of factors = 4 * 2 = 8
    product = 24^4


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    Number of Weighing

    If N is the number of balls and question doesn’t specify whether the ball weighs more or less, Then total number of weighing = m + 1 (where N < = 3^m + 3^(m-1) + 3^(m-2) + …. + 3^0)
    If the question specifies that ball weighs more or less then total number of weighing = m (where N < = 3^m)

    Examples :

    (a) There are 47 identical coins. All the coins have same weight except one coin which has a different weight. What's the minimum number of weighing required to be certain of identifying the coin with different weight?

    1 + 3 + 9 + 27 = 40
    1 + 3 + 9 + 27 + 81 = 121 > 47
    So here m = 4
    Number of weighing required m + 1 = 5

    (b) You have 80 pearls . One is lighter than remaining.Minimum number of weighing required on a common balance to ensure that odd pearl is identified ?

    80 < = 81 so m = 4
    Number of weighing required =4


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    Match Sticks Or Coins concept

    Suppose two players A and B play a game of coins in which the person picking up the last coin from table wins the game. Let us consider there are 50 coins on a table and a person can pick-up maximum 7 coins and minimum one coin of his chance. So if question says A starts the Game what should be the number of coins he must pick-up to ensure he wins

    If a starts his targets will be 50, 42, 34, 26, 18, 10, 2. So he must pick-up two coins to ensure Win.

    Suppose in the same question if the player picking up last coin loses the game.
    Add minimum and maximum = 1+ 7 = 8
    Closest multiple of 8 which is less than 50 is 48. So 50-48 = 2

    If last person who is picking the coin will win , a would have picked 2 . But here he will pick1 so that in the end 1 is left for b. Now whatever b picks up a will match with a number that will make sum 8.

    A ----- B
    1 ------ 4
    4 ------ 7
    1 ------ 7
    1 ------ 6
    2 ------ 6
    2 ------ 5
    3 ------ 1

    This is an example and here B is helpless.


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    Remainder theorem :

    Euler’s theorem

    Euler’s function E[n] denotes number of positive integers that are coprime to and less than a certain positive integer ‘n’ .

    Prime divisors of 100 are 2 and 5.
    So number of integers less than 100 and not divisible by 2 or 5 are
    100 - 100/2 - 100/5 + 100/2*5
    = 100(1-1/2) ( 1-1/5)

    In general a positive integer N having prime divisors p1, p2, p3 … pn
    E[N] = N(1-1/p1)(1-1/p2) ... (1-1/pn)
    Euler’s theorem states that:
    If P and N are positive coprime integers then p ^E[N] mod N = 1
    Where E[N] is the Euler’s function for N.

    9^101 mod 125
    E[125] = 125 * 4/5 = 100
    Since 9 and 125 are coprime , 9^100 mod 125 = 1
    so remainder is 9 * 1 = 9

    Sum of all coprimes of N which are less than N = N * E(N)/2

    Sum of coprimes of 377 which are less than 377?
    E[377] = 377 * 28 * 12/29 * 13 = 336
    Sum = 377 * 336/2 = 63336
    another way also this can be done
    Sum = [1+2+3….376] – [sum of all multiples of 29] - [sum of all multiples of 13]

    Wilson Theorem

    If P is a prime number, (P-1)! +1 is divisible by P
    (P-1)! Mod P = -1 or (P-1)
    (P-2)! Mod P = 1
    (p-3)! Mod P = (p-1)/2

    16! = (16!+1)-1 = (16!+1)+16-17
    16! Mod 17 = 0 + 16 - 0 = 16

    15! Mod 17 => 16!mod 17 = 16
    15! * 16 mod 17 = 16
    15! Mod 17 =1

    14! Mod 17 -> 15! Mod 17 = 1
    14! * 15 mod 17 = 1
    14! * -2 mod 17 = -16
    14! Mod 17 = 8

    When both dividend and divisor have a factor in common

    Step 1 : Take out the common factor (k)
    Step 2 : Divide the resultant dividend by resulting divisor and find out remainder (r)
    Step 3 : The real remainder is remainder (r) multiplied by common factor (k)

    Example : Remainder when 2^96 is divided by 96
    96 = 2^5 * 3
    2^96 = 2^5 * 2^91 -> common term 32
    2^91 mod 3 = 2
    Remainder = 2*32 = 64

    Negative remainder

    When a number N < D gives a remainder R When divided by D , it gives a negative remainder of R-D

    Remainder when 7^52 is divided by 2402
    7^52 mod 2402 = (7^4 )^13 mod 2402 = 2401^13 mod 2402 = -1^13 = -1
    remainder 2402-1 = 2401

    Fermat’s Theorem

    If P is a prime number and N is co prime to P , Then N^p – N is divisible by P.

