Number System questions 1

1. When x is a prime number                      a^x - a is always divisible by x.


Q1. What is remainder when 5¹⁰¹ - 5 is divided by 101.

(1) 0   (2)  100   (3) 5 (4)  50  (5) 20 

Solution :- As concept states, it is divisible by 101. Hence remainder is zero.



Q2. 91¹⁰¹ -91 is divisible by numbers except ?



(1) 91   (2) 13   (3) 7   (4) 101   (5) Divisible by all above


Solution :- As concept states, it is divisible by 101 hence option 5 is eliminated.

91¹⁰¹ -91 = 91(91¹⁰⁰ -1), hence divisible by 91

91 = 13 * 7 hence option 5.
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2. When a base of a number increases, the number decreases and vice-versa.
Q3. (2345)₆ is equal to

(1) (569)₁₀ (2) (3458)₁₀  (3)  (4286)₁₀  (4) (569)₅ 

Solution :- Only option 1 follows concept. No need to solve.


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 3. Relation between a two digit number and number formed using its digits.

    ab = 10a +b                     &
    ba = 10b + a

hence ab - ba = (10a +b) - (10b + a) = 9(a-b)
  &     ab + ba = (10a +b) + (10b + a) = 11(a+b)


Q4. The difference between a two digit number and number formed by reversing its digits is 45. What is difference between the digits of its number ?

(1)  9  (2)  5  (3)  3  (4)  4  (5) None of these


Solution :- According to concept 9(a-b) = 45 hence (a-b)  5. Hence option 2.
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4.IMP - Divisible count in a range.
Steps to follow.
I. Subtract extreme numbers & divide by number (x)
II. Check remainder (n)
III. Check first n digits of range, if one of the number is divisible then add 1 to x.


Q5. How many three digit numbers are divisible by 7 ?

(1)  128  (2)  129  (3)  127  (4)  142  (5) 140

Solution :-
Extreme three digit numbers are 100 - 999
I. 999 - 100 = 899 the 899 / 7 = 128 * 7 + 3
II. Remainder = 3
III. None of the first three digits (100,101 & 102) are divisible by 7 hence answer is option 1.


Q6. How many four digit numbers are divisible by 7 ?

(1)  1284  (2)  1285  (3)  1286 (4)  1428  (5) 1287

Solution :-
Extreme four digit numbers are 1000 - 9999
I. 9999 - 1000 = 8999 the 8999 / 7 = 1285 * 7 + 4
II. Remainder =4
III.Out of the first four digits (1000,1001,1002 & 1003) number 1001 is divisible by 7 hence answer = 1285 + 1 = 1286. Hence option 3.
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5. Product of three consecutive numbers is always divisible by 6.
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6. To find highest power of a number in factorial .
I. Factorize the number.
II. Find the power of Individual factor in factorial by simply dividing number continuously with ignoring remainder. Add all the numbers.


Q7. Find highest power of 15 in 30!

(1)  2  (2)  6  (3)  21  (4)  7  (5) 14

Solution: Prime Factors of 15 = 3 * 5.
To find highest power of 15 just find highest power of 5.
Highest power of 5 = 30/5 + 6/5  (ignore remainders for next calculation)
= 6 + 1 = 7.
As proved highest power of 15 in 30! is 7.Hence answer option 4.


Q8. Find highest power of 12 in 30!.

(1)  3  (2)  26  (3)  14  (4)  13  (5) 12

Solution: Prime Factors of 12 = 2² * 3.

Highest power of 2 = 30/2 + 15/2 + 7/2 + 3/2  (ignore remainders for next calculation)
=15 + 7 + 3 + 1 = 26 => 13 pairs of 2.
Highest power of 2² = 13.


Highest power of 3 = 30/3 + 9/3 + 3/3   (ignore remainders for next calculation)
= 10 + 3 + 1 = 14.

As proved highest power of 12 in 30! is 13.Hence answer option 4.
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 7. Sum of numbers of numbers formed using n digits.

Sum = Sum of digits * (n-1)! * 1111.... n times.
                    provided no digit is zero & digits are not repeated. 


Q9. Find sum of all digits formed using 2,3,4 & 6 such then digits are not repeated.

(1)  99990  (2)  62525  (3)  99900  (4)  98900  (5) None of these.

Solution :-
Since zero is not one of digits & digits are not repeated, we can use formula

Sum = Sum of digits * (n-1)! * 1111.... n times.

sum =  15 * 3!* 1111 = 99990. Hence option 1.

