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08 Jul 2025, 07:02
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x^3 -> multiple of 81=3^4
If x=a*3^2, x^3=a^3 * 3^6, divisible by 81. So x has two threes (3*3=9) in its factorization x=9a
Dividend = Divisor*Quotient + Remainder
x=36*Quotient + Remainder
9a=36*Quotient + Remainder
a must be also an integer, so:
a=4*Quotient + Remainder/9
Remainder/9 must be an integer.
I. 0/9=0 integer
II. 3/9 non-integer
III. 27/9=3 integer
I and III only
Correct answer is D
If x=a*3^2, x^3=a^3 * 3^6, divisible by 81. So x has two threes (3*3=9) in its factorization x=9a
Dividend = Divisor*Quotient + Remainder
x=36*Quotient + Remainder
9a=36*Quotient + Remainder
a must be also an integer, so:
a=4*Quotient + Remainder/9
Remainder/9 must be an integer.
I. 0/9=0 integer
II. 3/9 non-integer
III. 27/9=3 integer
I and III only
Correct answer is D
08 Jul 2025, 07:19
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x is a positive integer and x^3 is a multiple of by 81
We need to know remainder when x is divided by 36 and the given options are
1. 0
2. 3
3. 27
So,
x is divisible by 3^⌈4/3⌉ = 3^2 = 9
1. Remainder = 0
x ≡ 0 mod 36
=> x is divisble by 36, also by 9
Satisfied
2. Remainder = 3
x≡3 mod 36
=> x is not divisble by 9
Thus x is not divisble by 81
Not satisfied
3. Remainder = 27
x≡27 mod 36
Say,
x = 36k + 27
x mod 9 = (36k + 27) mod 9 = 0+0 = 0
=> x≡ 0 mod 9
Satisified
=> Only 0 and 27 possible
C. I and III only
We need to know remainder when x is divided by 36 and the given options are
1. 0
2. 3
3. 27
So,
x is divisible by 3^⌈4/3⌉ = 3^2 = 9
1. Remainder = 0
x ≡ 0 mod 36
=> x is divisble by 36, also by 9
Satisfied
2. Remainder = 3
x≡3 mod 36
=> x is not divisble by 9
Thus x is not divisble by 81
Not satisfied
3. Remainder = 27
x≡27 mod 36
Say,
x = 36k + 27
x mod 9 = (36k + 27) mod 9 = 0+0 = 0
=> x≡ 0 mod 9
Satisified
=> Only 0 and 27 possible
C. I and III only
08 Jul 2025, 07:31
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x^3 = int*81 = int*3^4, that implies that x must have at least 3^2=9 as a divisor, because x^3 would have 3^6 as a divisor (so also 3^4).
On the other hand
x=36*q + remainder
9*integer = 36*q + remainder
integer = (36*q + remainder)/9 = 4*q + remainder/9
remainder must a multiple of 9
I. 0 multiple of 9
II. 3 non multiple of 9
III. 27 multiple of 9
I and III could be possible
Answer D
On the other hand
x=36*q + remainder
9*integer = 36*q + remainder
integer = (36*q + remainder)/9 = 4*q + remainder/9
remainder must a multiple of 9
I. 0 multiple of 9
II. 3 non multiple of 9
III. 27 multiple of 9
I and III could be possible
Answer D