    Chinese Remainder theorem

    If a number N = a * b where HCF (a,b) = 1 and S is a number such that S mod a = r1 and S mod b = r2 then remainder S mod N =ar2x+ br1y where ax+by = 1

    A number which when divided by 7,11 and 13 leaves remainder 5, and 8 respectively. IF all such numbers less than 10000 are added what will be the remainder when the sum is divided by 11?

    7x+5 = 11y+7 = 13z+8
    7x+5 = 11y+7
    7x = 11y+2
    Solutions are (5,3)(16,10)(27,17)……
    Comparing Second and third
    11y+7 = 13z+8
    11y = 13z+1
    solutions are (6,5)(19,16)(32,27)…..
    so general form of Y
    y = 7a+3 = 13b+6
    7a = 13b+3
    So numbers of form (6,3),(19,10),(32,17)……
    So lowest y = 7 * 6 + 3 = 45
    Lowest number = 11 * 45 + 7 = 502
    next Y = 19 * 7 + 3 = 136
    Next number = 11 * 136 + 7 = 1503
    numbers are 502,1503,2504,3505,4506,5507,6508,7509,8510,9511
    Sum = 50065
    50065 mod 11 = 4


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    Finding Digit(s) :

    Last Two digits of Number ending in 1

    Unit digit will be 1.
    Multiply the tens digit of number with the last digit(unit digit) of exponent to get tens digit.

    Last two digits of 31^786 -> 3 * 6 = 18 hence last two digits 81
    Last two digits of 41^2789 -> 4 * 9 = 36 hence last two digits 61

    Last two digits of Number ending in 3 or 7 or 9

    Try to convert it in to the form of digits ending in 1.

    Last two digits of 19^266 = (19^2)^133 = 361^133 => hence last two digits 81
    Last two digits of 33^288 = (33^4)^72 = 21^72 -> hence last two digits 41
    Last two digits of 87^474 = (87^4)^118 * 87^2 = >61^118 * 87^2 => 81*69 =>89

    Last two digits of Number ending in 2, 4, 6 or 8

    Only one even two digit number which ends in itself(last two digits() is 76.
    i.e 76^ any power -> last two digits will be 76.
    24^2 ends in 76 and 2^10 ends in 24 . 24^even power always ends in 76 and 24^odd power ends with 24.
    Last two digits of 2^543 = (2^10)^54 * 2^3 = 24^54 * 2^3 = 76 * 8 => 08
    If you multiply 76 with 2^n last two digits will be last two digits of 2^n
    64^236 -> (2^6)^236 = 2^1516 = (2^10)^151 * 2^6 = 24 * 64 -> 36
    56^283 = (2^3)^283 * 7^283 = 2^849 * 49^140 *7^3 = (2^10)^84 *2^9 * 01^70 *43 = 76 * 12 *01 * 43 => 16

    Last two digits of Number ending in 5

    There are two cases

    1. Numbers where previous digit of 5 is 0 or any even number
    2. Numbers where previous digit of 5 is odd number

    In first case if number raised to any power -> last two digits 25
    In second case if number raised to even power -> last two digits 25
    In second case if number raised to odd power -> last two digits 75.

    Last non zero digit of n!

    Z(n) denotes last non zero digit in n!
    Z(n) = 4 * Z(n/5) * Z (unit digit), if ten's digit is odd
    = 6 * Z(n/5) * Z (unit digit), if tens digit is even

    Last non zero digit of 36!
    Z(36) = 4 * Z(7) * Z(6) = 4 * 4 * 2 = 32. so answer 2

    Last non zero digit of 15!
    Z(15) = 4 * Z(3) * Z(5) = 4 * 6 * 2 -> 48. hence answer 8

    To calculate unit digit of A^B

    Case 1) When B is not a multiple of 4
    B = 4X + Y Where 0 < Y < 4
    So here unit digit of A^B will be unit digit of A^Y

    Case 2) When B is a multiple of 4
    Even numbers (2,4,6,8) raised to powers which are multiples of 4 give the unit digit as 6
    For 1 and 5 ,any power of this will give same unit digit
    Other odd numbers (3,7,9) raised to any powers which are multiples of 4 give the unit digit as 1.