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 8. All numbers which are perfect squares of odd numbers are of the form 8k+1 but vice-versa not true.
OR a perfect square of a odd number leaves a remainder of 1 when divided by 8.
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9. A four digit number formed with its 2 digits repeating (abab) = ab * 101
     A six digit number formed with its 3 digits repeating (abcabc) = abc * 1001 = abc * 11 * 13 * 7


(Very frequently asked question pattern in any exams. Concept saves a lot of time)


Q10. Number 2828 is divisible by except

(1)  2  (2)  7  (3)  14  (4) 101   (5) 11


Solution :- I know your reaction would be "Wow i can answer this without calculations, answer is option 5.



Q11. Number 123123 is divisible by except

I. 123
II. 11
III. 77
IV. 143
V. 91


(1) Statement I, II but not III, IV.
(2) Statement II, III but not I.
(3) Statement I, II, IV but not III.
(4) Statement I, II, III, IV but not V.
(5) Divisible by all



Solution :
123123 = 123 * 1001 = 123 * 7 * 11 * 13 = 123 * 7 * 143 = 123 * 77 * 13 = 123 * 91 * 11
hence answer option 5.
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10.Number of Factors of a number is determined by power of its prime factors.


Number form in prime factors = a^x * b^y * c^z  ....
where a, b , c ... are prime numbers. then

Number of factors = (x+1)(y+1)(z+1)....


Q12. Find the number of factors of number 96 ?

Solution:  N = 96 = 2⁵ *3¹   
 Number of factors = (5+1)(1+1) = 6 * 2 = 12


Note:
1. In a perfect square power of prime factors are always even and they always have odd number of factors.
2. In a perfect cube power of prime factors is always multiple of three.


Q13. How three many numbers with odd number of factors are divisible by 3 ?

(1)  7  (2)  21  (3)  149  (4) 150   (5) 299

Solution : Only perfect squares have odd number of factors. And three digit perfect squares divisible by 3 are 144, 225, 324, 441, 576, 729 & 900. Hence option 1.

Q14. Which of the following can not be number of factors of a number which is a perfect cube ?
(1)  4  (2)  16  (3)  7 (4) 28   (5)9

Solution : Power of prime factors of a perfect cube are always multiple of 3.



let a & b any prime factors. then



a³ then number of factors = 3 + 1 = 4

a⁶ then number of factors = 6 + 1 = 7
a³ * b³ then number of factors = 4 * 4 = 16
a⁶ * b³ then number of factors = 7 * 4 = 28
hence option 5.

Or We know that only perfect square can have odd number of factors so now we are only left with option 3 & 5. We can easily find the number a⁶ is a perfect cube as well as perfect square. Hence option 5.


Q15. Find the number of factors in 6⁶ - 5⁶
 (1)  4  (2)  8  (3)  16 (4) 36   (5) 49


Solution:
6⁶ - 5⁶  =  (6³)²-  (5³)²
 =  (216)²-  (125)²   =   91 * 341 = 13 * 7 * 31 * 11

Number of factors = 2 * 2 * 2 * 2 = 16. Hence answer option 3.
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11. Number of ways in which a number can be written as a product of two co-prime numbers = 2^M-1. where M is number  of prime factors.

Q16. In how many number of ways 540 can be written as a product of two co-prime numbers ?
(1)  1  (2)  2  (3)  3 (4) 4   (5) 6


Solution:- 540 = 2² * 3³ * 5¹
 Number of ways in which a number can be written as a product of two co-prime numbers
= 2^{M-1 =} 2^M-1  = 2^3-1 = 4. Hence option 4.
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12. A perfect square will never end in 2,3,7 &8.

CET & SNAP Pattern
 Q17. Which of the following is not a perfect square ?
 (1)  2025  (2)  3844  (3)  6568 (4)  7056  (5) 3025


Solution: - As concept states answer is option 3. ( Don't try to calculate in exams.)
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13. A square of n digit number will have either 2n or 2n-1 digits.

Q18. Which of the following number , with which some of digits expressed as 'x' can be a perfect square of 5 digit number ?

(1) 30xxxxxxxxx8 
(2) 30xxxx25
(3) 4xxxxxx6
(4) 1xxxxxxx0
(5) 4xxxxx25


Sol:- Square of a 5 digit number will have 9 digits or 10 digits. Hence option 2, 3 & 5 are eliminated.
A square can not end with 8 so option 1 is eliminated. Hence option 4 is answer.
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14. Series  type.

