Maths - Lesson 1
Decimal - II
2/3 = 0.66 3/4 = 0.75 4/5 = 0.8 5/6 = 0.833 6/7 = 0.857 7/8 = 0.875 8/9 = 0.888 9/10 = 0.9
Divisibility by 11
Difference of Sum of Digits at Odd place and Even place is 0 or divisible by 11
Finding the last 3 Digit of an expression
Divide by 1000
Q. Attached
c = s Seconds k = ? seconds Cross Multiplying ? = sk/c Seconds ? = sk/60c Minutes
Even roots of Negative Numbers are not asked in GRE
√(-2) -> Cannot be determined ¹⁰√(-2) -> Cannot be determined
Find Maximum and Minimum factorial where there are 13 Zeroes
13/5 = Quotient is 2 13 - 2 = 11 Hence Factorials will be 5*11 = 55! to 59! will have 13 Zeroes
Find number of Zeroes in 173!
173/5 = 34 , 34/5 = 6 , 6/5 = 1 Number of Zeroes = 34+6+1 = 41 Zeroes
Number of Zeroes in an expression
Find number of pairs of 5 and 2
Prime Numbers
Group 0 2, 3, 5, 7 Group 1 ( 1, 1+3 = 4, 4+3 =7) 11, 13, 17, 19 41, 43, 47 71, 73, 79 Group 2 ( 2, 2+3 = 5, 5+3 =8) 23, 29 53, 59 83, 89 Group 3 ( 3, 3+3 = 6, 6+3 =9) 31, 37 61, 67 97
Divisibility by 2 or 5
Last Digit of Number should be divisible by 2 or 5
Number of Zeroes from 10! to 14! is same and Number of Zeroes from 15! to 19! will be same
The same logic works for all of them
(ax + 1) power of n /a where x is any positive integer
Will always give remainder 1 . Ex - 7⁹/6 = (6*1+1)⁹/6 = 1
Formulaes
(a + b)³ = a³ + b³ + 3ab(a + b); (a − b)³ = a³ − b³ − 3ab(a − b); a³ − b³ = (a − b)(a² + ab + b²) a³ + b³ = (a + b)(a² - ab + b²)
Q. Find the number of numbers between 100 to 200 *if Neither 100 nor 200 is counted/included*
200 - 100 - 1 = 99
Number of Prime Number between 1 and 100
25 Numbers
When you are dividing EVEN and ODD ODD ÷ ODD = ODD EVEN ÷ EVEN = May be ODD or EVEN ODD ÷ EVEN = Not Integer
Keep in Mind that ODD is 2n + 1 hence would not have factor of 2 Even is 2n and will definitely have 2 in factor
Difference between any number and the number obtained by writing its digit in reversed will always be divided by 9
True
Odd roots of Negative Numbers
³√(-27) = -3 Possible ⁵√(-32) = -2 Possible
Cube Roots
³√2 = 1.3 ³√3 = 1.4 ³√4 = 1.5 ³√5 = 1.7 ³√6 = 1.8 ³√7 = 1.9 ³√8 = 2 ³√9 = 2.1 ³√10 =2.2
Square Root
√2 = 1.414 √3 = 1.732 √5 = 2.2 √6 = 2.4 √7 = 2.6 √8 = 2.8 √10 = 3.162 √11 = 3.3
Factorial
2! = 2 3! = 6 4! = 24 5! = 120 6! = 720 7! = 5040
Cube
2³ = 8 3³ = 27 4³ = 64 5³ = 125 6³ = 216 7³ = 343 8³ = 512 9³ = 729
Q. Find the number of numbers between 140 to 259 both included which are divisible by 7
= (259 - 140) / 7 + 1 = 119/ 7 + 1 = 17 + 1 = 18
Q. Attached
Lets Say Number of Boys = B Number of Girls = G Yesterday = B/G = 1 Today = (B - 18) / G = 3/4 Solving Second Equation B = 72 and hence number of Boys Today will be (72-18 ).
