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What is the largest positive integer that will divide 398, 436 and 542 leaving remainder 7, 11 and 15 respectively?
- a)16
- b)18
- c)17
- d)14
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
What is the largest positive integer that will divide 398, 436 and 542...
To find the largest positive integer that will divide 398, 436, and 542 and leave remainders of 7, 11, and 15 respectively, we can use the concept of congruence.
Let's denote the unknown integer as 'x'. We can set up congruence equations for each given condition:
398 ≡ 7 (mod x) -- (1)
436 ≡ 11 (mod x) -- (2)
542 ≡ 15 (mod x) -- (3)
To solve these congruences, we can start by finding the least common multiple (LCM) of the remainders (7, 11, 15) which is 105.
Now, we can rewrite the congruences as:
398 ≡ 7 (mod x)
436 ≡ 11 (mod x)
542 ≡ 15 (mod x)
Using the LCM, we can rewrite these congruences as:
398 ≡ 7 (mod 105)
436 ≡ 11 (mod 105)
542 ≡ 15 (mod 105)
By subtracting the remainders from each side, we get:
391 ≡ 0 (mod 105)
425 ≡ 0 (mod 105)
527 ≡ 0 (mod 105)
These congruences indicate that 105 is a divisor of 391, 425, and 527.
To find the largest positive integer 'x' that satisfies all three congruences, we can find the greatest common divisor (GCD) of 391, 425, and 527 with 105.
GCD(391, 425, 527) = 17
Therefore, the largest positive integer that will divide 398, 436, and 542 leaving remainders of 7, 11, and 15 respectively is 17.
Hence, the correct answer is option C: 17.
Let's denote the unknown integer as 'x'. We can set up congruence equations for each given condition:
398 ≡ 7 (mod x) -- (1)
436 ≡ 11 (mod x) -- (2)
542 ≡ 15 (mod x) -- (3)
To solve these congruences, we can start by finding the least common multiple (LCM) of the remainders (7, 11, 15) which is 105.
Now, we can rewrite the congruences as:
398 ≡ 7 (mod x)
436 ≡ 11 (mod x)
542 ≡ 15 (mod x)
Using the LCM, we can rewrite these congruences as:
398 ≡ 7 (mod 105)
436 ≡ 11 (mod 105)
542 ≡ 15 (mod 105)
By subtracting the remainders from each side, we get:
391 ≡ 0 (mod 105)
425 ≡ 0 (mod 105)
527 ≡ 0 (mod 105)
These congruences indicate that 105 is a divisor of 391, 425, and 527.
To find the largest positive integer 'x' that satisfies all three congruences, we can find the greatest common divisor (GCD) of 391, 425, and 527 with 105.
GCD(391, 425, 527) = 17
Therefore, the largest positive integer that will divide 398, 436, and 542 leaving remainders of 7, 11, and 15 respectively is 17.
Hence, the correct answer is option C: 17.
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Community Answer
What is the largest positive integer that will divide 398, 436 and 542...
398 - 7 = 391;
436 - 11 = 425;
542 - 15 = 527
Required number = H.C.F. of 391, 425 and 527
= 17
436 - 11 = 425;
542 - 15 = 527
Required number = H.C.F. of 391, 425 and 527
= 17
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What is the largest positive integer that will divide 398, 436 and 542 leaving remainder 7, 11 and 15 respectively?a)16b)18c)17d)14Correct answer is option 'C'. Can you explain this answer?
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