Question

Find the largest number that will divide 382 and 710 and leaves a remainder 13 in each case

Answer 1

Answer:

41

Step-by-step explanation:

382-13=369

710-13=697

369=3^2*41

697=17*41

the greatest no.(HCF or GCF)=41

cheak:- 382/41=quotient=9,remainder=13

            710/41==quiotient=17,remainder=13

Answer 2

The largest number required is 41.

Given,

Number 1 = 382

Number 2 = 710

To Find,

The largest number that will divide the given numbers and leave a remainder of 13 in each case =?
Solution,

Let the number be x

When number 1 and number 2 are divided by x, they leave a remainder of 13.

That means, if 13 is subtracted from the numbers then numbers are completely divisible by x leaving no remainder behind.

Therefore,

Number 1 = 382 - 13 = 369

Number 2 = 710 - 13 = 697

By factorization of 369 and 697,

369 = 3 * 3 * 41

697 = 17 * 41

The highest common factor in both these numbers is 41.

The H.C.F of 369 and 697 is 41

Hence, the largest number that will divide 382 and 710 and leaves a remainder of 13 in each case is 41.