Q. Attached
Lets Say Weekly Allowance is W After Spending at the Arcade Amount Left was = 1 - 3/5 = 2/5W And We know that he spent 1/3 of this remaining amount at the Toy store and then the remaining amount was $0.80. There 2/5W * (1-1/3) = 0.80 or 4/15W = 8/10 -> W = $3
Q. Attached
Lets Say the Admission Price is $1 . As per Question , Admission Price was the only source for Revenue If 148 People Paid then Total Revenue = $148 Also as Revenue = Three times Cost of Sponsoring Event ∴Cost = $148/3 = ~$49 However Only 50 people paid then Total Revenue = $50 So Cost of Event( ~$49) < Total Revenue ($50)
Q. The ratio of two positive numbers is 3 to 4 . If k is added to each number the new ratio will be 4 to 5 , and the sum of the number will be 117 . What is the value of k .
Lets say two numbers are a and b Equation 1 - a/b = 3/4 -> a = (3/4)b Equation 2 - (a+k)/(b+k)= 4/5 ->5a +5k = 4b +4k -> 5a -4b = - k -> 5*(3/4)b - 4b = -k -> -b = -4k -> b = 4k Equation 3 - a + k +b +k = 117 -> a + b + 2k = 117 -> (3/4)b + b +2k = 117 -> 7b +8 k = 4*117 -> 7(4k) + 8k = 4*117 -> 36k = 4*117 -> k = 13
No. Of Integers between n , m
m-n+1 Ex: 28 to 45 = 45 - 28 +1= 18 Ex: -38 to -47 = -47 +38 +1 = 10
Rules for finding HCF of two fractions
( HCF Of Nummerator) ÷ (LCM of Denominator ) Ex - 1/2 ,1/5 and 4/1 (HCF of 1 , 1 and 4 is 1) ÷ (LCM of 2 , 5 and 1 is 10) =1/10
Co prime
Co prime are any two numbers for which HCF is 1 or called Relatively prime to each other .
Q. Find the number of Zeroes in 1¹! * 2²! * 3³! * ..10¹⁰!
Count number of 5s =5⁵! * 10¹⁰! =5(⁵×⁴׳ײ׹) * 10(¹⁰×⁹×⁸×⁷×⁶×⁵×⁴׳ײ׹)
Finding LCM - Shortcut
Ex - Find 9 , 10 , 12 and 15 Step 1 : Find Set of Co Prime Number - In above example 9 and 10 are Co prime Number hence Start off writing LCM by writing 9 * 10 . Step 2 : For other number consider what part of them have already been taken into the answer . Multiply the part which is missing . In 12 (2*2*3) and 15(3*5) , 9 (3*3) is already there in above step and 10 (2*5), so there is one 2 which is missing . so 9*10*2 = hence LCM is 180
Find the Number of factors/Divisors of 1200 such that the factors are divisible by 15
Factors of 1200 = 4*3*25*4 = 2⁴*3*5² factors of 15 = 5*3 2⁴*3*5² / 5*3 = 2⁴*5 i.e total of 5 * 2 = 10 factors
√x² = ±x and not x
True
Rules for Finding LCM of two fractions
(LCM of Numerators) ÷ (HCF of Denominators) Ex - 1/2 , 1/5 and 4/1 (LCM of 2, 5 and 1 is 10) ÷ (HCF of 1, 1, and 4 is 1 ) =10
Number of Zeroes in Factorial Value
*In Any factorial value the numbers of 5's will always be lesser than the number of 2's. Hence just count number of 5's. *
Decimals - I
1/2 = 0.5 1/3 = 0.33 1/4 = 0.25 1/5 = 0.2 1/6 = 0.1616 1/7 = 0.1428 1/8 = 0.125 1/9 = 0.111 1/11=0.09090 1/12=0.083 1/13=0.077 1/14=0.072 1/15=0.066
Shortcut to find whether number is Prime
0 - 169 : Check for Divisibility by 3 Just Remember - 77 , 91 , 119 , 133 , 143 and 161 are not Prime Numbers
Changing Impure Recurring Decimals
0.43542542542 = (43542 - 43)/99900 0.43543444444 = (435434 - 43543 ) /900000
Unit Conversion
1 Foot = 12 Inches 1 meter = 100 Cms
Square
11² = 121 12² = 144 13² = 169 14² = 196 15² = 225 16² = 256 17² = 289 18² = 324 19² = 361 21² = 441 22² = 484 23² = 529 24² = 576 25² = 625
Q. Find the Maximum value of n such that 157! is perfectly divisible by 12 power n .
12 is 2²*3 Occurrence of 2 in 157 = 157/2 = 78 , 78/2 = 39 , 39/2 = 19 , 19/2 = 9 , 9/2 = 4 , 4/2 = 2 , 2/2 = 1 -> 78+39+19+9+4+2+1 =152 or Occurrence of 2² = 152/2 = 76 Occurrence of 3 in 157 = 157/3 = 52, 52/3 = 17 , 17/3 = 5 , 5/3 = 1 -> 52+17+5 +1 = 75 Hence Maximum value of n will be 75 .
