Tougher problems on HCF and LCM

In the HCF of two numbers which were prime to each other, the quotients are 2,2,7,2 on finding HCF by method of division. Find the two numbers.

 Answer:

Since the numbers are prime, the HCF is 1.

Now following the method of finding HCF by the method of division, we get

Final Dividend = 1x2 = 2

In the pen ultimate division, the divisor is 2, the quotient is 7 and the remainder is 1.

So, the dividend is = 2x7 + 1 = 15

In the prior division, the divisor is 15, the quotient is 2 and the remainder is 2.

So, the dividend is = 15x 2 + 2 = 32

Therefore, the smaller of the two numbers is 32.

The next number is the divisor = 32 x 2 + 15 = 79

So, the two numbers are 32 and 79.

                                                               

Check:

                               32) 79 (2

                                     64

                                       15) 32 (2

                                              30

                                                2)15 (7

                                                   14

  1) 2 (2

2

x

HCF by method of division

In finding the HCF of two numbers, the last divisor is 49 and the quotient 17, 3 and 2. Find the numbers?

Answer:

Since the last divisor is 49 and the quotient is 2, we have the dividend = 49 x 2 = 98. (Because the last divisor will not leave any remainder)

For the pen-ultimate, 98 will become the divisor and 3 will be the quotient and 49 will be the remainder. Therefore, the dividend will be 98 x 3 + 49 = 343

Now, 343 will become the divisor and 17 will be the quotient and 98 will be the remainder.

Dividend = 343 x 17 + 98 = 5929.

So, the two numbers are 5929 and 343.

Finding HCF by the method of Division

Finding HCF by the method of Division

Step 1: Divide the bigger number with the smaller number.

Step 2: divide the divisor by the remainder

Step 3: divide the remainder by the next remainder, and proceed till no remainder is left.

Example:

Find the HCF of 13281 and 15844

The smaller of the two numbers is 13281, so, we will divide 15844 by 13281

13281 ) 15844 ( 1

               13281

                 2563 ) 13281 ( 5

                             12815

                                 466) 2563 (5

                                         2330

                                           233 ) 466 ( 2

                                                    466

                                                     x x

Therefore, 233 is the HCF of the 2 numbers.

tougher problems in HCF and LCM

                                                                                                                                                                                               

The numbers 11284 and 7655, when divided by a certain number of three digits, leave the same remainder. Find the number and the remainder.

                                                                                                                                                                                               

Answer:

Let, Y be the common remainder. So, we have 11284 – y , 7655 – y as the perfectly divisible numbers.

If 2 numbers are exactly divisible by a certain number, then there difference is also exactly divisible.

So, (11284 – y) – (7655 – y), is exactly divisible.

Or, 3629 is divisible by the required number. Let us find the factors of 3629.

We have, 3629 = 191 x 19

Therefore the 3 digit number is 191.

Now let us divide 11284 by 191 to find the remainder

11284/191 = 59 ¹⁵/₁₉₁.

Therefore, the remainder is 15.

Tougher problems on HCF and LCM

What is the greatest number that will divide 2930 and 3250 and will leave a remainder 7 and 11 respectively?  

    

Answer:

(2930 -7)= 2923 is perfectly divisible.

Also,( 3250 -11) 3239 is perfectly divisible.

Now, let us find the HCF of 2923 and 3239

2923 = 37 x 79

3239 = 41 x 79

Therefore, the HCF is 79.              

Find the greatest number which is such that when 12288, 19139 

and 28200 are divided by it, the remainders are all the same.

Answer:

Let, Y be the common remainder. So, we have 12288 – y , 19139 – y , 28200 – y as the perfectly divisible numbers.

If 2 numbers are exactly divisible by a certain number, then there difference is all exactly divisible.

So, (19139 – y) – (12288 – y), (28200 – y) – (19139 – y) and (28200 – y) – (12288 – y) are exactly divisible.

Or, 6851, 9061 and 15912 are divisible by the required number.

Therefore let us find the HCF of  6851, 9061 and 15912.

6851 = 17 x 13 x 31

9061 = 17 x 13 x 41

15912 = 17 x 13 x 2 x 3 x 3 x 2 x 2

Therefore, HCF = 17 x 13 = 221.

So, the required number is 221.