    Unit digit of 9^46
    46 = 11*4 + 2
    Hence unit digit of 9^46 is same as unit digit of 9^2
    hence answer 1

    Find the unit digit of 7^11^13^17
    Here we have to find out whether power is a multiple of 4
    so 11^13^17 mod 4 = (12-1)^13^17 = (-1)^13^17 = -1^(odd number) = -1
    Hence remainder will be 4-1 =3
    So unit digit of 7^11^13^17 will be same as 7^3 which is 3


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    Power of a natural number contained in a factorial

    Highest power of prime number P In n! = [n/p]+[n/p^2]+[n/p^3]…

    Highest Power of 3 in 50! = [50/3]+[50/9]+[50/27] = 16+5+1 =22

    Highest power 30 in 70!
    30 = 2 * 3 * 5
    Since 5 is the largest prime factor power of 5 will be less than that of 2 and 3.
    Hence power of 30 will be equal to power of 5
    70/5+70/25 = 14+2 = 16

    Find the number of zeros present at the end of 90!
    Number of zeros present at the end means we have to calculate highest power of 10.
    since 10 = 2*5
    Highest power of 10 means highest power of 5
    90/5+90/25 = 18+3 =21
    Hence number of zeros present at the end 21

    Find the highest power of 24 in 100!
    24 = 3 * 2^3
    Power of 2 =100/2+100/4+100/8+100/16+100/32+100/64 = 50+25+12+6+3+1 = 97
    Power of 2^3 = [97/3] = 32
    Power of 3 = 100/3+100/9+100/27+100/81 = 33+11+3+1 = 48
    Since power of 2^3 is less , power of 2^3 in 100! is 32


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    A number in base N is divisible by N - 1 when sum of digits of number is in base N is divisible by N - 1

    Example : The number 28A65432 is in base 8. The number is divisible by 7.Then value of A
    2 + 8 + A + 6 + 5 + 4 + 3 + 2 = 30 + A
    (30 + A) mod 7 = 0
    A = 5

    A number has even number of digits in base N, the number is divisible by N + 1 if the number is palindrome

    For two given numbers, HCF * LCM = Product of numbers

    If numbers N1, N2 and N3 giving remainders R1, R2 and R3 when divided by the same number P. Then P is the HCF of (N1 - R1), (N2 - R2) & (N3 - R3)

    If numbers N1, N2 and N3 give same remainder when divided by same number P then P is a factor of (N1 - N2) and (N2 - N3)

    Let N be a composite number such that N = (x)^a * (y)^b * (z)^c where x, y and z are prime factors.
    a) If N is not a perfect square then number of ways N can be written as a product of two numbers => number of divisors /2
    b) If N is a perfect square then number of ways N can be written as a product of two numbers => (number of divisors +1)/2


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    Concept 1 : Find last non zero digit of N!
    Represent N = 5a + b form then last non zero digit will be 2^a x a! x b!

    Concept 2 : Find last two non zero digits of N!
    Express N = 5a + b then last two non zero digits will be 12a x a! x (5a + 1)(5a + 2) ... (5a + b)

    Examples :

    Find last non zero digit of 21!
    21 = 5 x 4 + 1
    so a = 4 and b = 1
    so last non zero digit =2^4 x 4! x 1!
    16 x 24 x 1 so last non zero digit = 4

    Find last non zero digit of 34!
    34 = 5 x 6 + 4
    So a = 6 and b = 4 so 2^6 x 6! x 4!
    Now we need to find the last non zero digit in 6! so 6 = 5 x 1 + 1
    so a = 1 and b = 1
    so 2^1 x 1! x 1! = 2
    now we know that 6! = 720 so last non zero digit is 2 so no need to do this
    so 2^6 x 2 x 4!
    2^7 x 24 = 128 x 24 so last digit = 2

    Find last two non zero digits of 23!
    23 = 5 x 4 + 3
    here a = 4 and b = 3
    so 12^4 x 4! x (5 x 4 + 1)(5 x 4 + 2) (5 x 4 + 3)
    12^4 x 4! x 21 x 22 x 23
    now multiply and find last two digits