Q. Find the number of numbers between 100 to 200 *if Both 100 and 200 are counted/included*
200 - 100 + 1 = 101
Q. Find the number of numbers between 100 to 200 *if Only one of 100 and 200 is counted/included*
200 - 100 = 100
Finding the Last 2 digit of an Expression
22 * 31 * 44 * 27 * 37 * 43 Step 1 :Divide by 100 to get last 2 digit Step 2 : (22 * 31 * 44 * 27 * 37 * 43 )/100 -> (22 * 31 * 11 * 27 * 37 * 43 )/25 Step 3 : Use Remainder theorem (-3*6*11*2*12*-7)/25 -> (-18 *22* -84)/25 Step 4 : ( -(-7)*-3*-9)/25 -> (21*9)/25 -> (-4*9)/25 -> -36/25 -> -11/25 Step 5 : Remainder is -11 or +14 when divided by 25 Step 6 : Remainder when divided by 100 is 14*4 = 56 Hence 56 is the last two digit of above expression .
Power of 2
2⁴ = 16 2⁵ = 32 2⁶ = 64 2⁷ = 128 2⁸ = 256 2⁹ = 512 2¹⁰ = 1024
Q. Find all five digit number of form 34X5Y that are divisible by 36 .
36 is 9*4 hence 5Y should be divisible by 4 , so Y can be 2 or 6 for 9 - Case I - 3+4+X+5+2 should be divisible by 9 then X can be 4 - Hence our first number is 34452 Case II - 3+4+X+5+6 should be divisible by 9 then X can be 0 or 9 - Hence our numbers are 34056 or 34956 Three Number - 34452 , 34056 ,34956
Q. Find the number of Zeroes in 1¹ * 2² * 3³ * 4⁴ * 5⁵ * 6⁶ * 7⁷......*49⁴⁹ :Q 3 Pg 33 , Arun Sharma
As we know that the 5s will occur more than 2s .. So Lets Count 5s to predict number of Zeroes 5⁵ will have 5 five 10¹⁰ will have 10 five 15¹⁵ will have 15 five . . 25²⁵ will have 50 five . . 45⁴⁵ will have 45 five Hence (5+10+15+20+25+30+35+40+45 ) + 25 Zeroes Total Number of Terms = (45 - 5)/5 + 1 = 9 Terms Sum = [ (5 + 45)/2 ] * 9 = 225 Hence Total Number of 5's = 225 + 25 = 250 Zeroes
Compound Interest
Case I - Interest Calculated Annually A = P(1 + r/100)∧n Where P = Principal r = rate of Interest per year in percentage n = number of years and A = Total Amount after N year Case II - Interest Calculated half yearly If the annual rate is r% per annum and to be calculated for n years . Then in this case , rate = (r/2)% half yearly Time = 2n (half yearly) A = P(1 + (r/2)/100)∧2n A = P(1+r/200)∧2n Where P = Principal r = r % rate of Interest per year in percentage n = number of years Case III - Interest Calculated Quarterly If the annual rate is r% per annum and to be calculated for n years . Then in this case , rate = (r/4)% quarterly Time = 4n (quarterly ) A = P(1 + (r/4)/100)∧4n A = P(1+r/400)∧4n Where P = Principal r = r % rate of Interest per year in percentage n = number of years
Divisibility
I = QN + R where Q = Quotient R = Remainder N = Divisor I = Dividend
Q. Find the number of numbers between 140 to 259 both included which are divisible by 7 none of of 140 and 259 is included
= (259 - 140) / 7 - 1 = 119/ 7 - 1 = 17 - 1 = 16
Test Strategy
Trap Answers 1. Answer choices that use extreme or categorical words such as "only, all, always, every, never and exclusively" 2. Remember, you should never use common sense on reading comprehension passages 3. Answer Choices that ask you to make Judgements: Any answer choice that asks you to affirm that one method/approach/thought is "better, or more successful, or more efficient" than another, can be considered wrong without second consideration. Passages on the GRE never ask the reader to judge anything, and the judgment will have already been made by the author. Hence, any answer option that needs your judgment, is wrong. Ex - Women in East Africa are more easily prone to suffer from the viral fever than men. (asks you to make a comparison/judgment without sufficient proof)
