Gyan Room  Number Theory  Concepts & Shortcuts

Gyan Room  There's always room to learn more!
This thread is reserved for sharing concepts, short cuts and good questions from Number Theory topic.Happy Learning, Stay MBAtious!

Knowing squares & square roots also will help to save time in many scenarios.
Calculation of squares of numbers ending with 5
If the number is of the form X5 then (X5)^{2}= X * (X + 1) and append 25 at the end.
65^{2}= 6 * (6+1) and append 25 at the end = 4225
(135)^{2}= 13 * (13+1) and append 25 at the end = 18225This comes handy in scenarios like 85 * 87
= 85 * 85 + 85 * 2 = 7225 + 170 = 7395
This method is actually derived from another useful Vedic math concept. It gives an easy way to find the product of two numbers if their unit digits add up to 10 and rest of the digits are same.
Say, 72 * 78
Units digits adds up to 10 (2 + 8 ) and other digits are same (7).
Here product can be found out as 7 * (7 + 1) (2 * 8 ) = 5616.Similarly 144 * 146 = 14 * 15 4 * 6 = 21024
Many more neat tricks are there to easily find the square of numbers…
To calculate squares of two digit numbers
Let the number be xy then (xy)^{2 }= [x^{2}][y^{2}] + 20 * x * y
Take 58, (58)^{2 }= [5^{2}][8^{2}] + 20 * 5 * 8 = 2564 + 800 = 3364
Take 81, (81)^{2 }= [8^{2}][1^{2}] + 20 * 8 * 1 = 6401+ 160 = 6561 ( note: we wrote 1^{2 }as 01 ( 2 digits) )
To calculate squares for numbers closer to 100
Take 94, 94 is 6 less than 100, so 94^{2 }= [94 – 6][6^{2}] = 8836 (note: we subtracted the difference as the given number is lesser than 100)
Take 108, 108 is 8 more than 100, so 108^{2 }= [108 + 8][8^{2}] = 11664 (note: we added the difference as the given number is greater than 100)
Take 119, 119 is 19 more than 100, so 119^{2} = [119 + 19] [19^{2}] = [138][361] as the second part is more than 2 digit take the most significant number (here 3) and add it to the first part (138 ).
So 119^{2 }= [138 + 3][61] = 14161Take 125, 125 is 25 more than 100, so 125^{2} = [125 + 25]... we dont need to do that... we have a better trick, right? 125^{2} = [12 * 13][25] = 15625 :)
To calculate squares of numbers closer to 50
58 = 50 + 8
58^{2} = [25 + 8] [8^{2}] = 3364 [Note: Add the difference to 25 to get the first 2 digits as the given number is greater than 50]63 = 50 + 13
63^{2 }= [25 + 13] [13^{2}] = [38] [169], as the second part is more than 2 digit take the most significant number (here 1) and add it to the first part (38 ).
63^{2 }= [38 + 1][69] = 396951 = 50 + 1
51^{2 }= [25 + 1][1^{2}] = 2601 [Note: we wrote 01 as we need 2 digits in each bracket]47 = 50 – 3
47^{2 }= [25 – 3][3^{2}] = 2209 [Note: Subtract the difference to 25 to get the first 2 digits as the given number is lesser than 50]33 = 50 – 17
33^{2 }= [25 – 17][17^{2}] = [8][289], as the second part is more than 2 digit take the most significant number (here 2) and add it to the first part (8 ).
33^{2 }= [8+2][89] = 1089So now we know how to find squares faster. If just knowing was enough, life would have been much easier… right? We should be able to apply our strengths to our advantage in solving a situation… our confidence in calculating squares can help us in some quick multiplication… How?
We all know the basic formula (a + b) (a  b) = a^{2 }– b^{2}
87 * 93 = ( 90 – 3 ) * (90 + 3) = 90^{2 }– 3^{2} = 8091
similarly, 43 * 47 = 45^{2 }– 4 = 2025 – 4 = 2021
Ok.. Now solve 35 * 45 in 5 seconds :)
Multiplication
9 * 12 = ____ Ok, you got 108; question is how you did it?
Most of you might have done it using the logic 9 * 10 + 9 * 2 = 108.27 * 23 = ___
27 * 20 + 27 * 3 = 540 + 81 = 621634 * 126 = ___
634 * 126 = 634 * 100 + 634 * 26 = 63400 + 634 * 20 + 634 * 6
= 63400 + 12680 + 3804 = 79884
To find 634 * 6 we used 634 * 5 + 634, an easy way to multiply by 5 is multiply by 10, i.e. put a zero at the end, and divide by 2; 636 * 6 = 3170 + 634 = 3804)Division
Easiest way is to represent numerator as sum/ difference of terms related to denominator.
183 / 16 = ___
183 = 160 (16 * 10) + 16 (16 * 1) + 8 (16 * 0.5) – 1
183 / 16 = 10 + 1 + 0.5  x (some small value) ≈ little less than 11.5 (approx)3840 / 23 = ____
3840 = 2300 + 1150 + 230 + 115 + 45
3840 / 23 = 100 + 50 + 10 + 5 + (little less than 2) ≈ little less than 167There will be a lot of calculations based on fractions waiting for you, mostly in data interpretation section. It’s very important to have a high comfort level with such numbers. Keep below fractions handy.
Let’s play!
Which is greater, 10/9 or 7/6?
10/9 = 10 * 0.111 = 1.11
7/6 = 7 * 1/2 * 1/3 = 7 * 0.5 * 0.333 = 3.5 * 0.333 = 1.1655
7/6 > 10/9Another method is 10 = 9 + 0.9 + 0.09 + …
10/9 = 1 + 0.1 + 0.01 + … = 1.1117 = 6 + 0.6 + 0.36 + 0.03 + …
7/6 = 1 + 0.1 + 0.06 + 0.005 +... = 1.165 (approx)Personally I feel second method is better as it’s a generic way of tackling fractions and percentages. You just need to brush up your addition skills on two and three digit numbers. Represent numerator in terms of denominator. That’s all it takes.
321/562 = ____
321 = 281 (562/2) + 28.1(562/20) + 11.24(562/50) + …
321/562 = 0.5 + 0.05 + 0.02 + … = 0.57(approx)562/321 = ___
562 = 321 + 160.5 (321/2) + 80.25 (321/4) + …
562/321 = 1 + 0.5 + 0.25 = 1.75 (approx)You will find these methods very useful while dealing with calculation intensive DI problems or number system questions in quant.If you feel this method is helping (or can help) to save some time in calculation, practice well so that these stuffs will come automatically when you solve a question. Reduce your pen usage just to write the sub values of an expression and not to write the steps.
Here we are not using any speed math techniques, but we are trying to do smart math. Forget the “magic trick” of finding the square of 4 digit numbers by drawing circles or finding the square root by connecting dots. Stick to the basics and make it strong which will be good enough for kind of calculations expected in our aptitude exams. Everything we need to win an aptitude exam is taught to us when we were in high school... We just need to brush it up.

