In this post we will understand question when the sum of number and their HCF is given. All the questions are solved step by step so that you can understand the concept better.

Also remember that HCF/GCD/GCF all are same thing so don’t get confused with the different terms used.

## HCF and Sum of Numbers

Its given that HCF of two numbers is 33

So the numbers can be written as 33x and 33y

Also, the sum of number is 528

so, 33x + 33y =528

==> x + y = 528/33

==> x+ y = 16 –eq(1)

Now we have the value of x and y such that there is no common factor involved between then (i.e x and y are co-prime)

The possible pair of x & y are**(1, 15), (3, 13), (5, 11), (7,9)**

Thus there are 4 possible pairs of x and y

The possible numbers are

(33, 33*15) ===> (33, 495)

(33*3, 33*13) ==> (99, 429)

(33*5, 33*11) ==> (165, 363)

(33*7, 33*9) ==> (231, 297)

The HCF of two number is 27

So the numbers can be written as 27x and 27y

Also the sum of the number is given as 216

==> 27x + 27y =216

==> x + y = 216/27

==> x + y = 8 ——– eq(1)

Now we have the value of x and y such that there is no common factor involved between then (i.e x and y are co-prime)

The possible pair of x & y are need to be found using hit and trial method

Putting the value of x and y to get the number value,

(27*1, 27*7) ==> (27, 189)

(27*3, 27*5) ==> (81, 135)

**Hence there are two sets of number possible**

The HCF of two number is 5

So the two number can be written as 5x and 5y

(where x & y are co-prime, which means there is no common factor between the)

Also the LCM of the two number is given as 495

Now keeping in mind of the two numbers 5x and 5y, the LCM can be expressed as:

==> 5 * x * y = 495

==> x * y = 99 —- eq(1)

Now from eq (1) , we can have to find the set of value of x * y such that the numbers are co prime.

The possible set of (x, y )numbers are :

(1, 99)

(9, 11)

Now the possible value of numbers are

(5, 99*5) ==> (5, 495)

(9*5, 11*5) ==> (45, 55)

In the question, we have been provided with the condition that the sum of number is 100.

So out of above two sets, the second set (45, 55) satisfy the condition fully**Hence (45, 55 ) are the two required numbers**

Its given that HCF of two numbers is 4

So the numbers can be written as 4x and 4y

Also, the sum of number is 36

so, 4x + 4y =36

==> x + y = 36/4

==> x+ y = 9 –eq(1)

Now we have the value of x and y such that there is no common factor involved between then (i.e x and y are co-prime)

The possible pair of x & y are**(1, 8), (2, 7), (7,9)**

Let the HCF be x

Now as per the question, the LCM will be 12x

The sum of HCF and LCM is 403

==> x + 12x =403

==> 13x = 403

==> x =31

Hence the HCF =31 —-(1)

and LCM =12 * 31 —-(2)

Now one number is given as 93, we have to find the other number

According to HCF & LCM formula

Using the above formula

==> 31 * 12 * 31 = 93 * y

==> y = 31 * 4 => 124

**Hence 124 is the other number**

The HCF of the number is 2

So the numbers are 2x+2 and 2x+4 (since they are consecutive even numbers) —(1)

LCM of the number is 84

According to HCF & LCM Formula

HCF * LCM = Product of two numbers

=> 2 * 84 = (2x+2) * (2x+4)

=> (x+1) (x+2) = 84

Hence we get x=> 5

Putting the value of x in (1) we get

=>((2*5)+2)) and ((2*5)+4))

=> 12 and 14

**Hence 12 and 14 is the right answer**

Let the HCF of two numbers be **a**

So the two numbers will be **ax** and **ay** (out of both let’s suppose ax is the larger number)

Given the HCF and two numbers, the LCM can be written as

LCM = a * x * y

Given that LCM is twice the largest number

LCM = a *x * y = 2 * a * x

we get y==> 2 —- eq(1)

Its given that difference of smaller number and GCD =4

==> ay – a =4

Putting value of y=2

==> 2a – a =4

==> a = 4 —eq(2)

We know that the smaller number is a * y

Putting the value of a & y we get

==> 4 * 2

==> 8**Hence 8 is the smaller number we are looking for**.