Find the largest positive integer that will divide 122, 150, and 115 leaving the remainder 5, 7, and 11 respectively.

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answer is C.

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Detailed Solution

Concept- We will now apply the remainder theorem, which is N = pq+r, to all three of the numbers that have been presented to us. When N is divided by p, q and r are the quotient and remainder, respectively. We will obtain the three multiples of the necessary number from the remainder theorem, and from those multiples, after determining the HCF of those numbers, we will proceed to our solution.
Given data: when 122, 150, and 115 are divided by the necessary number, 5, 7, and 11 correspondingly are left over. Let A be the integer.
According to the remainder theorem, if a number N is divided by p, then q and r are the quotient and remnant, respectively, and their relationship is N = pq+r. When 122 is divided by A, the facts indicate that there is a 5-digit remainder.
Let q1 be the quotient

5 is subtracted from both sides to get,

And if 150 divided by A gives a 7-point leftover, let

be the quotient.

When we subtract 7 from both sides, we get,

Furthermore, when 115 is divided by A, 11 is the result.
Let

be the quotient.

Using the prime factorization approach, we can now state that the HCF of

, and

will be the highest value of A.
Question Image

Thus the HCF is 13
Hence, the correct option is 3.

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