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.