    NOTE 1 -last two non zero digits of
    5! = 12
    25! = 84
    125! = 88

    Last non zero digits
    (10!) = 8
    (20!) = 4
    (30!) = 8
    (40!) = 2
    (50!) = 2
    (60!) = 6
    (70!) = 8
    (80!) = 8
    (90!) = 2
    (100!) = 4

    Example : If you want to find last non zero digit of 21!
    You know last non zero digit of 20! is 4
    so 21! = 21 x 20!
    21 x 4 = > 4 is the last non zero digit


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    @kamal_lohia

    Writing A Number As Sum Of Two Or More Consecutive Positive Integers

    In how many ways can 2010 be written as sum of two or more than two consecutive positive integers?

    For example: 2010 = 669 + 670 + 671.

    Now every number can be written as a sum of consecutive integers.

    Ex (-3) + (-2) + (-1) + 0 + 1 + 2 + 3 + 4 = 4

    But not all numbers can be written as sum of consecutive positive integers.

    A number can be written as the sum of consecutive positive integers if, and only if, its prime factorization includes an odd prime factor; then, since 2 is the only even prime number, the conclusion is that the powers of 2 are the only numbers that cannot be written as the sum of consecutive positive integers.

    Number of ways to write a natural number, N as sum of two or more than two consecutive positive integers is given by number of odd positive integral divisors of N - 1.

    So in this particular question, number of odd divisors of 2010 is 8. hence the required number of ways = 7.

    Now take an example of 36.

    now 36 - 1 = 35 has four positive odd integral divisors. So 36 can be expressed as sum of two or more positive consecutive integers in four possible ways. But how to construe those ways if the number is large?

    Ex 60; 60 - 1 = 59; 59 has two odd divisors, but the number of ways to express 60 as sum of consecutive positive integers is three:

    60 = 19 + 20 + 21 = 10 + 11 + 12 + 13 + 14 = 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11

    Also in how many ways can any natural number be expressed as sum of consecutive positive even integers or consecutive possible odd integers?

    Number of ways to write a natural number,N as sum of two or more than two consecutive natural numbers = (number of odd divisors of N) - 1.
    If N = 36, then the required number of ways are = 3 - 1 = 2.
    If N = 60, then required number of ways are = 4 - 1 = 3.

    Now if question is to find the number of ways to write a natural number, N as sum of two or more than two consecutive even natural numbers, then first condition is that n must be an even number. Number of ways remains same as above. Let's check for 60.

    60 = 3 * 20 = 5 * 12 = 4 * 15
    So, 60 = 18 + 20 + 22 = 8 + 10 + 12 + 14 + 16 = 12 + 14 + 16 + 18.

    Now if question is to find number of ways to write a natural number,N as sum of two or more than two consecutive odd natural numbers, then N has to be an odd number or a multiple of 4. Number of ways can be calculated accordingly.


  • QA/DILR Mentor | Be Legend


    NUMBER OF WAYS OF EXPRESSING A COMPOSITE NUMBER AS A PRODUCT OF TWO FACTORS

    Let us consider an example of small composite number say, 90
    Then 90 = 1 × 90
    Or = 2 × 45
    Or = 3 × 30
    Or = 5 × 18
    Or = 6 × 15
    Or = 9 × 10

    So it is clear that the number of ways of expressing a composite no. as a product of two factors =1/2 × the number of total factors

    Example: Find the number of ways of expressing 180 as a product of two factors.
    Solution: 180 =2^2×3^2×5^1
    Number of factors = (2+1)(2+1)(1+1)=18
    Hence, there are total 18/2=9 ways in which 180 can be expressed as a product of two factors.

    Perfect Squares: As you know when you express any perfect square number 'N' as a product of two factors as √N x √N, and you also know that since in this case √N appears two times but it is considered only once while calculating the no. of factors so we get an odd number as number of factors so we can not divide the odd number exactly by 2 as in the above formula. So if we have to consider these two same factors then we find the number of ways of expressing N as a product of two factors=((Number of factors+1))/2 .