Positive Integers / Natural Numbers
All Integers greater than 0 1, 2, 3, 4,
Integers
All numbers - Doesn't Include Decimals -5, -4, -3, 0 , 1 , 2, 3
Even Numbers
All the Integers which are Divisible by 2 0 , 2, 4, 6, 8
Odd Numbers
All the Integers which are Not Divisible by 2 (2n + 1 , 2n -1)
Divisibility by 7 , 11 and 13
All the numbers are divisible by 7 , 11 and 13 if the difference of the number of its thousands and the remainder of its division by 1000 is divisible by 7 , 11 and 13 . For Ex - 473312 is divisible by 7 coz , 473 -312 = 161 is divisible by 7
Numbers / Real Number
All the numbers on Number Line 0 , -1 , -2 , 1, 2 , 1/2 , -pi
Divisibility by 6
All the numbers which are divisible by 3 and 2 are divisible by 6
Divisibility by 12
All the numbers which are divisible by 3 and 4 are divisible by 12
Divisibility by 8
All the numbers whose last three digit are divisible by 8 are divisible by 8
Divisibility by 4
All the numbers whose last two digits are divisible by 4 are divisible by 4
Divisibility by 3 or 9
All the numbers whose sum are divisible by 3 or 9 are divisible by 3 or 9 .
Product of -ive Integers if total is Even count of Number
Always +ive -2*-2*-4*-4 =64
Product of -ive Integers if total is Odd count of Number
Always -ive -2*-2*-4* = -16
Composite Number
Any numbers which are divided by 1 , itself and any other Integer i.e Its not a Prime Number .
Simple Interest
Simple Interest = (P * R * T )/100 where p is the principal or money deposited r is the rate of interest in Percentage t is time Total Amount = Principal + Simple Interest Half yearly rate of Interest is half the Annual rate of Interest . So if the interest is 10% per annum to be charged six monthly , we have to add Interest every six months @5%
Q. Find the greatest 4 digit number which when divided by 10 , 11 , 15 and 22 Leaves 3, 4 , 8 and 15 as remainder respectively :Q 19 , Pg 27 , Arun Sharma
Step 1 - LCM of 10 , 11 , 15 and 22 is 330 Step 2 - Greatest 4 Digit Number is 9900 Step 3 - ->N ÷ 10 gives 3 as remainder when N is added with 3 or Subtracted with 7 . ->N ÷ 11 gives 4 as remainder when N is added with 4 or Subtracted with 7 . ->N ÷ 15 gives 8 as remainder when N is added with 8 or Subtracted with 7 . ->N ÷ 22 gives 15 as remainder when N is added with 15 or Subtracted with 7 . Step 4 - In all cases when we Subtract N with 7 we will get our desired result . Hence 9900 - 7 is the answer . 9893 .
Q. Three runners running around a circular Track can complete one revolution in 2 , 4 and 5.5 hours respectively . When will they meet at Starting Point . : Q13 , Pg 27 , Arun Sharma
Step 1 - LCM of 2 , 4 and 5.5 -> LCM of 2/1 , 4/1 , and 11/2 LCM = (LCM of 2 , 4 and 11 ) ÷ (HCF of 1 , 1 , 2) LCM = 44/1 Hence they will meet after 44 Hours.
Q. What's the largest 4 Digit number when added to 7249 gives the number which will be divisible by 12 , 14, 21, 33 and 54 . : Q16 , Pg 27 , Arun Sharma
Step 1 - To find LCM of 12 , 14 21, 33 and 54 . i.e 8316 Step 2 - 8316 *1 is the first number divisible by all 4 numbers 8316*2 is the second number divisible by all 4 Numbers 8316*3 is the third number divisible by all 4 Numbers Step 3 - In order to find max 4 digit number for x 7249 + x = N ( where N is multiple of 8316 ) x = N - 7249 Step 4 - x = 8316 - 7249 = 1067 x can be = (8316*2) - 7249 = 9383 Hence 9383 is the Largest 4 Digit Number when added to 7249 returns number divisible by 12 , 14 , 21 , 33 and 54 .
