Question

3.3 The largest possible number
with which
when 38, 66 and 80 are divided
The remainders remain the same as-​

Answer 1

Answer:

Let the common remainder be 10

From the given numbers we now subtract the common remainder

i.e 38 − 10 = 28,66 −10 = 56, 80 −10 = 70

Now we will find the HCF of 28, 56, 70

28 = 2 x 7 x 2

56 = 2 x 2 x 2 x 7

70 = 2 x 5 x 7

So HCF of 28, 56 and 70 = 2 x 7

So the required number is 14

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Answer 2

Answer:

14

Step-by-step explanation:

Let the common remainder be 10

From the given numbers we now subtract the common remainder

i.e 38 − 10 = 28,66 −10 = 56, 80 −10 = 70

Now we will find the HCF of 28, 56, 70

28 = 2 x 7 x 2

56 = 2 x 2 x 2 x 7

70 = 2 x 5 x 7

So HCF of 28, 56 and 70 = 2 x 7

So the required number is 14