I. Sum of first 'n' natural numbers = n(n+1) / 2
II. Sum of first 'n' odd numbers = n²
III. Sum of first 'n' even numbers = n² + n
IV. Sum of the square of first 'n' natural numbers = n(n+1)(2n+1)/6
V. Sum of the cube of first 'n' natural numbers = (n(n+1) / 2)²

Q19. What is sum of all three digit numbers lesser than 251 ?


(1)  4950  (2)  14825  (3)  26250 (4)  26425  (5) 31375

Sol:-
Sum of all digits from 1 - 250(Series formula I) = 250 * 251 / 2 = 31375
Sum of all digits from 1-99 = 99 * 100 / 2  = 4950
Sum of all three digits numbers less than 251 i.e. 100-250 = 31375 - 4950 = 26425
Hence answer option 4.


Or
Total number between 100 - 250 = 151
Avg of numbers = 175
sum of numbers = 151 * 175 = 26425


Q20. What is sum of all numbers which are perfect cube and less than 8001  ?
(1) 21000  (2)  32600  (3)  36800 (4) 42100   (5) 44100
Solution :-
8000 is cube of 20. Hence question is asking of sum of all first 20 cubes. Which can be calculated using 5th formula.
sum = (20(20+1) / 2)²
= (20(21) / 2)²
= 210 * 210
= 44100. Hence answer option 5.
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Indices and Surds

Indices

1. If x^N is given then x is the base and N is the index or power or exponent
2. A³ means A multiplied with itself 3 times i.e. A x A x A
3. A ^p X a ^q = a ^(p+q)
4. A ^p / A ^q = A ^(p-q)
5. (A ^p)^q = A ^pq
6. A ^–p = 1 / A^p
7. p√a = a ^1/p i.e. pth root of a
8. (ab)^p=a^p x b^p
9. A⁰ = 1 (provided a ≠0)
10.  A¹ = a
11. _A mⁿ = a ^p, where p = mⁿ ie. A raised to the base m raised to the power
12. If a ^p = b ^p, then if p is ≠0, then a = b, if p is odd and [a = b or a = (-b) if p is even]
13. If a ^p = a ^q and a ≠ 0 or -1, then p = q
14. A ^-1 = 1/A
15. (A/B)^-1 = B/A
16.  (A)^m/n = (n√A)^m
17.  √A X √B = √(AB)

Surds

1. They are irrational numbers
2. When an irrational number is simplified, the remainder which cannot be simplified and is normally expressed in the form of square root is called a surd.
3. Normally for exam questions, number whose square root cannot be further found out as a perfect rational number are surds. 4 is not a surd, as square-root of 4 is 2, where 2 is a surd as square root of 2 is 1.414… which is not a rational number.
4. To solve simplification problems regarding surds, square the numbers
5. For 1 / (p + √q) or 1 / (p + √q + √r) kind of problems, to simplify, multiply by the conjugate, which is (p – √q) or (- p + √q) for 1^st case and (p + √q – √r) or (p – √q + √r) for the 2^nd case
6.  = .555555 hence, whenever in a decimal form there is a repeated number; a dot is mentioned over it.
7. Rationalising Surds:
   When you have a fraction where both the nominator and denominator are surds, rationalising the surd is the process of getting rid of the surd on the denominator. To rationalise a surd you multiply top and bottom by fraction that equals one. Take the example shown below
   1/√2
   To rationalise this multiply by effectively 1
   1/√2 * √2 /√2
   Can you see why √2 /√2 was chosen? This is because √2 * √2 = 2 so the denominator becomes surd free.
   
   For a more complex term
   
   
   
   
   Rationalizing the surd now
   
   
   
   

Practice-Divisibility Rules 1

 Practice-Divisibility Rules (10 Qs)

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q1. A number when divided by 13 gives a natural number, but when divided by 23 gives a real number. The smallest such number is

(1) 0         (2)   1     (3)   437    (4)  13   (5)   23

Sol:-Option 4. Very easy one. To get a natural number it has to be least 11. Real number can be a fraction hence answer option D.

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q2. The  number 276A621B is divisible by 9. How many possible values can A+B take ?