Negative Integer
All Integers Less than 0 -1, -2 , -3, -4
Remainders
*Remainders are Always Positive * - Arun Sharma Quant - Pg 10 a) 7n + 1 and 7n - 6 when divided by 7 will give 1 as remainder Ex : 27/7 gives 6 as remainder -27/7 gives -6 as remainder but ( 7 - 6 = 1 ) hence 1 is the final remainder
Changing Pure Recurring Decimals
0.55555 = 5/9 3.242424 = 3 + (24/99) 5.362362 = 5 + (362/999)
1 Million
1 Million - 10⁶ Zeroes 1 Billion - 10⁹ Zeroes
Rules for Co - Primes ( 2 Numbers)
1. *Two Consecutive Natural numbers* are always Co prime ( 5, 6 ; 84, 85) 2. * Two Consecutive Odd Numbers * are always Co Prime ( 33, 35 ; 45 , 47) 3. * Two Prime Numbers * are always Co prime (2, 3; 5, 7 ; 11, 13 ) 4. * One prime Number and Other Composite Number * such that Composite Number is not Multiple of Prime Number (17,38 ; 23, 49)
Rules for Co - Primes ( 3 Numbers)
1. OOO - *Three Consecutive Odd Numbers * are always Co Prime ( ex - 15 , 17, 19) 2. OEO - *Three Consecutive Natural Numbers with first one being Odd * ( 15 , 16, 17) 3. EOO - *Two Consecutive Natural Number with next Odd Number such as the first one is Even ( 22,23, 25) 4. PPP -* Three prime Numbers ( 17 , 23, 29)
Absolute Value
Absolute Value - Distance from Origin |x| = Distance from 0 |x-5| = Distance from 5 |x+3| = Distance from -3 |x+3| > 5 = Distance from -3 and should be greater than 5 (x + 3) > 5 or - (x + 3) > 5 x > 2 or -x > 8 i.e x > 2 or x < -8
Test Strategy - Maths
1. Read Question and try to understand in Hindi . Sort out the strategy you are going to implement to solve it . 2. Note all the important things which question is asking , Spend 15 -30 seconds on understanding question . For example a) Check whether its asking in Kg , m/s , miles/hours or any other format . b) In Data Analysis , Pay special attention to Years and Metric System c) In Probability check whether its asking for number of ways or Probability . As probability is always less than 1 and number of ways are more than 1. 3. *Check whether Answer gives Clue* - If Answers has big gaps then put option C in the question and see whether it satisfies the Question . This technique is helpful in Quant , Profit And Loss , Numbers , Time and Work etc . 4. *Clean Calculation* - While Solving Question , use a clean space and don't cut the numbers , try to write it as Clean as Possible . 5. In geometry ..Look for Big Triangles , Big trapezoid, Big quadrilaterals ..its all about Big Picture . 6. *Recheck Calculation* - Revisit the calculation. 7. *Read Question* Read the question again and then try to match it with answer . 8. *Mark the Option* - Mark the option with care .
One
1. Smallest Odd Number
Prime Number Properties
1. The Lowest Prime Number is 2 , 2 is also the only even Prime Number . 2. The Lowest Odd Prime Number is 3. 3. 1 is not a Prime Number . 4. p/6 gives remainder either 1 or 5 if P is a prime number and p ≥ 5. 5. p²/12 will give remainder 1 if p ≥ 5 and p is prime number . 6. p²/24 will give remainder 1 if p ≥ 5 and p is prime number . 7. (p² -1)/24 give remainder 0 if p > 3 and p is prime number
Number of Factors
1. Total Number of Factors for Prime Number is 2 2. Total Number of Factors for Perfect Square is ODD 3. Total Number of factors for Square of Prime Number is always 3 . Ex - 3² = 9 , 9 has three factors 1, 3 and 9 .
