For a positive integer k, Is k/200 an integer

Solution

Step 1: Analyse Question Stem

• k is a positive integer.
• We need to find if \(\frac{k}{200}\) is an integer.

o i.e. \(\frac{k}{2^3*5^2}\) is an integer.

 \(\frac{k}{2^3*5^2} \) will be an integer only if prime factorization of k contains powers of 2 and 5 more than or equal to 3 and 2 respectively.

Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE

• \(\frac{√k}{10} = I\), where I is an integer.
• Squaring both sides of the above equation, we get

o \(\frac{k}{100} = I^2\), since I is an integer so I\(^2\) must be an integer.
o Or, \(\frac{k}{2^2*5^2} = I^2\)

 It means prime factorization of k must contains \(2^2*5^2\), but we do not know if it contains \(2^3\) or not.

• If I (hence \(I^2\)) is even, then \(\frac{k}{200}\) will be an integer. For example: \(\frac{k}{2^2*5^2} = 4 ⟹ \frac{k}{2^3*5^2 }= 2 \)or, \(\frac{k}{200} = 2\)
• However, if I (hence \(I^2\)) is odd then \(\frac{k}{200}\) will not be an integer.

• \(\frac{k^2}{80} = \frac{k^2}{2^4*5} \)

o Since, k is an integer, \(k^2\) must be a perfect square, hence \(k^2 \)must contains even powers of all its’ prime factors.

 Therefore, \(\frac{k^2}{2^4*5^2}\) will also be an integer i.e. \(\frac{k^2}{2^4*5^2} = N ……Eq(i)\)
• Since both numerator and denominator of \(\frac{k^2}{2^4*5^2} \)is a perfect square, the resultant integer will also be a perfect square.
 Taking square root on both side of Eq.(i), we get,
 \(\frac{k}{2^2*5} = integer\) or √N,
• So, \(\frac{k}{200}\) will be an integer, only if √N is a multiple of 10. But we don’t know the value of √N (or N)

Step 3: Analyse Statements by combining.

• From statement 1: \(\frac{k}{2^2*5^2} = Integer\)
• From statement2 : \(\frac{k}{2^2*5 }= integer\)
• On combing the above two statements we get; \(\frac{k}{2^2*5^2} = Integer\)