(1) 0         (2)  1     (3)   2    (4)  3   (5)  Infinite

Sol:- Option 3.  To be divisible by 9, sum of digits must be divisible by 9.

sum of digit = 2+7+6+A+6+2+1+B = 24+B

hence A+B can be 3 or 12. Can't be more as A & B can max be 9.

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q3. A number 2524232221..........109876543210 is divided by 8. What is remainder ?

(1) 0         (2)  2     (3)   4    (4)  6   (5) 1

Sol:- Option 2. To check divisibility by 8, we are only concerned about last 3 digits. Remainder of 210 divided by 8 is 2.

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q4.  A number is divisible by 12 and 10.Which of the following is/are false ?

 I. Number is also divisible by 15.

II. Number is also divisible by 30

III. Smallest such number is 120.

IV. Number is always co-prime with 7.

(1) Statement I & II

(2)  Statement III

(3) Statement IV

(4)  Statement III & IV

(5) None of the statement.

Sol:-Option 4.

Number is divisible by 12 & 10 hence divisible by 2,3,4,5,6 and their product.

Statement I - True as it is divisible by 3*5.

Statement II - True as it is divisible by 2*3*5. (Also If Statement 1 is true, this one has to be true)

Statement III - False, smallest such number is 60.

Statement IV - False, 7 is not factor of 12 or 10 but still a number can be formed eg. 840

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q5. 14MN27 is divisible by 9. What can be maximum value of M if (M+N) is at its minimum value ?

(1) 0         (2)  2     (3)  4    (4)  8   (5) 9

Sol:- Option 3. Number is divisible by 9 hence M+N can be 4 or 13. As M+N is at its minimum,  M can maximum be 4 when N is zero. __________________________________________________________________________________

q6. How many four digit numbers can be formed using  the digits 2,3,5,7 exactly once, such that number is divisible by 25 ?

 (1) 0         (2)  1     (3) 2    (4)  3   (5) 4

Sol :- Option 5. Number should have last two digits as 00, 25, 50 or 75 to be divisible by 25. Such possible numbers are 3725, 7325, 2375 & 3275.

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q7.  If a positive integer n is divided by 7, the remainder is 4. Which of the numbers below yields a remainder of 0 when it is divided by 7?

(1) n + 3         (2)  n + 2     (3) n - 1    (4)  n - 2   (5) n + 1

Sol:-  n divided by 7 yields a remainder equal to 4 is written as follows
n = 7 k + 4 , where k is an integer.
add 3 to both sides of the above equation to obtain
n + 3 = 7 k + 7 = 7(k + 1)
The above suggests that n + 3 divided by 7 yields a remainder equal to zero. The answer is Option 1.

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q8.If n is an integer, when (2n + 2)² is divided by 4 the remainder is

  (1) 0         (2)  1     (3) 2    (4)  3   (5) 4

Sol:-
We first expand (2n + 2)²
(2n + 2)² = 4n ² + 8 n + 4
Factor 4 out.
= 4(n ² + 2n + 1)
(2n + 2)² is divisible by 4 and the remainder is equal to 0. The answer is Option 1.

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q9.Which of these numbers is not divisible by 3?

 (1) 339      (2)  342     (3) 552    (4)  1111   (5) 672

Sol:-   One may answer this question using a calculator and test for divisibility by 3. However we can also test for divisibility by adding the digits and if the result is divisible by3 then the number is divisible by 3.

3 + 3 + 9 = 15 , divisible by 3.

3 + 4 + 2 = 9 , divisible by 3.

5 + 5 + 2 = 12 , divisible by 3.

1 + 1 + 1 + 1 = 4 , not divisible by 3.

The number 1111 is not divisible by 3 the answer is Option 4.

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q10.  How many 4 digit numbers can be formed using 5, 8, 3, 4 exactly once, such that the number is divisible by 11 ?

 (1) 16     (2) 5     (3) 1     (4) 22     (5) None of these

Sol:- 

Dividing by 11
The difference between the sum of the odd numbered digits (1st, 3rd, 5th...) and the sum of the even numbered digits (2nd, 4th...) is divisible by 11.

No such number can be formed. Hence Option 5.

Cyclicity & Remainders (for CAT)

Number Cyclicity:-

Number cyclicity is very important for every chapter of the quant. Any exams will definitely need the basics of number cyclicity. Questions can directly be asked on this topic and you will need this to solve many questions on other topics.

Don’t forget time is biggest constrain for any entrance exam and cyclicity will have one to save time.

