Adapted from various sources of BeatTheGMAT
#1. How to test whether a number is prime or composite:
A Prime Number is a positive integer that is divisible by ONLY 2 numbers (1 and itself).
Whereas, A composite number is a positive integer which has divisor(s) other than the 2 numbers (1 and itself).
Following are the steps to test whether a number is a prime or composite,
1. Identify the perfect square (P.S) closest to the n < n.
2. Compute the square root of P.S
3. List all prime numbers upto the computed square root
4. Check if all listed prime numbers divide n equally. If not, then n is a prime. Even if atleast one of the listed prime numbers divide n, then n is a composite.
Example:
Take n as 113. To test whether 113 is a prime,
1. 100 is the closest perfect square to 113 (Remember that you take a closest perfect square that is smaller than n itself!)
2. Square root of! 100 ==> 10
3. Prime numbers upto the square root (10) ==> 2,3,5,7.
4. Check whether 2,3,5,7 divides 113. None of the numbers divide 113. So, 113 is a prime.
#2. How to calculate LCM and HCF of fractions:
L.C.M of 2 fractions = L.C.M of NUMERATORS / H.C.F of DENOMINATORS
H.C.F of 2 fractions = H.C.F of NUMERATORS / L.C.M of DENOMINATORS
Product of any 2 numbers = Product of LCM and HCF of those 2 numbers
Product of any 2 fractions = Product of LCM and HCF of those 2 fractions
Highest common factor of two or more than two numbers is the greatest number that divides each of them exactly (without remainder).
Ex. Find HCF of 18, 20 and 36
18 = 2 × 3^2
20 = 2^2 × 5
36 = 2^2 × 3^2 Therefore HCF = 2
The smallest number divisible by all the numbers is the Least Common Multiple of the given numbers.
Ex What is the LCM of 4, 8 and 12?
4 = 2^2, 8 = 2^3, 12 = 2^2 x 3!
Therefore LCM = 2^3 x 3 = 24
Q1: Find the L.! C.M. of 36, 48, 72 and 108. = 432
Q2: Find the greatest number which on dividing 277, 757 and 3307 leaves the same remainder 7 in each case. (Hint: Find HCF of numbers minus 7 = 30)
Q3: The L.C.M. of two number is 252 and their ratio is 6:7. Find the numbers. = 36,42
Q4: Find the greatest number that will divide 63, 45 and 69 so as to leave the same remainder (options 6,9,8,10). Hint find HCF = 6
Q5: A room is 2.47m long and 2.09m broad. It is required to pave the floor with minimum square slabs. Find the number of slabs required for floor. - 143
Q6: Find the least perfect square number which is divisible by 2,3,4,5,6. - 900
Q7: Find the least five digit number which on divided by 12, 18, 21, and 28 leaves the same remainder - 11019, 12089, 10119, 10059. - 10119
Q8: Which smallest number must be added to 605329 so that it becomes divisible by 9? - 2
Q9; Find the greatest number of four digits which is exactly divisible by each of 12, 18, 40 an! d 45. - 9720
Q10: Find the L.C.M. of 6/7, 12/13, 7/15 - 84
Q11: Find the least number which is exactly divisible by 18, 15, 25 and 30. - 450
Q12: The sum and the differences of the L.C.M. and the H.C.F. of two numbers are 312 and 264, respectively. Find the numbers if their sum is 168. - 96,72
Source: http://www.quizmoz.com/quizzes/GMAT-Practice-Tests/g/GMAT-Practice-Test-Math-LCM-and-HCF.asp
Solutions:
A1. 36=2^2.3^2, 48=2^4.3^1, 72=2^3.3^2, 108=2^2.3^3
LCM = 2^4.3^3 = 27*2*2*2*2 = 54*2*2*2 = 108*2*2 = 216*2 = 432
A2.
277-7 = 270 = 2^1.3^3.5^1
757-7 = 750 = 2^1.3^1.5^3
3307-7 = 3300 = 2^2.3^1.5^2.11^1
HCF = 2^1.3^1.5^1 = 30
A3.
