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If X is a positive integer, what is the remainder when X^2 is divided

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firas92
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If X is a positive integer, what is the remainder when X^2 is divided [#permalink] New post   19 Jun 2019, 07:17
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If \(X\) is positive integer, what is the remainder when \(X^2\) is divided by 6?

1. \(X≥5\)

2. \(X\) has no factor P such that \(1<P<X\)
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If X is a positive integer, what is the remainder when X^2 is divided [#permalink] New post   Updated on: 19 Jun 2019, 15:38
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Statement 1 is not sufficient - the answer is 0 if x = 6, but is 1 if x = 7.

Statement 2 tells us x is prime, which is not sufficient, since the answer is 4 if x=2, but is 1 if x=7.

Together, if x is a prime bigger than 5, then x must be odd and cannot be divisible by 3. So it must equal 6q + 1 or 6q + 5 for some integer q (since 6q, 6q+2 and 6q+4 are even, and 6q+3 is divisible by 3). So we either have a remainder of 1 or 5 when we divide x by 6, and thus a remainder equivalent to the remainder of 1^2 = 1, or 5^2 = 25 when we divide x^2 by 6. Each gives a remainder of 1. So the answer is C.

Originally posted by IanStewart on 19 Jun 2019, 09:43.
Last edited by IanStewart on 19 Jun 2019, 15:38, edited 1 time in total.
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Re: If X is a positive integer, what is the remainder when X^2 is divided [#permalink] New post   19 Jun 2019, 10:04
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firas92 wrote:
If \(X\) is positive integer, what is the remainder when \(X^2\) is divided by 6?

1. \(X≥5\)

2. \(X\) has no factor P such that \(1<P<X\)


1. \(X≥5\)
X^2 can be 5^2, 6^2, 7^2, . . .
= 25, 36, 49, 64, . . .
Remainders when divided by 6
= 1, 0, 1, 4 . . .
—> Insufficient

2. \(X\) has no factor P such that \(1<P<X\)
—> Numbers with no factors between themselves and 1 are only PRIME NUMBERS.
Possible Values of x^2 are 2^2, 3^2, 5^2, 7^2, 11^2 . . .
= 4, 9, 25, 49, 121 . . .
Remainders = 4, 3, 1, 1, 1 . . . .
—> Insufficient

Combining (1) & (2)
For \(X≥5\)
X^2 values are 25, 49, 121, 169 . . .
Remainders when divided by 6 = 1, 1, 1, 1, . . . . A Unique value
—> Sufficient

IMO Option C

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Re: If X is a positive integer, what is the remainder when X^2 is divided [#permalink] New post   19 Jun 2019, 15:13
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Statement 1 and 2 are clearly insufficient alone

Combining both equations
X is a prime number greater than or equal to 5

X=6k±1

\(X^2=36k^2±12k+1\)
Hence \(X^2=1 (mod 6)\)

firas92 wrote:
If \(X\) is positive integer, what is the remainder when \(X^2\) is divided by 6?

1. \(X≥5\)

2. \(X\) has no factor P such that \(1<P<X\)
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Re: If X is a positive integer, what is the remainder when X^2 is divided [#permalink] New post   15 Aug 2022, 13:38
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Re: If X is a positive integer, what is the remainder when X^2 is divided [#permalink]
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