Cyclicity is basically use to find the unit digit or tens digit of the number.

Unit Digit Cyclicity:-

Q1. Find the unit digit of 2^2548.

Sol: - You will need more than 6 hours to solve this problem if you don’t use cyclicity theorem.

We notice that

2^1 end with 2

2^2 end with 4

2^3 end with 8

2^4 end with 6

2^5 end with 2

2^6 end with 4

2^7 end with 8

2^8 end with 6

We notice that 5^th power end in 2 and number repeats after 4 powers. Hence cyclicity for 2 is 4. It will always end with 2, 4, 6 and 8.

So Divide 2548 by 4 and we get remainder = 2

Hence unit digit of 2^2548 with be 4.

Remember:-

When exponent is 1 number ends with 2

When exponent is 2 number ends with 4

When exponent is 3 number ends with 8

When exponent is 0 or 4 number ends with 6.

Similarly we can find of all other numbers.

Number	Unit digit of Cyclicity
0,1,5,6	1
4,9	2
2,3,7,8	4

Tens Digit Cyclicity:-

Similarly we can arrive for tens digit cyclicity.

Number Tens Digit Cyclicity

Number	Unit digit of Cyclicity
0,1,5,6	1
7	4
6	5
4,9	10
2,3,8	20

Base System

Numbers : Basic Operations on Numbers

Let a and b any two numbers then find the results of basic operations below.

Points to remember:-
1. Sum and Difference of any two odd or two even numbers is always even.
2. a^b always depends on a.
ie Even raise to any number is even and Odd raise to any number is odd.
eg 2^ 3 = 8 and 3^ 2 = 9.
3. If atleast one number is even - Product is always even.

Numbers : Converting a recurring decimal to vulgar fraction

A decimal with recurring value is called recurring decimal.
E.g: 2/9 will give 0.22222222...... where 2 is recurring number.

Method:
a) Separate the recurring number from the decimal fraction.
b) Annex denominator with "9" as many times as the length of the recurring number.
c) Reduce the fraction to its lowest terms.

Example: Consider 0.2323232323

Step a: The recurring number is 23
 Step b: 23/99 [the number 23 is of length 2 so we have added two nines] 
Step c: Reducing it to lowest terms : 23/99 [it can not be reduced further].

How to Convert a mixed-recurring decimal to vulgar fraction ? A decimal with both recurring and non-recurring value is called mixed recurring decimal.
E.g: 28/25 will give 1.1199999999...... where 11 is non-recurring number and 9 is recurring number.

Method:
a) Separate the recurring number, non recurring number from the decimal fraction.
b) Round the decimal after point to the first recurring value.
c) Result of step b - non recurring number.
d) Annex as many "0" as non-recurring number length and as many "9" as recurring number length.
e) Step c / Step d
f) Add the fraction with the number before decimal point.
Example: Consider 1.11999999...

Step b and c: 119-11 [rounded value of number after decimal point - non recurring value]Step d: 900
Step e: 108/900 [c/d]
Step f: 1+¹⁰⁸/₉₀₀ [adding with number before decimal point ]

Reducing it to lowest terms : 900+108 / 900 = 1008/900 = 28/25.Step a: The recurring number is 11, non-recurring number is 9

Special Number - Ramanujam’s Number

It is the smallest number which can be written as the sum of two cubes in two different ways. It is a very famous number, so if you dare to proclaim maths as your hobby, you must know it.
Number is 1729=10^3+9^3=1^3+12^3.

Ramanujam Number is 1729.

But allowing negative perfect cubes (the cube of a negative integer) gives the smallest solution as 91 (which is a factor of 1729):

91 = 6^3 + (−5)^3 = 4^3 + 3^3

There is one interesting story on this. Do google search if you have enough time.

Set theory for cat

27) n(A U B U C) = n(A) + n (B)+ n(C) – n(A n B) – n(A n C) – n(B n C) + n(A n B n C)
28) n(A_only) = n(A) – n(A n C) – n(A n B) + n(A U B U C)
To refresh, the union of sets is all elements from all sets. The intersection of sets is only those elements common to all sets. Let’s call our sets A, B, and C. If n = intersection and u = union. The need-to-know formulas:
P(A u B u C) = P(A) + P(B) + P(C) – P(A n B) – P(A n C) – P(B n C) + P(A n B n C)
To find the number of people in exactly one set:
P(A) + P(B) + P(C) – 2P(A n B) – 2P(A n C) – 2P(B n C) + 3P(A n B n C)
To find the number of people in exactly two sets:
P(A n B) + P(A n C) + P(B n C) – 3P(A n B n C)
To find the number of people in exactly three sets:
P(A n B n C)
To find the number of people in two or more sets:
P(A n B) + P(A n C) + P(B n C) – 2P(A n B n C)
To find the number of people in at least one set:

P(A) + P(B) + P(C) – P(A n B) – P(A n C) – P(B n C) + 2 P(A n B n C)
For questions involving set theory, it may be helpful to make a Venn diagram to visualize the solution.
To find the union of all set: (A + B + C + X + Y + Z + O)
Number of people in exactly one set: (A + B + C)
Number of people in exactly two of the sets: (X + Y + Z)
Number of people in exactly three of the sets: O
Number of people in two or more sets: (X + Y + Z + O)

Number of Divisors of number

Number of divisors/factors of the number
Let the number be n = a^m * b^n * c^p
 (where a, b, and c are n's prime divisors and m,n, and p are the number of times that divisor is repeated) then the total count for all of the divisors is (m + 1) * (n + 1) * (p + 1).

Sum of factors

 Sum = (a ₀ + a¹ + ...+a ^m-1)(b ₀ + b¹ + ...+b ^n-1)(c ₀ + c¹ + ...+c ^p-1)

i.e.
= (a^m+1 - 1)/(a-1) * (b ⁿ⁺¹ - 1) / (b-1) *(c ^p+1 - 1)/ (c-1)

HCF and LCM

Prime Factorization:

A prime factorisation of a natural number can be expressed in the exponential form.
For example:(i) 48 = 2 x 2 x 2x 2 x3 = 24 x 3(ii) 420 = 2 x 2 x 3 x 5 x 7 = 22 x 3 x 5 x 7

Highest Common Factor(H.C.F) or Greatest Common Meaure(G.C.M) or Greatest Common Divisor(G.C.D) :

The H.C.F of two or more than two numbers is the greatest number that divides each of them exactly.
There are two methods :


Method 1 - Factorization method: Express each one of the given numbers as the product of prime factors.

The product of least powers of common prime factors gives HCF.

Example : Find HCF of 2⁶ * 3²*5*7⁴ , 2² *3⁵*5² * 7⁶ , 2²*5 *7²

Sol: The prime numbers given common numbers are 2,5,7

Therefore HCF is 2² * 5 *7² .


Method 2- Division Method : Divide the larger number by smaller one. Now divide the divisor by remainder. Repeat the process of dividing preceding number last obtained till zero is obtained as
number. The last divisor is HCF.


Example: Find HCF of 513, 1134, 1215


Sol:

1134) 1215(1
1134
----------
81)1134(14
81
-----------
324
324
-----------
0
-----------
HCF of this two numbers is 81.

81)513(6
486
--------
27)81(3
81
-----
0
---

HCF of 81 and 513 is 27.

Least common multiple[LCM] : The least number which is divisible by each one of given numbers is LCM.
A common multiple is a number that is a multiple of two or more numbers. The common multiples of 3 and 4 are 0, 12, 24, ....



The least common multiple (LCM) of two numbers is the smallest number (not zero) that is a multiple of both.



Method 1:Simply list the multiples of each number (multiply by
2, 3, 4, etc.) then look for the smallest number that appears in each
list.

Example: Find the least common multiple for 5, 6, and 15.

Multiples of 5 are 10, 15, 20, 25, 30, 35, 40,...

Multiples of 6 are 12, 18, 24, 30, 36, 42, 48,...

Multiples of 15 are 30, 45, 60, 75, 90,....

Now, when you look at the list of multiples, you can see that 30 is the
smallest number that appears in each list.Therefore, the least common
multiple of 5, 6 and 15 is 30.



Method 2

To use this method factor each of the numbers into primes. Then for
each different prime number in all of the factorizations, do the
following...

1. Count the number of times each prime number appears in each of the factorizations.

2. For each prime number, take the largest of these counts.

3. Write down that prime number as many times as you counted for it in step 2.

The least common multiple is the product of all the prime numbers written down.



Example: Find the least common multiple of 5, 6 and 15.

Factor into primes

Prime factorization of 5 is 5

Prime factorization of 6 is 2 x 3

Prime factorization of 15 is 3 x 5

· Notice that the different primes are 2, 3 and 5.

