Question

Find the least number that divides 626, 3127 and 15628 leaving remainders 1, 2 and 3.​

Answer 1

Answer:

The required number is 625

Step-by-step explanation:

626 , 3127 and 15628 leaves remainders 1,2 and 3

So,we have to subtract the following number with their remainders respectively

(1) 626-1=625

(2) 3127-2=3125

(3) 15628-3=15625

Now, we have to find the HCF of thes following resultant number

1) 625=5×5×5×5

2) 3125=5×5×5×5×5

3) 15625=5×5×5×5×5×5

The HCF = 5×5×5×5

= 625

So, the least number that divides the following number is 625.

Answer 2

Answer:

The required least number is 625.

Step-by-step explanation:

From the question it’s understood that,

626 – 1 = 625, 3127 – 2 = 3125 and 15628 – 3 = 15625 has to be exactly divisible by the

number.

Thus, the required number should be the H.C.F of 625, 3125 and 15625.

First, consider 625 and 3125 and apply Euclid’s division lemma

3125 = 625 x 5 + 0

∴ H.C.F (625, 3125) = 625

Next, consider 625 and the third number 15625 to apply Euclid’s division lemma

15625 = 625 x 25 + 0

We get, the HCF of 625 and 15625 to be 625.

∴ H.C.F. (625, 3125, 15625) = 625

So, the required number is 625.