    Perfect Squares as product if 2 distinct factors: Again if it is asked that find the no. Of ways of expressing N as a product of two distinct factors then we do not consider 1 way (i.e.,N=√N×√N) then no. Of ways = (Number of factors-1)/2

    Example: Find the number of ways of expressing 36 as a product of two factors.
    Solution: 36 =2^2×3^2
    Number of factors = (2+1)(2+1)=9
    Hence the no. Of ways of expressing 36 as product of two factors = (9+1)/2=5.
    As 36=1×36,2×18,3×12,4×9 and 6×6

    Example: In how many of ways can 576 be expressed as the product of two distinct factors?
    Solution: 576 =2^6 × 3^2
    ∴ Total number of factors = (6+1)(2+1)=21
    So the number of ways of expressing 576 as a product of two distinct factors = (21-1)/2 = 10.
    Note since the word ‘distinct’ has been used therefore we do not include 576 = 26 × 26.


  • QA/DILR Mentor | Be Legend


    Finding the Last Digit

    Last digit of any number raised to a power is decided by the last digit of the base
    For example last digit of 5763^67 is same as the last digit of 3^67
    So we just need to know the concept of last digit for single digit number

    Now for single digit numbers, the whole thing can be divided in 3 categories

    1. 0, 1, 5, 6: last digit is always same, raised to any power.
      Example 5^113 will end in 5’
      6^239 will end in 6
      2346 ^ 5732 will end in 6

    2. 4, 9: Here there is a cycle of 2;
      For 4 it is 4 and 6 and for 9 it is 9 and 1
      So odd powers of 4 will end in 4 and even power in 6
      Odd powers of 9 will end in 9 and even power in 1
      Example:
      564 ^ 231 will end in 4
      75689 ^ 568 will end in 1

    3. 2, 3, 7 and 8
      Here there is a cycle of 4
      Divide the power by 4 and take the remainder as the power
      For example: last digit of 2347 ^2347 is same as 7^2347
      Divide the power by and take the remainder; to find the remainder on dividing by 4, we just need to take the last 2 digits
      So last digit of 2347^2347 is same as 7^47
      Divide 47 with 4; remainder is 3
      So last digits is same as that of 7^3 = 3
      Example: 5748 ^5748
      Same as 8^48
      On dividing 48 with 4 the remainder is 0
      When the remainder is 0, we take it as 4
      So 8^4; last digit is 6

    Find the last digit of : (101^101 + 102^102……109^109)

    101^101: 1
    102^102: 2^2 : 4
    103^103: 3*3: 7
    104^104: 6
    105^105: 5
    106^106:6
    107:107:7^3: 3
    108^108: 8^4: 6
    109^109: 9
    Last digit: 7
    So , 1 4 7 6 5 6 3 6 9 adds up to 47 so last digit is 7

    Find the last digit of 3^(2^2^2….)

    We know that is the base is 3, we need to divide the power by 4 and take the remainder
    Now when you divide a number with 4, possible remainders are 1, 2, 3 and 0
    Remainder of 3 can also be taken as remainder of -1; for example 19 divide by 4 is 4x4+3 (remainder 3) but it can also be written as 4x5-1 (remainder -1)
    Remainder of 2 will only be there when the given power is even. Any even power raised to any number when divided by 4 will give the remainder as 0
    Remainder of 0 will be taken as 4; for finding the last digit
    So when we divide (2^2^2..) with 4, the remainder is 0
    Any even digit ^ any number will always be divisible by 4
    So the last digit of 3^(2^2^2….) is same as 3^4 = 1

    Find the last digit of (1002 ^ 1003 ^ 1004…..2000)

    1002 ^ 1003 ^ 1004…..2000)
    (2 ^ 1003 ^ 1004…..2000)
    2 ^ ( (-1) ^1004…..2000)
    -1^even number = 1
    2^ (1^…..) = 2

    Find the last digit of: 2468^ (4682^6824^8246)

    8^(2^even number..)
    8^4 = 6
    So 6


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    Finding the second last digit

    Cyclicity

    There lies the cyclicity of tens' place digit of all the digits. This is given below:

    DigitsCyclicity
    2, 3, 820
    4, 910
    51
    65
    74

    Example 2367^2367
    Take 7^2366; this will be same as 7^2 (as 7 has a cyclicity of 4 so we divide the power with 4)
    7^2 = 49
    Now, Multiply the 2nd last digit of the given number with the last digit of power and place the last digit of the original number in the last place
    Last digit of 6 x 7 = 2
    So 27
    Now multiply 27 with 49; last 2 digits are 23