Now, we do

Step #1 - Count the number of times each prime number appears in each of the factorizations...

The count of primes in 5 is one 5

The count of primes in 6 is one 2 and one 3

The count of primes in 15 is one 3 and one 5



Step #2 - For each prime number, take the largest of these counts. So we have...

The largest count of 2s is one

The largest count of 3s is one

The largest count of 5s is one



Step #3 - Since we now know the count of each prime number, you simply -
write down that prime number as many times as you counted for it in
step 2.

Here they are...2, 3, 5



Step #4 - The least common multiple is the product of all the prime numbers written down.

2 x 3 x 5 = 30 Therefore, the least common multiple of 5, 6 and 15 is 30.

So there you have it. A quick and easy method for finding least common multiples.

 Co-prime numbers : Two natural numbers are called co-prime numbers if they have no common factor other than 1.
in other words, two natural numbers are co-prime if their H.C.F. is 1.

Some examples of co-prime numbers are: 4, 9; 8, 21; 27, 50.

Relation between L.C.M. and H.C.F. of two natural numbers

The product of L.C.M. and H.C.F. of two natural numbers = the product of the numbers.

Note. In particular, if Two natural numbers are co-prime then their L.C.M. = The product of the numbers.
Important
For LCM
1. LCM >= the largest number of a set.
2. For Co-prime numbers - LCM is product of co-prime numbers.
3. LCM is multiple of all the numbers of a set as well as HCF of a set of numbers.


For HCF
1. HCF =< the largest number of a set.
2. For Co-prime numbers - HCF is equals to 1.
3. HCF is factor of all the numbers of a set as well as LCM of a set of numbers.

 

Prime, Coprime and Composites

Prime Numbers
A natural number larger than unity is a prime number if it  does not have other divisors except for itself and unity.
Note:-Unity i e,1 is not a prime number.

 Properties Of Prime Numbers:
->The lowest prime number is 2.
->2 is also the only even prime number.
->The lowest odd prime number is 3.

–> 2 and 5 are the only primes that end in 2 or 5.  

Remember :  

The remainder when a prime number p>=5 s divided by 6 is 1 or 5.
However, if a number on being divided by 6 gives a remainder 1 or 5 need not be prime. 

The remainder of the division of the square of a prime number p>=5 divided by 12 is 1. 
 

Remember 

 >If a and b are any 2 odd primes then a²-b² is composite. Also a²+b² is composite.

Remember:
If x is a prime number then number of the form (a^x – a) is always divisible by x.



Remember:
A composite number n which is not a perfect square has one factor
greater than √ n one less than √ n , product of these factors will give
the number.



Remember:
If p is a prime number such that  the

p^2 = 6K +1  where p>3
p^2 = 24k + 1  where p >= 5
In short p^2 - 1 is divided by 6 and 24


One just needs to remember the above rule. Still for deriving that formula one can Google to get it.

->Goldback's Conjecture
- if n is an even number greater than 2, then there are always 2 prime numbers whose sum is n
 

Process to Check A Number s Prime or not:
Take the square root of the number.
Round of the square root to the next highest integer call this number as Z.
Check for divisibility of the number N by all prime numbers below Z. 
If there is no numbers below the value of Z which divides N then the number will be prime.

Example 239 is prime or not?
√239 lies between 15 or 16.Hence take the value of Z=16.
Prime numbers less than 16 are 2,3,5,7,11 and 13.
239 is not divisible by any of these. Hence we can conclude that 239 
is a prime number.


Composite Numbers
The numbers which are not prime are known as composite numbers.

Except for 0 and 1, a number is either a prime number or a composite number. A composite number is defined as any number, greater than 1, that is not prime.

4 is first composite number.

Co-Primes (Relatively Prime Numbers):
Two numbers a an b are said to be co-primes,if their H.C.F is 1.
Example (2,3),(4,5),(7,9),(8,11).....

For instance (9,16)
Factors of 9 -> 1, 3, 9
Factors of 16 -> 1, 2, 4, 8, 16

Remember:-
1. One is co-prime with any other number. (1,2) , (1,3) …
2. Two Consecutive numbers are always co-prime.
3. A prime number is co-prime with all numbers except its multiple.
4. Two prime numbers are always co-prime. 

Twin Prime Numbers
Pair of prime numbers with the difference of 2 is called twin prime number.

Eg. (3,5) , (5,7) , (11,13) and go on.