Zero
1. Zero is neither Positive Nor Negative Integer 2. Zero is not a Prime Number 3. Zero is the smallest Even Number
Roots comparison
2 < (2)² < (2)³ 1/2 > (1/2)² > (1/2)³ -1/2 < (-1/2)² > (-1/2)⁴ > (-1/2)⁶ -1/2 < (-1/2)³ < (-1/2)⁵ For Roots - Case - for a > 1 2 > √2 > ³√2 > ⁴√2 > ⁵√2 > ⁶√2 ... > ¹⁰⁰√2 > 1 Case - for a < 1 1/2 < √(1/2) < ³√(1/2) < ⁴√(1/2) < ..¹⁰⁰√(1/2) < 1
n Power
3⁴ = 81 3⁵ = 243 3⁶ = 729 3⁷ = 2187 4⁴ = 256 4⁵ = 1024 5⁴ = 625 5⁵ = 3125 6⁴ = 1296 6⁵ = 7776 7⁴ = 2401 7⁵ = 16807 8⁴ = 4096 8⁵ = 32768 9⁴ = 6561 9⁵ = 59049
Q. Find the highest power of 42 which divides 122!
42 = 2*3*7 , So to find maximum power of 42 which can divide 122! , we need to find which out of 2 , 3 and 7 divides least number of times . 2 will occur -> 122/2 = 61, 61/2 = 30 , 30/2 = 15 , 15/2 = 7 , 7/2 = 3 , 3/2 = 1-> 61+30+15+7+3+1 =117 Times 3 will occur -> 122/3 =40, 40/3 = 13 , 13/3 = 4, 4/3=1 -> 40+13+4+1 = 58 Times 7 will occur -> 122/7 = 17 , 17/7 = 2 -> 17+2 = 19 Times As 7 will occur least number of times hence we can say that Maximum power of 42 which can divide 122! is 19 .
Find Maximum and Minimum factorial where there are 43 Zeroes
43/5 = Quotient is 8 43 - 8 = 35 Hence factorial should be equal to or greater than 5*35 = 175! to 179! 5*35 = 175! -> 175/5 = 35 -> 35/5 = 7 -> 7/5 =1 -> 35 + 7 +1 = 43 Zeroes Right Answer
Q. Find the number of numbers between 140 to 259 which are divisible by 7 if Only one of 140 and 259 is included
= (259 - 140) / 7 = 119/ 7 = 17
√a < a when a > 1 √a > a when a < 1
Case I - √a < a when a > 1 √2 = 1.414 which is less than 2 Case II - √a > a when a < 1 a = 1/2 = 0.5 then √a = 1/√2 = 1/1.414 = 0.707 Hence √a > a for decimals
Q. Find the number of Zeroes in 100¹ * 99² * 98³ * 97⁴ * ........* 1¹⁰⁰ :Q4 Pg 33 , Arun Sharma
Count number of 5s 100¹ * 95⁶ * 90¹¹ * ..........5⁹⁶ (1+6+11+16+21..+96) + (1 +26 + 51 + 76) / As 25 , 50 , 75 and 100 will have 2 5s each) Total Number of Terms in (1+6+11+16+21..+96) = [ (96 -1)/5 ] + 1 = 20 Terms Sum = [(96 + 1)/2] * 20 = 970 Total Number of Terms in (1 +26 + 51 + 76) = [ (76 -1)/25 ] + 1 = 4 Terms Sum = [(76 + 1)/2] * 4 = 154 Total after using AP = 970 + 154 = 1124
If you are *Multiplying any number with Even* it will result in *Even*
EVEN * EVEN = EVEN ODD * EVEN = EVEN ODD * ODD = ODD
If you are *Adding Similar Numbers ( Even or Odd ) * it will always give you *Even *
EVEN + EVEN = EVEN ODD + ODD = EVEN ODD + EVEN = ODD
Finding HCF - Shortcut
Ex - Find HCF of 39 , 78 and 195 Step 1 :: Take the difference between two closest Number . 78 and 39 = 78 - 39 =39 Step 2 :: HCF of 39 and 78 has to be factor of 39 . Now in order to find Highest Common Factor of 39 , find all the divisor less than Square root of 39 . 1*39 3*13 Step 3 :: Hence one of the above 4 number has to be HCF of 39 , 78 and 195 . As we want to find Highest common factor then search from highest number . Step 4 : Check whether 39 Divides all the number . As 39 is a factor of 78 and 39 then we just need to check whether 39 divides 195 and yes it divides that . If not then we have to go to next highest level ( which is 13 in this case). Step 5 : 39 is the HCF of 38 , 78 and 195 .
