Arithmetic aptitude :: HCF and LCM :: General Questions

• HCF and LCM - General Questions
• HCF and LCM - Important Formulas and Facts

Factors and Multiples:
If number x divided another number y exactly, we say that x is a factor of y.
and y is called a multiple of x.
Highest Common Factor (H.C.F.) or Greatest Common Measure (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 them exactly.
Mehods To Find HCF
Factorization Method:
Step 1) Find the prime factors of each number
Step 2) Find the product of all common prime factors, which is the HCF of given numbers
e.g. To find the HCF of 14 and 8
14 = 2 x 7
8 = 2 x 2 x 2
HCF = 2
Division Method:
To find the H.C.F. of two given numbers
Step 1)Divide the larger by smaller one.
Step 2)Divide the divisor by the remainder.
Step 3)Repeat the process of dividing the preceding number by the remainder last obtained till zero is obtained as remainder.
The last divisor is the required H.C.F.
Finding the H.C.F. of more than two numbers:
Step 1)H.C.F. of [(H.C.F. of any two) and (the third number)] gives the H.C.F. of three given number.
Similarly, we can find the H.C.F. of more than three numbers.
Least Common Multiple (L.C.M.):
LCM is the least number which is exactly divisible by each one of the given numbers.
Methods to find LCM
Factorization Method:
Step 1)Find prime number factors of each given number.
Step 2)Find the product of highest powers of all the prime number factors, which gives the LCM of given numbers
e.g. To find the LCM of 14 and 8
14 = 2 x 7
8 = 2 X 2 x 2 = 2 ³
LCM = 2 ³ x 7 = 56
Division Method (short-cut):
Step 1)Arrange the given numbers in a rwo in any order.
Step 2)Divide by a number which divided exactly at least two of the given numbers and carry forward the numbers which are not divisible.
Step 3)Repeat the above process till no two of the numbers are divisible by the same number except 1.
LCM = The product of the divisors and the undivided numbers
Product of two numbers = (HCF x L.C.M) of two numbers.
Co-primes: Co-primes means, if their H.C.F. is 1.
HCF and LCM of Fractions:
HCF = (HCF of Numerators) / (LCM of Denominators)
LCM = (LCM of Numerators) / (HCF of Denominators)
H.C.F. and L.C.M. of Decimal Fractions:
In a given numbers,
Step 1)Make the same number of decimal places by annexing zeros in some numbers, if necessary.
Step 2)Considering these numbers without decimal point, find H.C.F. or L.C.M.
Step 3)In the result, mark off as many decimal places as are there in each of the given numbers.
Comparison of Fractions:
Step 1)Find the L.C.M. of the denominators of the given fractions.
Step 2)Convert each of the fractions into an equivalent fraction with L.C.M as the denominator, by multiplying both the numerator and denominator by the same number. The resultant fraction with the greatest numerator is the greatest.

The smallest number which when diminished by 7, is divisible by 12, 16, 18 and 21 ?.

Answer:

Explanation:

Let the smallest number be X,
Lcm of 12,16,18,21,28 = 1008.
Here in question,
Diminished by 7, so, x-7 = 1008.
x = 1008+7 = 1015.

Let the least number of six digits, which when divided by 4, 6, 10 and 15 leaves in each case the same remainder of 2, be N. The sum of the digits in N is :

Answer:

Explanation:

LCM of 4, 6, 10, 15 = 60
Least number of 6 digits = 100000
The leats number of 6 digits which is exactly divisible by 60 = 100000 + (60 – 40) = 100020
Required Number = 100020 + 2 = 100022
Hence, the sum of digits = 1 + 0 + 0 + 0 + 2 + 2 = 5

The least number which is a perfect square and is divisible by each of the numbers 10, 20 and 24, is 

Answer:

Explanation:

Take LCM of 10, 20 and 24 = 120

So, 120 is the lowest number, which is divisible by 10, 20 and 24

So, from the options, 3600 is a multiple of 120 and its a pefect square,

also it is least value which satisfies all the conditions

So, the answer is 3600

The least number of five digits which is exactly divisible by 12, 15 and 18, is :

Answer:

Explanation:

LCM of 12, 15 and 18 = 180

Least 5 digit number = 10000

Now find the exact multiple of 180 near to smallest digit of 10000;

10000 / 180 = 55. 55

So by taking round of this number 55 or 56 may be the nearest number to 10000 which is divisible by all above given numbers.

55 x 180 = 9900

56 x 180 = 10080

So, clearly 10080 is the right answer

Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case Then sum of the digits in N is :

Answer:

Explanation:

N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)

  = H.C.F. of 3360, 2240 and 5600 = 1120.

Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

Note:

We are taking hcf of (4665 - 1305), (6905 - 4665) and (6905 - 1305) because to cancel the remainder.

take an example..
12 and 7, both give us remainder 2 when divided by 5.
now when we substact (12-7) the differance 2 is cancelled.

• HCF and LCM - General Questions