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When dividing each of the numbers $$3759$$ and $$4139$$ by some positive integer number, the remainder happened to be equal to $$16$$. What is the smallest possible value of the divisor?

User Zymus
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1 Answer

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Answer:

The smallest possible value of the divisor is 19

Explanation:

The given numbers for division are;

3759 and 4139, the remainder = 16

Let x represent the number

Therefore, we have;

x > 16

x is a factor of 3759 - 16 = 3743, and x is also a factor of 4139 - 16 = 4123

The factors of 3743 obtained from an online resource are 1, 19, 197, 3759

The factors of 4123 obtained from an online resource are 1, 7, 19, 31, 133, 217, 589, 4123

The common factors of 3743 and 4123 are 1, 19

Given that x > 16, we have, x = 19

Therefore, the smallest possible value that will divide both 3759 and 4139 whereby the remainder is 16 = 19

User Nareshkumar
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