    Example 3438 ^ 126
    8 has a cyclicity of 20, so 8^6
    8^3: 12
    So 8^6: 44
    Last digit of 3 x 6 = 8; so 88
    44 x 88 = 16

    SHORT CUT FOR LAST 2 DIGITS OF NUMBERS ENDING WITH 1:
    EXAMPLE: 2341 ^ 678
    Multiply the 2nd last digit of base with last digit of the power: 4 x 8 = 32
    Last digit was 2
    So last digit of overall expression is 21


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    Algebraic representation of numbers

    Representing numbers in algebraic form is very useful while we chase X. Sharing some useful Fundas

    Consecutive numbers: n, n+1, n+2 …

    Even number: 2n

    Consecutive even numbers: 2n, 2n+2, 2n+4 …

    Odd number: 2n+1

    Consecutive Odd numbers: 2n+1, 2n+3, 2n+5 …

    2 digit number (say ab) = 10a + b (a and b can take values from 0 to 9)

    3 digit number (say abc) = 100a + 10b + c

    n digit number =10(n-1)digit1+ 10(n-2)digit2+… + digitn (Digits taken left to right)

    Three consecutive numbers: n-1, n, n+1 (sums to zero)

    (a + b)2= a2+ 2ab + b2

    (a - b)2= a2- 2ab + b2

    (a + b)(a - b) = a2- b2

    a3+ b3= (a + b) (a2- ab + b2)

    a3- b3= (a - b) (a2+ ab + b2)

    (a + b)3= a3+ 3a2b+ 3ab2+ b3

    (a - b)3= a3- 3a2b+ 3ab2- b3

    (a + b + c)2= a2+ b2+ c2+ 2ab + 2bc + 2ca

    (a + b + c) (a2+ b2+ c2- ab - bc - ca) = a3+ b3+ c3- 3abc

    a3+ b3+ c3= 3abc, if a + b + c = 0

    a0= 1

    a1= a

    am× an= a (m + n)

    am/ an= a ( m – n )

    (am)n= am.n

    a- m= 1 / am

    (ab)m= ambm

    Symbol √ is called as radical sign or radix.

    Need a case to solve? Here we go…

    When you reverse the digits of the number 13, the number increases by 18. How many other two digit numbers increase by 18 when their digits are reversed? (CAT 2006)

    Here we are not just asked to find X but the whole X Gang! What are the clues?

    1. X Gang contains only two digit numbers

    2. If X Gang members are reversed they are increased by 18 than the original value.

    We can write any 2 digit number xy as 10x + y (x and y are the first and second digit).
    After reversing the number is yx represented as 10y + x.
    Given 10y + x = 10 x + y + 18, solving y = x + 2. y should be 2 more than x.
    Now y cannot be 2 as then x = 0 and the number 02 is not a 2 digit number!
    y can take values from 3 to 9
    when y = 3, x = 3 – 2 =1; y=4, x = 4 – 2 = 2; and so on…
    Our gang is (10x + y) for all the above (x, y) which are 13, 24, 35, 46, 57, 68 and 79.

    Gang busted! :)


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    Divide and Conquer

    We will now learn an interesting and equally important concept, divisibility of numbers. These concepts will be used extensively during prime factorization and while calculating HCF/LCM. Here we will just focus on how we can easily check whether a given number is divisible by some common divisors or not.

    Divisible by 2: If the last digit is divisible by 2.
    12, 142, 68…

    Divisible by 3: Sum of digits of the number is divisible by 3.
    15672, sum of digits = 1+5+6+7+2 = 21 = 3 * 7, hence divisible by 3.

    Divisible by 4: If the last 2 digits are divisible by 4.
    724, Last 2 digits (24) gives a number divisible by 4.

    Divisible by 5: If the last digit is 5 or 0.
    E.g. 625, 310 etc…

    Divisible by 7: Subtract twice the unit digit from the remaining number.
    If the result is divisible by 7, the original number is.
    14238, 2 * 8 – 1423 = -1407 = -201 * 7, hence divisible by 7

    Divisible by 8: If the last 3 digits are divisible by 8.
    1040, Last 3 digits (040) gives a number divisible by 8.
    A number is divisible by 2nif the last n digits are divisible by 2n.

