Question

find the greatest number which divides 2706, 7041, 8250 leaving remainder 6,21 and 42 respectively

Answer 1

$2706 \times 6 = 10236$
so the first and is here
$7041 \times 21 = 147861$
the second one

Answer 2

The greatest number which divides 2706, 7041, and 8250 leaving remainder 6, 21, and 42 respectively is 108

Given:

Numbers = 2706, 7041, 8250

Their remainders = 6, 21, 42

To find:

Greatest number dividing "2706, 7041, and 8250" leaving remainder "6, 21 and 42" respectively.

Solution:

(i) If the number 2706 is divided by the specific number, it gives a remainder of 6.

That means 2706 - 6 = 2700, 2700 is accurately divisible by that number.

(ii) If the number 7041 is divided by the specific number, it gives a remainder of 21.

That means 7041 - 21 = 7020, 7020 is accurately divisible by that number.

(iii) If the number 8250 is divided by the specific number, it gives a remainder of 42.

That means 8250 - 42 = 8208, 8208 is accurately divisible by that number.

Then taking H.C.F for the three numbers,

$2700=3 \times 3 \times 3 \times 5 \times 5 \times 2 \times 2$

$7020=5 \times 2 \times 2 \times 3 \times 3 \times 3 \times 13$

$8208=2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$

The highest number that divides all the three numbers (2700, 7020, and 8208) is 108.