Question

FIND THE LEAST NUMBER WHICH WHEN DIVIDED BY 48,72 AND 96 WILL LEAVE THE REMAINDERS AS 42,66 AND 90.
PLEASE QUICKLY WITH PROPER CALCULATIONS PLEASE

Answer 1

Answer:

Take HCF of 48,72 and 60

You will get 12 , right.

Which is your desired result

There are two methods of finding HCF.

1- by division method

2- by prime factorization method

So by first method, solution goes like this

Take any two numbers from above

Say 48,72

Now divide 72 by 48 without solving it to decimal.

You will get here-

Remainder- 24. Quotient-1

Now you have to divide the first divisor(48) by this remainder(24) .

Now you get remainder zero.

Now this last divisor 24 that makes your remainder zero is the HCF OF THESE TWO numbers.

We repeat this process of division till we zero as remain

Give Amazon a try.

You will need to find the greatest common factor. You can do this multiple ways, but one way is to find the prime factorization of each number and handpick the GCF.

48= 6x8= (2x3)(2^3)=2^4 x 3

72= 9 x 8= 3^2 x 2^3

60=10x 6= 5 x 2 x 3 x 2= 5 x 2^2 x 3

First notice that you can pull out a 2^2 from each one

Dividing by 2^2, or 4 gives:

12= 2^2 x 3

18= 3^2 x 2

15= 5 x 3

Now it’s clear to see that you can also take out a 3 from each number

Dividing by 3 gives

4= 2^2

6= 3 x2

5=5 x 1

The remaining numbers are 4, 6, 5, which are the original numbers with the common factors taken out.

The GCF is 4x3=12

Hope this is clear.

Step-by-step explanation:

Answer 2

Answer:

When we divide 487 by 84 we get 67 as Remainder. So, 487 is Required No.