252 = 2^2.3^2.7 = 6*6*7
Thus, use 6*6 and 6*7 = 36, 42
A4: Try choices, remainders are
(1) 6: 3, 3, 3
(2) 9: 0, 0, 6
(3) 8: 7, 5, 5
(4) 10: 3, 5, 9
A5: Area = 2! .47*2.09 m^2 = 247*209 cm^2 = n*x^2 where x is the HCF in 247 ! and 209< br />Q: Is 247 or 209 prime
Prime #s - 2,3,5,7,11,13,17,19,23,27
247: Closest square = 225 = 15^2 => Check 2,3,5,7,11,13 => 13 divides 247 => 247 is a composite number = 13*19
209: Closest square = 196 = 14^2 => Check 2,3,5,7,11,13 => 11 divides 209 => 209 is also a composite number = 11*19
Thus, Area = 11*13*19^2 => x=19 cm and n = 11*13
Therefore, # of slabs = 11*13 = 143 with each having 19 cm side
A6: Want to find least perfect square number which is divisible by 2,3,4,5,6
Number have the same prime factor two times
2 = 2^1.3^0.5^0
3 = 2^0.3^1.5^0
4 = 2^2.3^0.5^0
5 = 2^0.3^0.5^1
6 = 2^1.3^1.5^0
2^2.3^2.5^2 = 4.9.25 = 900
#3. How to find unit digit of powers of numbers:
All numbers from 0 to 9 have a repetitive pattern for powers
Pattern 1:
Unit's place that has digits - 2/3/7/8
Then, unit's digit repeats every 4th value.! Divide the power (or index) by 4.
After dividing,
If remainder is 1, unit digit of number raised to the power 1.
If remainder is 2, unit digit of number raised to the power 2.
If remainder is 3, unit digit of number raised to the power 3.
If remainder is 0, unit digit of number raised to the power 4.
Pattern 2:
Unit's place that has digits - 0/1/5/6
Then, all powers of the number have same digit as unit's place.
For e.g., 6^1 = 6, 6^2 = 36, 6^3 = 216, 6^4 = 1296
Pattern 3:
Unit's place that has digit - 4
Then,
If power is odd --> unit's digit will be '4'
If power is even --> unit's digit will be '6'
Similarly,
Unit's place that has digit - 9
Then,
If power is odd --> unit's digit will be '9
If power is even --> unit's digit will be '1'!
Ex. OG-12 PS190
190. What is the units digit ! of (13)^ 4(17)^2(29)^3 ?
powers of 3
3
9
7
1
7
9
9
1
9
1*9*9 = 1
#4. Divisibility Tests:
To check whether a number (say n) is divisible
By 2: unit's place of n must be 0 (OR) unit's place of n must be divisible by 2.
By 3: Sum of the digits of n must be divisible by 3.
By 4: Last 2 digits (Unit's place and ten's place) of n are 0's (OR) Last 2 digits of n must be divisible by 4.
By 5: Unit's digit must be a 5 (OR) a 0.
By 6: n must be divisible by both 2 and 3 (Follow the method used for 2 and 3).
By 8: Last 3 digits (units, tens and hundredth place) of n are 0's (OR) Last 3 digits of n is divisible by 8.
By 9: Sum of the digits of n must be divisible by 9.
By 11: (Sum of the digits of n in odd places) - (Sum of the digits of n in even places) ==> Either 0 (OR) divisible by 11.
By 12: n must be divisible by ! both 3 and 4 (Follow the method used for 3 and 4).
By 25: Last 2 digits (units and tens place) of n are 0's (OR) Last 2 digits of n must be divisible by 25.
By 75: n must be divisible by both 3 and 25 (Follow the method used for 3 and 25).
By 125: Last 3 digits of n are 0's (OR) are divisible by 125.
#5. Whether a number is NOT a perfect square or MIGHT BE
The perfect squares always end in 1, 4, 5, 6, 9 or 00 (i.e. Even number of zeros).
If they end in even zeros, then remove the zeros at the end of the number and apply following tests:
If unit digit ends in 5, ten’s digit is always 2.
If unit digit ends in 6, ten’s digit is always odd
otherwise it is always even.
Digital roots are 1, 4, 7 and 9.
To find digital root of a number, add all its digits.
If this sum is more than 9, add the digits of this sum. The single digit obtained at the! end is the digital root of the number.
If a number! is divi sible by 4, its square leaves a remainder 0 when divided by 8. Square of an even number not divisible by 4 leaves remainder 4 while
square of an odd number always leaves remainder 1 when divided by 8.
Total numbers of prime factors of a perfect square are always odd.
Examples:
1. 5329 = sum of digits = 19 = 1 => Digital root is in (1,4,7,9) => MAYBE
2. 1280 = NO because does not end in 1,4,5,6,9,00
3. 1800 = Ends in 00, sum of digits = 9, unit digit not in (5,6), tens
4. 3600 = Ends in 00, sum of digits = 9, unit digit = 6, tens digit = odd => MAYBE
5. 729 = sum of digits = 18 = 9 => Digital root is in (1,4,7,9)
6. 38910 = No because does not end in 1,4,5,6,9,00
7. 14798678562 = No because does not end in 1,4,5,6,9,00
8. 15763530163289 = sum of digits = 58 = 13 = 4 digital root ends in (1,4,7,9) MIGHT BE
Ex. 1800 = 9 YES =>
dividing three digit single digit number
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