Previously, we learned the concept of LCM and HCF. Also, we saw how to calculate LCM and HCF. It’s time we proceed further to see where does LCM and HCF find application.

**Application of LCM:**

** ****TYPE 1: (When remainder is same)**

In this type, a number *N* is divided by two or more divisors such that they give same remainders. Let a number *N* be divided by two divisors *n*_{1} and *n*_{2} such that remainder in each case is *r* then the number *N* must be in the form of k (LCM of *n*_{1}*n*_{2}) + *r *,

So,

*N* = *k*(LCM of *n*_{1}*n*_{2}) + *r*

**Example 1:** Find the largest 3-digit number that gives remainder 3 when divided by 5 or 8.

**Solution:**

Step 1:- As per the given condition the number will be in the form of *N* = *k*(LCM of *n*_{1}*n*_{2}) + *r i.e *

N= 40*k* + 3, (LCM of 5 and 8 is 40 ).

Step 2:- The largest 3 digit number in the form of 40*k* i.e. 960.

Step 3:- So the largest 3 digit number that satisfy this condition = 960 + 3 = 963.

**Example 2:** Find the smallest number which when divided by 5 or 3 leaves a remainder 3 in each case.

**Solution:**

Step 1:- As per the given condition the number will be in the form of *N* = *k*(LCM of *n*_{1}*n*_{2}) + *r i.e *

N= 15 *k* + 3, ( LCM of 5 and 3 is 12 ).

Step 2:- we will get smallest number when *k* = 1 i.e., smallest number is 18.

**TYPE 2: (When remainders are different) **

** **In this type, a number *N* is divided by two or more divisors such that they give different remainders but the difference between the divisor and remainder is same in each case. Let a number N be divided by two divisors n_{1} and n_{2} that remainders are *r*_{1} and *r*_{2} respectively, and *n*_{1}–r_{1}=n_{2}–r_{2} =x then the

*N* = *k*(LCM of *n*_{1}*n*_{2}) –*x *, where* K* is any +ve integer.

**Example 3:** Find the largest 3-digit number that gives 2 and 5 when divided by 5 and 8 respectively.

**Solution:**

Here in this case the difference between divisor and remainder i.e. 5 –3 = 8 –6 = 3. Since this difference is same, we can apply the Type 2 formula.

The number must be in the form of 40*k* – 3 since LCM of 5 and 8 is 40.

The largest 3 digit number will be in the form of 40*k* –3

25th multiple of 40 will be 1000; reducing 3 will give smallest 3-digit number fulfilling condition.

(40*25)-3 =997.

Number will be 997.

** ****Example 4:** Find the smallest 4 digit number that when divided by 3, 5 or 7 gives remainder 2, 4, and 6 respectively.

**Solution:**

Here 3 –1 = 5 – 3 = 7 – 5 = 1; that means the difference between divisor and remainder is same. Hence, any such number that satisfies the above condition must be in the form of

*N* = *k*(LCM of *n*_{1}*n*_{2}) –*x ; n*_{1}–r_{1}=n_{2}–r_{2} =x

*k*(LCM of 3, 5 & 7) – 1, or 105 *k* – 1.

Now we need to find smallest 4-digit number that is multiple of 105.

When smallest 4-digit number i.e. 1000 is divided by 105 we get a remainder 55 that means 1000 – 55 = 945 is divisible by 105.

But that is the largest 3-digit number, and the next multiple of 105 is smallest 4-digit number that is divisible by 105.

So smallest 4-digit number that is divisible by 105 will be 945 + 105 = 1050

But the required form is105 *k* – 1.

So, the required number is 1050 – 1 = 1049.

* ***Type 3 (****neither the remainder nor the difference is same)**

Let us understand this with the help of examples.

* ***Example 5**: Find the largest 3 digit number that gives remainder 1 and 5 when divided by 5 and 8 respectively.

**Solution:**

We know that dividend =divisor *quotient + remainder

Dividend = N

Divisor = d

Remainder = r

Quotient = q

So we can say that N= d*q + r

Now, according to question

N=5*a+1 …… (1)

N= 8*b+5 ……. (2)

As we know N/5 leaves remainder 1, so N must be in the form of 5*a +1

When the same number is divided by 8 then it leaves remainder 5 i.e. (5*a +1 )*/N * will give a remainder of 5

So we can say if we subtract 5 from the number .the number will be completely divided by 8

So (*5x *+ 1) –5 = 5*x* – 4 must be divisible by 8.

Now we have to do trial and error method to find the minimum value of *x *such that *5x – *4 divisible by 8.

For *x *= 1, *5a *– 4 = 1 (not divisible by 8)

For *x = 2, 5a – 4 = 6*

For *x = 3, *5a –4 =11

For *x = 4, 5a – 4 *= 16, it is divisible by 8

Now at *x = 4 ; *N=5*a+1 will give us Smallest value of 21 that satisfies these conditions.

Now any number in the form of *k*(LCM of 5 and 8) + smallest Number

or *40k + *21 will satisfy this condition.