Concept: f(x) = x^1/x is maximum at x = e ( e = 2.71) and f(x) will decrease if you go either side of x = e
We can use the same trick to compare a^b and b^a
Closer the base (a or b) is to e, higher the value will be.example:
9.1 ^ 8.9 and 8.9 ^ 9.1
both these values are greater than e, and 8.9 is closer to e than 9.1, so 8.9^9.1 is higher than 9.1^8.9another classic one,
which is greater, e^PI or PI^e ?
we know e = 2.7 and PI = 3.14 => e^PI is greater.This trick is especially handy when all the given numbers belongs to the same side of e, where we can do the comparison in no time.

Some useful properties of numbers
Sum of first n natural numbers = n * (n + 1) / 2
Sum of first n odd natural numbers is equal to n^{2}
Sum of first n even natural numbers is equal to n (n + 1)
Sum of squares of first n natural numbers = n (n + 1) (2n + 1) / 6
Sum of cubes of the first n natural nos. = [n (n + 1) / 2] ^{2}
Goldbach's conjecture Every even integer greater than 2 can be represented as a sum of two prime numbers ; 12 = 5 + 7
Any positive integer can be written as the sum of four or fewer perfect squares.
14 = 4 + 1 + 9 (Lagrange's foursquare theorem)Any prime number of the form (4n + 1) can be written as sum of two squares.
17 = 16 + 1, 41 = 25 + 16A perfect square can end only with digit 1, 4, 6, 9, 5 or even number of zeros.
Adding the digits of perfect square will always yield 1, 9, 4 or 7. A number is not a perfect square if its digit sum does not evaluate to one of these. Number may or may not be a perfect square if its digit sum is 1, 9, 4 or 7.
Digit sum (1936) = digit sum (1 + 9 + 3 + 6 = 19) = digit sum (1 + 9 = 10) = 1All prime numbers can be represented as (6n1) or (6n+1). So if we cannot represent a number in (6n+1) or (6n1) form we can say it is not a prime. The reverse may not be true, i.e. all numbers that can be represented as (6n1) or (6n+1) need not be a prime.
Any odd number is a difference of two consecutive squares
(k + 1)^{2}= K^{2}+ 2k +1^{2}, so (k + 1)^{2} k^{2}= 2k + 1 (an odd number)Any multiple of 4 is a difference of the squares of two numbers that differ by two
(k + 2)^{2}– k^{2}= 4k + 4 = 4 (k +1)3^{0}, 3^{1}… 3^{n} can produce all integers from 1 to 1 + 3^{1} + ... + 3^{n}
To find LCM and HCF of (a/b) and (c/d) the generalized formula will be:
H.C.F = H.C.F of numerators / L.C.M of denominators
L.C.M = L.C.M of numerators / H.C.F of denominatorsa x b = HCF (a, b) x LCM (a, b)
If a and b are co primes then HCF (a, b) = 1
So for co prime numbers, a x b = LCM (a, b)

