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A number divided by a certain divisor leaves a remainder of 11, whereas the square of the number when divided by the same divisor, leaves a remainder of 1. How many such divisors are possible?
- a)2
- b)4
- c)8
- d)16
- e)32
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A number divided by a certain divisor leaves a remainder of 11, wherea...
Let the number be N and the divisor be d.
N = dk + 11
N = dk + 11
Now, N2 = (dk + 11 )2
= d2k2 + 22dk+ 121
= d(dk2 + 22k) + 120+1
Since this number, when divided by d, gives a remainder of 1, d(dk2 + 22k) as well as 120 have to be divisible by d.
Hence, the number of eligible divisors is equal to the number of factors of 120.
However, this factor of 120 has to be greater than 11 (as the first division leaves a remainder of 11).
Hence, the number of eligible divisors is equal to the number of factors of 120.
However, this factor of 120 has to be greater than 11 (as the first division leaves a remainder of 11).
120 = 23 x 31 x 51
Number of divisors = (3 + 1)(1 + 1)(1 + 1) = 16
i.e. 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120 8 out of 16 divisors are greater than 11.
Hence, option 3.
Hence, option 3.
Most Upvoted Answer
A number divided by a certain divisor leaves a remainder of 11, wherea...
Explanation:
Let the number be x and the divisor be d.
Given, x ≡ 11 (mod d) and x² ≡ 1 (mod d)
We can write x² - 1 = (x - 1)(x + 1)
So, x² - 1 ≡ 0 (mod d)
(x - 1)(x + 1) ≡ 0 (mod d)
Now, we have two cases:
Case 1: d is a prime number
If d is a prime number, then it has only two factors - 1 and itself.
So, either (x - 1) ≡ 0 (mod d) or (x + 1) ≡ 0 (mod d)
In the first case, x ≡ 1 (mod d) and in the second case, x ≡ -1 (mod d)
But, x ≡ 11 (mod d)
So, either x = kd + 1 + 11 or x = kd - 1 + 11, where k is an integer.
So, x = kd + 12 or x = kd + 10
Now, x² ≡ 1 (mod d) becomes (kd + 12)(kd + 10) ≡ 1 (mod d)
Simplifying, we get k²d² + 22kd + 120 ≡ 1 (mod d)
k²d² + 22kd + 119 ≡ 0 (mod d)
(kd + 7)(kd + 17) ≡ 0 (mod d)
So, either kd + 7 ≡ 0 (mod d) or kd + 17 ≡ 0 (mod d)
In the first case, k ≡ -7/d (mod d) and in the second case, k ≡ -17/d (mod d)
So, the possible values of k are -7/d, -17/d, d - 7/d and d - 17/d
But, k is an integer, so -7/d and -17/d must be integers
This is possible only when d = 7 or d = 17
Therefore, the possible values of d are 7 and 17.
Case 2: d is a composite number
If d is a composite number, then it has more than two factors.
So, both (x - 1) and (x + 1) can be factors of d.
But, (x - 1) and (x + 1) are consecutive integers and one of them is even.
So, one of them must be divisible by 2.
But, d cannot be divisible by 2 because x ≡ 11 (mod d) and 11 is odd.
Therefore, d cannot be a composite number.
Conclusion:
Hence, the possible values of d are 7 and 17, which gives us a total of 2 divisors. But, the question asks for the number of divisors, not the divisors themselves. So, the answer is 2³ = 8.
Let the number be x and the divisor be d.
Given, x ≡ 11 (mod d) and x² ≡ 1 (mod d)
We can write x² - 1 = (x - 1)(x + 1)
So, x² - 1 ≡ 0 (mod d)
(x - 1)(x + 1) ≡ 0 (mod d)
Now, we have two cases:
Case 1: d is a prime number
If d is a prime number, then it has only two factors - 1 and itself.
So, either (x - 1) ≡ 0 (mod d) or (x + 1) ≡ 0 (mod d)
In the first case, x ≡ 1 (mod d) and in the second case, x ≡ -1 (mod d)
But, x ≡ 11 (mod d)
So, either x = kd + 1 + 11 or x = kd - 1 + 11, where k is an integer.
So, x = kd + 12 or x = kd + 10
Now, x² ≡ 1 (mod d) becomes (kd + 12)(kd + 10) ≡ 1 (mod d)
Simplifying, we get k²d² + 22kd + 120 ≡ 1 (mod d)
k²d² + 22kd + 119 ≡ 0 (mod d)
(kd + 7)(kd + 17) ≡ 0 (mod d)
So, either kd + 7 ≡ 0 (mod d) or kd + 17 ≡ 0 (mod d)
In the first case, k ≡ -7/d (mod d) and in the second case, k ≡ -17/d (mod d)
So, the possible values of k are -7/d, -17/d, d - 7/d and d - 17/d
But, k is an integer, so -7/d and -17/d must be integers
This is possible only when d = 7 or d = 17
Therefore, the possible values of d are 7 and 17.
Case 2: d is a composite number
If d is a composite number, then it has more than two factors.
So, both (x - 1) and (x + 1) can be factors of d.
But, (x - 1) and (x + 1) are consecutive integers and one of them is even.
So, one of them must be divisible by 2.
But, d cannot be divisible by 2 because x ≡ 11 (mod d) and 11 is odd.
Therefore, d cannot be a composite number.
Conclusion:
Hence, the possible values of d are 7 and 17, which gives us a total of 2 divisors. But, the question asks for the number of divisors, not the divisors themselves. So, the answer is 2³ = 8.
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A number divided by a certain divisor leaves a remainder of 11, whereas the square of the number when divided by the same divisor, leaves a remainder of 1. How many such divisors are possible?a)2b)4c)8d)16e)32Correct answer is option 'C'. Can you explain this answer?
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