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07 Jul 2025, 21:51
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For x^3 to be multiple of 81 x should have atleast 2 3s means minimum 9 should be there then only x^2 would be divisible by 81.
If we write as 9m for value of x^3 then it is a cyclic remainer of 9, 18 27,0 and 9 again.
So option D is right.
If we write as 9m for value of x^3 then it is a cyclic remainer of 9, 18 27,0 and 9 again.
So option D is right.
Bunuel
07 Jul 2025, 22:46
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If \(x\) is a positive integer and \(x^3\) is a multiple of by \(81\), which of the following could be the remainder when \(x\) is divided by \(36\)?
Given \(x^3\) is a multiple of \(81\)
We have \(x^3 = 81y = (3^4)y\)
This mean \(x^3\) should have a minimum of product of four 3's; 3*3*3*3 = 81
For this to be true, \(x\) must be a multiple of \(9\); \(x = 9a\)
Now considering the multiples of \(9\) and its remainders,
\(\\
9 - 9/36 -> R = 9\\
18 - 18/36 -> R = 18\\
27 - 27/36 -> R = 27\\
36 - 36/36 -> R = 0\\
45 - 45/36 -> R = 9\\
54 - 54/36 -> R = 18\\
63 - 63/36 -> R = 27\\
\)
We can see the remainders possible as 9, 18, 27 & 0; The options I & III are the only ones correct.
So the correct option is Option D
Given \(x^3\) is a multiple of \(81\)
We have \(x^3 = 81y = (3^4)y\)
This mean \(x^3\) should have a minimum of product of four 3's; 3*3*3*3 = 81
For this to be true, \(x\) must be a multiple of \(9\); \(x = 9a\)
Now considering the multiples of \(9\) and its remainders,
\(\\
9 - 9/36 -> R = 9\\
18 - 18/36 -> R = 18\\
27 - 27/36 -> R = 27\\
36 - 36/36 -> R = 0\\
45 - 45/36 -> R = 9\\
54 - 54/36 -> R = 18\\
63 - 63/36 -> R = 27\\
\)
We can see the remainders possible as 9, 18, 27 & 0; The options I & III are the only ones correct.
So the correct option is Option D
07 Jul 2025, 23:05
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Analyze the constraint
Since x3 is a multiple of 81 = 34, we need x3 to be divisible by 34.
For x3 to be divisible by 34, x must be divisible by at least 32 = 9. (Since if x = 3a × k where gcd(k,3) = 1, then x3 = 33a × k3, and we need 3a ≥ 4, so a ≥ 4/3, meaning a ≥ 2)
So x must be a multiple of 9.
Check each option
Option I: Remainder 0 If x ≡ 0 (mod 36), then x is a multiple of 36. Since 36 = 9 × 4, x is also a multiple of 9.
Option II: Remainder 3 If x ≡ 3 (mod 36), then x = 36k + 3 = 3(12k + 1). Since 12k + 1 is not divisible by 3, x is only divisible by 31, not 32. So x is not a multiple of 9.
Option III: Remainder 27 If x ≡ 27 (mod 36), then x = 36k + 27 = 9(4k + 3). So x is a multiple of 9. ✓
Verify with examples
Answer: D. I and III only
Since x3 is a multiple of 81 = 34, we need x3 to be divisible by 34.
For x3 to be divisible by 34, x must be divisible by at least 32 = 9. (Since if x = 3a × k where gcd(k,3) = 1, then x3 = 33a × k3, and we need 3a ≥ 4, so a ≥ 4/3, meaning a ≥ 2)
So x must be a multiple of 9.
Check each option
Option I: Remainder 0 If x ≡ 0 (mod 36), then x is a multiple of 36. Since 36 = 9 × 4, x is also a multiple of 9.
Option II: Remainder 3 If x ≡ 3 (mod 36), then x = 36k + 3 = 3(12k + 1). Since 12k + 1 is not divisible by 3, x is only divisible by 31, not 32. So x is not a multiple of 9.
Option III: Remainder 27 If x ≡ 27 (mod 36), then x = 36k + 27 = 9(4k + 3). So x is a multiple of 9. ✓
Verify with examples
- For remainder 0: x = 36, x3 = 46,656 = 81 × 576
- For remainder 27: x = 27, x3 = 19,683 = 81 × 243
Answer: D. I and III only