How many ways can you express 360 as product of 2 coprimes
Unordered : 2^(n1)
Ordered : 2^nHow many factor pairs of 360 exist which are coprimes to each other
Let me explain the concept of factor pair with a small example.
let say you need coprime 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 itFactor 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

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)  1Practice 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 < = 1000Should help at a lot of places. So practice well if you are starting your preparations.

How to find the Number of digits in a^b ?
Number of digits in a^b = [1 + log (a^b)]
We will calculate the value of log(a^b) in base 10Example : 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

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 )

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 = 16In the above case, sum of divisors is ((a^(m+1) )1)/(a – 1) * ((b^n+1)1)/(b1) * ((c^(p+1) 1 )/(c1)
Find the sum of divisors of 120
120 = 2^3 * 3 * 5
sum = 2^4 – 1 / 1 * 3^21/2 * 5^2 – 1 / 4 = 15 * 4 * 6 = 360The 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^(m1) 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

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^(m1) + 3^(m2) + …. + 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

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 pickup 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 pickup to ensure he wins
If a starts his targets will be 50, 42, 34, 26, 18, 10, 2. So he must pickup 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 5048 = 2If 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  1This is an example and here B is helpless.

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(11/2) ( 11/5)In general a positive integer N having prime divisors p1, p2, p3 … pn
E[N] = N(11/p1)(11/p2) ... (11/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 = 9Sum 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, (P1)! +1 is divisible by P
(P1)! Mod P = 1 or (P1)
(P2)! Mod P = 1
(p3)! Mod P = (p1)/216! = (16!+1)1 = (16!+1)+1617
16! Mod 17 = 0 + 16  0 = 1615! Mod 17 => 16!mod 17 = 16
15! * 16 mod 17 = 16
15! Mod 17 =114! Mod 17 > 15! Mod 17 = 1
14! * 15 mod 17 = 1
14! * 2 mod 17 = 16
14! Mod 17 = 8When 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 = 64Negative remainder
When a number N < D gives a remainder R When divided by D , it gives a negative remainder of RD
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 24021 = 2401Fermat’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

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 61Last 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 =>89Last 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 => 16Last two digits of Number ending in 5
There are two cases
 Numbers where previous digit of 5 is 0 or any even number
 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 evenLast non zero digit of 36!
Z(36) = 4 * Z(7) * Z(6) = 4 * 4 * 2 = 32. so answer 2Last non zero digit of 15!
Z(15) = 4 * Z(3) * Z(5) = 4 * 6 * 2 > 48. hence answer 8To 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^YCase 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 1Find 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 = (121)^13^17 = (1)^13^17 = 1^(odd number) = 1
Hence remainder will be 41 =3
So unit digit of 7^11^13^17 will be same as 7^3 which is 3

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 = 16Find 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 21Find 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

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 = 5A 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

hemant_malhotra last edited by
Director at ElitesGrid  CAT 2017  QA 100 Percentile / CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
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 = 4Find 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 = 2Find 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 digitsNOTE 1 last two non zero digits of
5! = 12
25! = 84
125! = 88Last non zero digits
(10!) = 8
(20!) = 4
(30!) = 8
(40!) = 2
(50!) = 2
(60!) = 6
(70!) = 8
(80!) = 8
(90!) = 2
(100!) = 4Example : 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

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.

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 × 10So 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 factors1)/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 6×6Example: 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 = (211)/2 = 10.
Note since the word ‘distinct’ has been used therefore we do not include 576 = 26 × 26.

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 numberNow for single digit numbers, the whole thing can be divided in 3 categories
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 64, 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 12, 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 7Find 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 4x51 (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 = 1Find 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^…..) = 2Find the last digit of: 2468^ (4682^6824^8246)
8^(2^even number..)
8^4 = 6
So 6

Finding the second last digit
Cyclicity
There lies the cyclicity of tens' place digit of all the digits. This is given below:
Digits Cyclicity 2, 3, 8 20 4, 9 10 5 1 6 5 7 4 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 23Example 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 = 16SHORT 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