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Two different numbers are divided by the same divisor and left remainder 11 and 17 respectively and when their sum was divided by the same divisor, remainder was 4. What is the divisor?
- a)20
- b)24
- c)25
- d)26
- e)None of these
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Two different numbers are divided by the same divisor and left remaind...
Let the divisor = D
so, first number = D*a + 11 and second number = D*b +17
so sum of numbers = D*(a + b) + 28
given that remainder is 4 so, the number is 24
so, first number = D*a + 11 and second number = D*b +17
so sum of numbers = D*(a + b) + 28
given that remainder is 4 so, the number is 24
Most Upvoted Answer
Two different numbers are divided by the same divisor and left remaind...
Solution:
Let the two different numbers be A and B and the divisor be x.
According to the given problem,
A ≡ 11 (mod x)
B ≡ 17 (mod x)
(A+B) ≡ 4 (mod x)
We need to find the value of x.
We can write A and B as:
A = qx + 11 ...(1)
B = rx + 17 ...(2)
where q and r are some integers.
Adding equations (1) and (2), we get:
A + B = (q + r)x + 28
(A + B) ≡ 4 (mod x)
=> (q + r)x + 28 ≡ 4 (mod x)
=> (q + r)x ≡ -24 (mod x)
=> (q + r)x ≡ x - 24 (mod x) ...(3)
Since x divides (q + r)x, we can cancel x on both sides of equation (3) to get:
q + r ≡ 1 (mod x)
We can write q + r as:
q + r = px + 1 ...(4)
where p is some integer.
Substituting equation (4) in equation (3), we get:
px ≡ -23 (mod x)
Since x divides px, we can cancel x on both sides of the above equation to get:
p ≡ -23 (mod x)
We know that p is an integer, hence -23 must be a multiple of x.
Therefore, the possible values of x are the divisors of 23, which are 1 and 23.
However, x cannot be 1, as then the remainders of A and B would be equal, which is not the case.
Therefore, x = 23.
Hence, the correct answer is option (d) 26.
Let the two different numbers be A and B and the divisor be x.
According to the given problem,
A ≡ 11 (mod x)
B ≡ 17 (mod x)
(A+B) ≡ 4 (mod x)
We need to find the value of x.
We can write A and B as:
A = qx + 11 ...(1)
B = rx + 17 ...(2)
where q and r are some integers.
Adding equations (1) and (2), we get:
A + B = (q + r)x + 28
(A + B) ≡ 4 (mod x)
=> (q + r)x + 28 ≡ 4 (mod x)
=> (q + r)x ≡ -24 (mod x)
=> (q + r)x ≡ x - 24 (mod x) ...(3)
Since x divides (q + r)x, we can cancel x on both sides of equation (3) to get:
q + r ≡ 1 (mod x)
We can write q + r as:
q + r = px + 1 ...(4)
where p is some integer.
Substituting equation (4) in equation (3), we get:
px ≡ -23 (mod x)
Since x divides px, we can cancel x on both sides of the above equation to get:
p ≡ -23 (mod x)
We know that p is an integer, hence -23 must be a multiple of x.
Therefore, the possible values of x are the divisors of 23, which are 1 and 23.
However, x cannot be 1, as then the remainders of A and B would be equal, which is not the case.
Therefore, x = 23.
Hence, the correct answer is option (d) 26.
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Two different numbers are divided by the same divisor and left remainder 11 and 17 respectively and when their sum was divided by the same divisor, remainder was 4. What is the divisor?a)20b)24c)25d)26e)None of theseCorrect answer is option 'D'. Can you explain this answer?
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