HCF of ( a¹²⁵ - 1 ) and ( a³⁵ - 1)
HCF ( a HCF of (125 and 35 ) - 1 ) i.e (a⁵ -1 )
HCF Usage
HCF is used when we need to *divide* sets of numbers in Equal Portion.
HCF Rule
HCF of x and y is G then 1. HCF of x , (x+y) is also G 2. HCF of x , (x-y) is also G Ex - HCF of 10 and 15 is 5 then 1. HCF of 10 and 25 is also 5 2. HCF of 10 and 5 is also 5
Number of Even Factors
Number of Even Factors of 40 (2¹*2²*2³)(5⁰*5¹) = 2³ * 5 = Product of (highest power of 2 ) * (Highest Power of other number + 1) = 3* (2) = 6 Factors We are just eliminating 2⁰
Finding Number of Factors
Number of Factors of 40 is 2*2*2*5 -> (2⁰*2¹*2²*2³)(5⁰*5¹) -> 4 factors of 2 * 2 factors of 5 -> Total 8 Factors. Shortcut - Product of (Highest Power + 1) of all the Prime Numbers . The factors Include 1 and Number itself ( 1 and 40 in this case )
Number of Odd factors
Number of Odd Factor of 40 (2⁰)(5⁰*5¹)= 2⁰ * (5⁰*5¹) = 2 Factors
Whole Numbers
Number which we use for counting . All Positive Integers Including 0 0 , 1 , 2 , 3, 4
Q. Find the last digit in Expression (36472)¹²³! * (34767)⁷⁶! Q 1.7 Pg 46 Arun Sharma
Step 1 - Unit digit for the power of 2 goes as 2 , 4, 8 , 6, 2, 4 , 8, 6 Hence 4n + 1 is 2 ( Lets have n = 0 then 2¹ = 2) 4n + 2 is 4 ( Lets have n = 0 then 2² = 4) 4n + 3 is 8 ( Lets have n = 0 then 2³ = 8) 4n is 6 ( Lets have n = 1 then 2⁴ = 6) Step 2 - 123!(Note its factorial and not just plane number ) is can be written in 4n form ( As when you divide 123! with 4 , you will not have any remainder ) , Hence the remainder is 6 Step 3 - Unit Digit for the power of 7 goes as 7 , 9 , 3 , 1 , 7 , 9 , 3 , 1 Hence 4n + 1 is 7 ( Lets have n = 0 then 7¹ = 7) 4n + 2 is 9 ( Lets have n = 0 then 7² = 9) 4n + 3 is 3 ( Lets have n = 0 then 7³ = 3) 4n is 1 ( Lets have n = 1 then 7⁴ = 1) Step 4 - 76!(Note its factorial and not just plane number ) can be written in 4n form , Hence the remainder is 1 Step 5 - Unit Digit = 6 * 1 = 6
Q. Find the remainder when 51²⁰³ is divided by 7
Step 1: (7*7 +2 )²⁰³ / 7 = (2)²⁰³/7 = (2³*⁶⁷/7) * (2²/7) = 1* 4/7 = Remainder is 4
HCF * LCM = Product of Numbers
True
(ax - 1) power of n/ a where x is any positive integer
Will give remainder 1 if n is EVEN and -1 = (a -1) if n is ODD 6⁸/7 = (7*1 -1)⁸/7 = 1 6³/7 = (7*1 -1)³/7 = -1 = (7-1) = 6
(a) power of n / (ax +1 ) where x is any positive integer
Will have remainder of a if n is ODD 1 if n is EVEN Ex - 5⁵/6 = 5⁵/(5*1+1) as 5 is ODD 5⁴/26 = 5⁴(5*5+1) =1 as 4 is EVEN
Relationship Between Remainder and Decimal
a) 42 ÷ 5 = Quotient is 8 and remainder is 2 42/5 = 8.4 -> 8 + 2/5 b) -42 ÷ 5 = Quotient is 8 and remainder is -2 And as remainder is never negative so ( 5 -2 ) = 3 will be the remainder -42/5 = -8.4 = -9 +0.6 ( As Remainder is never negative ) = -9 + 3/5