    Divisible by 9: If the sum of the digits is divisible by 9
    972036, sum of the digits = 9 + 7 + 2 + 0 + 3 + 6 = 27, divisible by 9.

    Divisible by 11: If the difference between the sum of digits at the odd place and the sum of digits at the even place is zero or divisible by 11.
    1639, (9+6) - (3+1) = 11, divisible by 11.

    Divisible by 13: If the difference of the number of its thousands and the remainder of its division by 1000 is divisible by 13.
    2184, 2 - 184 = -182, so divisible by 13.

    Divisible by 33, 333, 3333… & 99, 999, 9999…:

    As a general rule, to check a given number is divisible by 333…3 (n digits) just see whether the sum of digits taken n at a time from right is divisible by 333…3 (n digits). If yes then the original number is also divisible by 33…3 (n digits). It is easy to understand through examples.

    Is 627 divisible by 33?
    Take 2 digits from right at a time, and get the sum.
    27 + 06 = 33, hence divisible by 33

    Is 22977 divisible by 333?
    Take 3 digits from right at a time and find the sum.
    977 + 022 = 999, hence divisible by 333

    Same can be applied for checking the divisibility of a given number with 99…9 (n digits). Check if the sum of digits taken n at a time from right is divisible by 999…9 (n digits). If yes then the original number is also divisible by 99…9 (n digits)

    Is 6435 divisible by 99?
    35 + 64 = 99. As per the above rule, 6435 is divisible by 99.

    How much time you need to tell whether the number 1000000998 is divisible by 999?


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    Approximate

    Wherever possible, approximate. Most of the times we don’t need precision level more than 2 decimal points to uniquely differentiate the given options. If the question demands more precision than that, leave the question and come back if you have time.

    As we discussed earlier, an easy approximation technique is to write numerator with respect to denominator

    2136 / 17 =

    2136 ≈ 1700 (100 * 17) + 340 (20 * 17) + 85 (17 * 05) + 8.5 (17 * 0.5) + 1.7 (17 * 0.1) …
    2136 / 17 ≈ 100 + 20 + 5 + 0.5 + 0.1 +… ≈ 125.6 ( Actual value is 125.64 )

    There are various approximation techniques which are discussed and debated at multiple forums. But I feel most of them are complicated considering the level of approximation we need. Stick with methods that are simple to understand and easy to apply. I am not sure, but many approximation techniques are some where related to the aforementioned method. If you know some methods which are useful in approximation, do share.

    Use Options

    Number S is equal to the square of the sum of the digits of a 2 digit number D. If the difference between Sand D is 27, then D is (CAT 2002)

    (1) 32
    (2) 54
    (3) 64
    (4) 52

    Here we can form the algebraic equation and solve… but easiest way is to take options one by one and see which option satisfies the given condition.

    Option1: 32, S – D = (3 + 2)2– 32 = 7, Not the answer.
    Option2: 54, S – D = (5 + 4)2– 54 = 27… Oila!!! :)

    Options are not just useful in substitution, but will also help us in deduction also.

    ‘Solving’ this equation will take us ages. We can write the above question as

    Options can be written as

    (1) (n+1) – 1/ (n+1)
    (2) n – 1/n
    (3) n – 1/ (n+1)
    (4) (n+1) – 1/n
    (5) (n+1) – 1/ (n+2)

    For n =1, sum is √ (2+1/4) = √9/4 = 3/2

    Now substitute n=1 in the options

    1) 2 – 1/2 = 3/2

    2) 1 – 1/1 = 0

    3) 1- 1/2 = 1/2

    4) 2- 1/1 = 1

    5) 2 – 1/3 = 5/2

    Only option 1 satisfies. This holds true for any value of n. Coming back to the original question where n = 2007, sum should be equal to (n+1) – 1/ (n+1) = 2008 – 1/2008.


  • Senior Math Wizard Champion (2011, 2012) and Math Count Champion (2011)


    In how many equal parts should you divide 100 so that the continued product of those equal parts will be maximum?

    Formula: N =floor (Sum/e)

    where e is the Euler’s constant approximately equal to 2.71828

    Floor function means the greatest integer less than or equal to

    Solution:
    N = floor (100/e) ~ 36
    N = 36


 

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