Positive integer N has exactly 12 unique factors. What is the largest

N has 12 factors so it isn't a perfect square.

Total number of factors of N= a^m * b^n *... is calculated as-
(m+1)*(n+1)*..
To get 12 as total factors, largest possible number of unique prime factors is 3-
a^1*b^1*c^2

=> (1+1)*(1+1)*(2+1) = 12

Hence option B

12 can be written as 1*12, 2*6, 3*4 or 2*2*3.

Number of unique factors of N= a^x*b^y*c^z.... are (x+1)*(y+1)(z+1).., where a, b and c are prime numbers.

Hence, Positive integer N has exactly 12 unique factors can be written as
i) \(a^{11}\)
ii) \(a^1*b^5\)
iii) \(a^2*b^3\)
iv) \(a^1*b^1*c^2\)
where a, b and c are prime numbers.

Hence largest possible number of unique prime factors of N is 3

IMO B

Question: Positive integer N has exactly 12 unique factors. What is the largest possible number of unique prime factors that N could have?

Unique factors of a number can be calculated by finding all prime factors, adding 1 to each unique prime factor quantity, then multiplying the resulting numbers.
The number 12 can be broken down to (2)(3)(2), this break down means we need to find a value for N with two prime factors with a quantity of one each, and one prime factor with a quantity of two.

Any three prime factors will do.
For example...
\(\mathtt{(3^1)(2^2)(7^1) = 84} \to \mathtt{12 unique factors\)
OR
\(\mathtt{(7^1)(5^2)(11^1) = 1925} \to \mathtt{12 unique factors\)
OR
\(\mathtt{(5^1)(3^2)(2^1) = 90} \to \mathtt{12 unique factors\)

The following technique is used to determine prime factors and factor number totals...

\[\text{Prime Factorization Calculation }\\
\text{Example: N = 90} \\
\begin{pmatrix*}
&&90&& \\
&\swarrow&|&\searrow&\\
2&&\downarrow&&5\\
&&9&&\\
&\swarrow&&\searrow&\\
3&&&&3
\end{pmatrix*}
\rightarrow \text{Prime Factors if N are } 2,3,3,5\\
\]

\[\text{Unique Factor Calculation }\\
\text{Example: N = 90}\\
\begin{pmatrix*}
Prime\text{ Factor } &\text{ Add 1 }& \text{ # of Factors } \\
2^1 & 2^{1+1} & 2 \\
3^2 & 3^{2+1} & 3 \\
5^1 & 5^{1+1} & 2\\
\end{pmatrix*}
\rightarrow\text{ Total number of factors are (2)(3)(2) = 12}\\
\\
\text{Prime Factors of N are } 2,3,3,5\\
\text{Unique factors remove duplicates, leaving } 2,3,5 = \text{ 3 Total Unique Factors} \]

Correct Answer: B. 3

We know that if \(N\) is prime factorized, it can be written as \(a^x\) or as \(a^x*b^y\) or as \(a^x*b^y*c^z\) depending on the number of prime factors.

\(a, b,\) and \(c\) - prime factors of \(N\)

\(x, y,\) and \(z\) - powers of these prime factors of \(N\)

1. The least possible number of unique prime factors of \(N\) is \(1\) when \(N\) is \(11th\) power of a prime number:

If \(N=a^x\) and \(x=11\), then the unique factors of \(a^1\) is \((11+1)=12\)

2. \(N\) can have \(2\) unique prime factors:

If \(N=a^x*b^y\) and \(x=2, y=3,\) then unique factors of \(a^2*b^3\) is \((2+1)(3+1)=12\)
If \(N=a^x*b^y\) and \(x=5, y=1,\) then unique factors of \(a^5*b^1\) is \((5+1)(1+1)=12\)

3. The largest possible number of unique prime factors of \(N\) is \(3\):

If \(N=a^x*b^y*c^z\) and \(x=1, y=1, z=2\) then unique factors of \(a^1*b^1*c^2\) is \((1+1)(1+1)(2+1)=12\)

Powers can be in any order:
\(N=a^1*b^1*c^2\)
\(N=a^1*b^2*c^1\)
\(N=a^2*b^1*c^1\)

In any case \(N\) has at most \(3\) unique prime factors provided that \(N\) has exactly \(12\) unique factors.

Hence B

Seems troubling at first to look at but fairly easy-ish. Think about it.

No. of unique factors is determined by the power of prime numbers for N in this case. Ignore the prime numbers, because we only truly care about the exponent.

Formula for calculating unique factors if 3^x * 5^y then unique factors = (x+1) (y+1) (Ignore the fact that I've taken 3 and 5, just for representation)

Now, to reach 12 there are the following ways:
1 x 12
2 x 6
3 x 4

The question asks us for largest number of unique prime factors.

2x 6 can be written as 2 x 3 x 2 (Suggesting that --> (1+1) * (2+1) * (2+1) meaning 3 unique prime factors. Same for 3 x 4 (3 x 2 x 2)

I chose B this way.

We know how to obtain the total number of factors of N.

If \(N = a^x * b^y * c^z ...\)

Total number of factors = (x + 1)(y+1)(z+1)...
given a, b and c are distinct prime factors.


Prime factorise 12 to get 2*2*3 (we cannot split 12 into smaller factors)
So N can have 3 distinct prime factors e.g. \(N = 5*7*11^2\)
Total number of factor here = (1 + 1)(1+1)(2+1) = 12

Answer (B)

Check this video on factorisation: https://youtu.be/Kd-4cH4cqHw