Then largest 3 digit number that satisfy these conditions are 960 + 21 =981.

* ***Application of HCF:**

** ****TYPE 1:**

The largest number which divides the numbers a, b and c and gives the same remainder is given by HCF of (*a* – *b*) and (*b* – *c*). Here, we are concerned with positive value of the difference.

**Note:** HCF of (*a* – *b*) and (*b* – *c*) = HCF of (*a* – *b*) and (*a* – *c*) = HCF of (*b* – *c*) and (*a* – *c*).

**Example:** When 302, 752 and 1502 is divided by *N* gives same remainder in each case then find the largest possible value of *N*.

**Solution:**

In this case, since remainder is same hence required number is HCF of (752 – 302) and (1502 – 752) or HCF of 450 and 750 and it is 150. Hence, largest value of *N* is 150.

** ****TYPE 2:**

The largest number by which the numbers *a, b*, and *c* are divided giving remainders as *p*, *q*, *r* respectively, then the largest number is given by HCF of three number (*a* –*p*), (*b* – *q*) and (*c* – *r*).

**Example:** Find the largest number by which when 182, 228 and 275 are divided remainders are 2, 3 and 5 respectively.

**Solution:**

Let the number be *x.*

Hence 182 – 2 = 180 is divisible by *x *and similarly 228-3= 225 and 275-5=270 is divisible by *x *.

So, largest value of *x *is given by HCF of 180, 225 and 270 and that is 45.

Try out some more examples:

**Example 1:** On a traffic signal, traffic light changes its color after every 24, 30 & 36 in green, red & orange light. How many times in an hour only green & red light will change simultaneously.

**Solution:**

L.C.M of 24 & 30 =120

So in 1 hr both green & red light will change 3600/120 times = 30 times

L.C.M of 24, 30 & 36 is 360

Hence in 1 hr all three light will change together 3600/360 = 10 times

So in 1 hr only red & green light will change 30 – 10 = 20 times simultaneously.

**Example 2:** Students of a class are preparing for a drill. If 7 students are arranged in a row then, 4 students are left out. If 9 students are arranged in a row, then 6 students are left out. If 21 students are arranged in a row then how many students will be left out?

**Solution:**

This is the question of LCM Type 2 questions. Here the difference 7 – 4 = 9 –6 = 3, hence any number in the form of *63k – *3 will satisfy this condition. When *63k *– 3 is divided by 21 then remainder will be 21 – 3 = 18.

Let’s practice some more questions based on this concept.

**EXERCISE: **

**Question 1.** If the students of a class can be grouped exactly into 6 or 8 or 10, then the minimum number of students in the class must be

(1) 60

(2) 120

(3) 180

(4) 240

### Answer and Explanation

**Solution: (2)**

If we need to group total number of students into sub group, total number has to be divided by every given number.

Therefore, the required number of students = LCM of 6, 8, 10 = 120.

**Question 2.** A number which when divided by 10 leaves a remainder of 9, when divided by 9 leaves a remainder of 8, and when divided by 8 leaves a remainder of 7, is :

(1) 1539

(2) 539

(3) 359

(4) 1359

### Answer and Explanation

**Solution: (3)**

*Number will be of the form *

*N* = *k*(LCM of *n*_{1}*n*_{2}) –*x *, *n*_{1}–r_{1}=n_{2}–r_{2} =x

Here, Divisor – remainder = 1 e.g., 10 – 9 = 1, 9 – 8 = 1, 8 – 7 = 1

So Required number = (L.C.M. of 10, 9, 8) – 1 = 360 – 1 = 359.

**Question 3.** When a number is divided by 15, 20 or 35, each time the remainder is 8. Then the smallest number is

(1) 428

(2) 427

(3) 328

(4) 338

### Answer and Explanation

**Solution: (1)**

LCM of 15, 20 and 35 = 420

Number will be of the form *N* = *k*(LCM of *n*_{1}*n*_{2}) + *r*

So, the required least number = 420 +8 = 428.

**Question 4.** Four bells ring at intervals of 4, 6, 8 and 14 seconds. They start ringing simultaneously at 12.00 O’clock. At what time will they again ring simultaneously ?

(1) 12 hrs. 2 min. 48 sec.

(2) 12 hrs. 3 min.

(3) 12 hrs. 3 min. 20 sec.

(4) 12 hrs. 3 min. 44 sec.

### Answer and Explanation

**Solution: (1)**

LCM of 4, 6, 8, 14 = 168 seconds = 2 minutes, 48 seconds

They ring again at 12hours + 2 min. 48 sec. = 12 hrs. 2 min. 48 sec.

**Question 5.** 4 bells ring at intervals of 30 minutes, 1 hour, 17: hour and 1 hour 45 minutes respectively. All the bells ring simultaneously at 12 noon. They will again ring simultaneously at;

(1) 12 mid night

(2) 3 a.m.

(3) 6 a.m.

(4) 9 a.m.

### Answer and Explanation

**Solution: (4)**