Solutions: HCF & LCM, Part-A

The solutions here are posted for the questions in:
http://arithmophobia-elixir.blogspot.in/2014/07/questions-on-hcf-lcm.html

(1)  We know the formula:

HCF of Fractions = ( HCF of Numerators ) / ( LCM of Denominators )

LCM of Fractions = ( LCM of Numerators ) / ( HCF of Denominators )

Given Fractions:  2/3 , 8/9, 16/81 and 10/27, 

HCF of given fractions = [ HCF (2,8,16,10) ] / [ LCM ( 3,9,81,27 ) ]

LCM of given fractions =  [ LCM (2,8,16,10) ] / [ HCF ( 3,9,81,27 ) ]

Let's express all the given numbers as the product of prime factors to find HCF and LCM.

( Factorization Method - http://arithmophobia-elixir.blogspot.in/2014/07/hcf-and-lcm.html )

2 = 2¹ * 5⁰                           3 = 3¹

8 = 2³ * 5⁰                                 9 = 3²

16 = 2⁴ * 5⁰                        81 = 3⁴

10 = 2¹ * 5¹                        27 = 3³

Looking at above we see :

HCF (2,8,16,10) = 2¹ * 5⁰ = 2 * 1 = 2 ( To find the HCF we need to check the least power among factors)

LCM (2,8,16,10) = 2⁴ * 5¹ = 16 * 5 = 80 ( To find the LCM we need to check the highest power among factors)

HCF ( 3,9,81,27 ) = 3¹ = 3 ( To find the HCF we need to check the least power among factors)

LCM ( 3,9,81,27 ) = 3⁴ = 81( To find the LCM we need to check the highest power among factors)

Therefore,

HCF of given fractions = [ HCF (2,8,16,10) ] / [ LCM ( 3,9,81,27 ) ] = 2/81

LCM of given fractions =  [ LCM (2,8,16,10) ] / [ HCF ( 3,9,81,27 ) ] = 80/3

(2) Three numbers are given: 0.63, 1.05, 2.1, We need to find their HCF and LCM.

To find the HCF or LCM of decimal numbers.

Multiply all the numbers by 10 as many times until they become integer.

All the numbers should be multiplied by 10 same number of times and then once the LCM/HCF is found then divide it by same number of 10s as multiplied.

Here, we will multiply all the numbers by 100, as 100 will make all of them an integer. So the numbers become: 63, 105 and 210.

Let's express all the given numbers as the product of prime factors to find HCF and LCM.

( Factorization Method - http://arithmophobia-elixir.blogspot.in/2014/07/hcf-and-lcm.html )

63 = 7¹ * 3² * 5⁰ * 2⁰

105 = 7¹ * 5¹* 3¹ * 2⁰

210 = 7¹ * 3¹ * 5¹ * 2¹

Looking at above we see :

HCF (63,105,210) = 7¹ * 3¹ * 5⁰ * 2⁰ = 7 * 3 *1 * 1 = 21 ( To find the HCF we need to check the least power among factors)

LCM (63,105,210) = 7¹ * 3² * 5¹ * 2¹ = 630 ( To find the LCM we need to check the highest power among factors)

Now, we will divide the HCF and LCM by 100 to get the final answer.

Therefore,

HCF ( 0.63, 1.05, 2.1 ) = 0.21

LCM ( 0.63, 1.05, 2.1 ) = 6.30

(3) We know the formula:
Product of two numbers = Product of their HCF and LCM
http://arithmophobia-elixir.blogspot.in/2014/07/basic-facts-and-formula.html

Other number * 77 = ( 11 * 693 )
Therefore,
Other number = [ ( 11 * 693 ) ] / 77 = 99

(4) To find Greatest possible length which can be used to measure exactly the lengths 4m 95 cm, 9m and 16m 65 cm.
Greatest possible length will be the highest common factor of :

4m 95 cm, 9m and 16m 65 cm.

Lets first convert all of them to same unit.

4m 95 cm, 9m and 16m 65 cm = 495 cm, 900 cm, 1665 cm.

Hence the answer is HCF ( 495, 900, 1665 ) = 45 cm

( Use either factorization or division method to find HCF : http://arithmophobia-elixir.blogspot.in/2014/07/hcf-and-lcm.html )

(5) To find the greatest numbers which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively.
Let x be the number.

Therefore, when 1657 is divided by x, remainder that is left is 6 and,

when 2037 is divided by x, remainder that is left is 5,

That means (1657 – 6) and (2037 – 5) are completely divisible by x.

That means 1651 and 2032 are completely divisible by x.

Hence the greatest number x, which divides 1651 and 2032 completely is:

HCF ( 1651, 2032 ) = 127

Explanation of above HCF by long division method.

       _{------------------}

1651) 2032 ( 1

          1651

       -----------_{____________}

                381 )    1651        ( 4

                            1524

                             ---------_{________}

                                 127 )       381    ( 3

                                                381

                                            ---------

                                                   0

                                                          ^____________

(6) To find the largest number of four digits exactly divisible by 12, 15, 18 and 27

Largest 4 digit number which is divisible by each of the given numbers must be divisible by LCM (12,15,18,27) = 540 
largest 4 digit number = 9999 
So, 9999/540. we get 279 as remainder. 

Therefore, the largest 4 digit number exactly divisible by 12,15,18,27

= 9999 – 279 = 9720 

(7) To find the smallest number of five digits exactly divisible by 16, 24, 36 and 54
Smallest 5 digit number which is divisible by each of the given numbers must be divisible by LCM (16, 24, 36, 54) = 432
Smallest 5 digit number = 10000
So, 10000/432. we get 64 as remainder. 
Now we need to subtract 64 to 10000 to get the smallest 5 digit number exactly divisible by 16, 24, 36, 54, but this will make the number as 4 digit number. Hence
the four digit number (10000-64 is divisible by 432, but to make it five digit smallest number divisible by 432 we need to add 432 )
= 10000 - 64 + 432 = 10368  

(8) Given numbers are 99, 101, 176,182 . We need to find the number with greatest number of divisors.
Let’s express each numbers as product of prime factors:
99 = 3 * 3 * 11 = 3² * 11¹
101 = 101¹
176 = 2 * 2 * 2 * 2 * 11 = 2⁴ * 11¹
182 = 2 * 7 * 13 = 2¹ * 7¹ * 13¹

Number of divisors of an integer = Product of (exponents + 1) of all the prime factors.

Therefore,

Number of divisors of 99 = (2 + 1) * (1+1) = 6

Number of divisors of 101= 1 + 1 = 2

Number of divisors of 176 = (4 + 1) * (1 + 1) = 10

Number of divisors of 182 = (1 + 1) * (1 + 1) * (1 + 1) = 8

Therefore, 176 is the number with the greatest number of divisors.

(9) To find the number of number-pairs lying between 40 and 100 with their HCF as 15.
Numbers with HCF 15 must contain 15 as a factor.
multiples of 15 between 40 and 100 are; 45,60,75,90;
HCFof 45,60 = 15;
HCF of 45,75 = 15;
HCF of 45,90 = 45;
HCF of 60,75 = 15;
HCF of 60,90 = 30;
HCF of 75,90 = 15




Hence, number of such pairs are : 4

(10) To find the number of numbers pairs, whose product is 2028 and their HCF is 13

Let the numbers be 13a and 13b.

Then, 13a * 13b = 2028, or ab = 12

or, a * b = 12

or, a * b = 4 * 3 = 12 * 1

Thus, a = 12 or 4 and b = 1 or 3

Hence, the numbers are 13 *12 and 13 *1 or 13 *4 and 13 *3.

Therefore, there are 2 such pairs.