That's an Odd Property

Since $n$ is odd, and it is a multiple of its number of divisors, so must be its number of positive divisors. It is known that the number of positive divisors of $n$ will be odd if and only if $n$ is a perfect square. So, $n= k^2$ for some $k \in \mathbb{N}$ . We are immediately restricted to checking the values of $n$ in the set $\{1^2, 3^2, 5^2, \cdots , 31^2 \}$ We can readily verify $1$ satisfies the problem conditions. Now, if $k$ is a prime, $n= k^2$ will have $2+1=3$ positive divisors, so $n$ must be a multiple of $3$ , so we must have $k=3 \implies n= 9$ .

So we now just have to go through the squares of odd composite numbers, i.e. we have to check the set $\{9^2, 15^2, 21^2, 25^2, 27^2 \}$ Going through the list, we find out that $15^2, 21^2, 25^2$ satisfy the conditions, whereas $9^2$ and $27^2$ don't. The set of all $n$ satisfying the conditions, thus, is $\{1, 9, 225, 441, 625 \}$ . So, there are a total of $\boxed{5}$ values of $n<1000$ .

When we first see this problem, the eye-catching phrase should be _ number of positive divisors_ . In fact, the crux to this problem is to realise that suppose we have $n$ , with it's prime factorisation as follows : $p_1^{e_1} p_2^{e_2} \ldots p_n^{e_n}$ , where ${p_i}$ are primes and ${e_i}$ their respective exponents, then the number of prime factors it has is given by : $(e_1+1)(e_2+1) \ldots (e_n+1)$ .

With this important theorem/observation in mind, we delve in on the word divisors and the seemingly odd condition that $n$ is an odd integer (pun intended). Hmm, the only limitation on odd and even is that one can be divisible by 2 , so basically, the divisors of an odd number needs to be odd . By the problem condition, we would need $(e_1+1)(e_2+1) \ldots (e_n+1)$ to be odd .

The previous sentence gives some really strict restriction on ${e_i}$ . For instance, clearly if one of ${e_i}$ is odd, say $e_j$ , then $e_j+1$ is even, then the product $(e_1+1)(e_2+1) \ldots (e_n+1)$ is even, a contradiction to what we want. With this in mind, we see that we need all the ${e_i}$ to be even.

Well, clearly a number with an even exponent is a square , so out analysis above shows that only odd squares satisfies the conditions. Clearly, $1$ and $9$ works.

However, we realise that we cannot find any more squares of the form $p_1^2$ since then $n$ needs to be divisible by $2 + 1 = 3$ . With a such a scheme in mind, we can be more efficient. Suppose we need $p_1^4$ to satisfy the conditions, then $n$ needs to be divisible by $5$ , clearly the only such integer is $5^4 = 625$ . Slowly increasing the power, we suppose we need $p_1^2 p_2^2$ to satisfy the conditions, then $n$ needs to be divisible by $9$ , so clearly one of $p_1, p_2$ needs to be $3$ and $p_2$ can be any arbitrary odd integer $\neq 1 \ \text{or any multiple of 3}$ (we need $p_1, p_2$ to be distinct by definition) less than or equal to $\lfloor \sqrt{\frac{1000}{9}} \rfloor = 10$ , this gives $441, 225$ . Next, we consider $n$ of the form $p_1^4$ , then clearly $5 | n$ , so we have $5^4 = 625$ as the answer.

We observe that if the squares are of the form $p_1^6$ where the exponent can be larger, $7^6 > 1000$ and similarly for the rest. Also, if the squares are of the form $p_1^2 p_2^2 p_3^2$ need to be divisible by $27$ which will not work since we defined $p_1, p_2, p_3$ to be distinct so it can only be divisible by $9$ . Analogously, $p_1^2 p_2^2 \ldots p_n^2$ will also not work. And with a similar token, we cannot have squares of the form $p_1^4 p_2^2$ that satisfy the conditions since then $5|n$ and $3|n$ (because $(4+1)(2+1) = 15$ ), so $n \geq 3^4 \times 5^2 =2025 > 1000$ contradiction. Clearly, all other squares would be $> 1000$ and hence would not work.

In a nutshell, we can conclude that the possible values of $n$ are $1,9,225,441,625$ , and the answer that we want is $\fbox{5}$ .

First, we know that $n = 1$ satisfied the equation. Let the prime factorization of $n$ are $p_1^{a_1}p_2^{a_2}...p_k^{a_k}$ with $p_1, p_2,...., p_k$ are prime numbers except $2$ ( $n$ is odd number), then the number of positive divisor of $n$ is $(a_1+1)(a_2+1)...(a_k+1)$ . We know that $n$ is an odd integer less than $1000$ , then the number of positive divisor of $n$ should be an odd integer too, because the number of positive divisor divide $n$ if and only if the number of positive divisor is an odd number ( even number does not divide odd number ) The number of positive divisor of $n$ is $(a_1+1)(a_2+1)...(a_k+1)$ . If the number of positive integer of $n$ is an odd number, then $a_1, a_2, ...., a_k$ should be an even number and then we can conclude that $n$ is a perfect square.

Case 1 If $n$ have $3$ distincts prime factorization, then the minimum value of n is $3^{2}.5^{2}.7^{2} = 11025$ (more than $1000$ ).

Case 2 If the smallest value of prime factorization of $n$ is $3$ , then the value of $n$ that satisfied is $3^{2}$ , $3^{2}.5^{2}$ , and $3^{2}.7^{2}$ .

Case 3 If the smallest value of prime factorization of $n$ is $5$ , then the value of $n$ that satisfied is $5^{4}$ .

Case 4 If the smallest value of prime factorization of $n$ more than $5$ , then there is no value of $n$ that satisfied because the value of $n$ more than $1000$ .

The values of $n$ that satisfied is $1$ , $9$ , $225$ , $441$ , and $625$ . So, there are $5$ odd integer of $n < 1000$ that satisfied.

For any integer N, there exist distinct prime numbers $p_{1}$ , $p_{2}$ , $\ldots$ , $p_{m}$ such that $k$ can be expressed as

$N = p^{q_{1}}_{1}, p^{q_{2}}_{2}, p^{q_{3}}_{3}, \ldots p^{ q_{m} }_{m}$

and total number of positive divisors are $(q_{1} + 1) (q_{2} + 1) \ldots (q_{m} + 1)$ that includes $$ 1 and $N$ .

It is given that $n$ is an odd number less than 1000. This implies that $q_{1}, q_{3} \ldots q_{m}$ should be even such that $(q_{i} + 1)$ is odd.

That gives first number 1. Following numbers must be $N^{2}$ or $N^{4}$ . This observation gives

$9 = 3^{2}; 225 = 3^{2} \cdot 5^{2}; 441 = 225 = 3^{2} \cdot 7^{2}$ and $625 = 5^{4}$ .

Therefore, there are all in $\boxed5$ numbers.

We will minimize the number of such numbers $n$ that we have to check. Let us do it step by step.

1. Since $n$ is odd, we can neglect all even numbers.

2. Since $n$ is odd, it cannot have an even factor (because an even number can never divide an odd number). This means that $n$ has only odd factors. Since number of factors of $n$ divides $n$ , $n$ must have odd number of factors.

3. Since $n$ has odd number of factors, $n$ has to be a perfect square.

4. Combining points 1 and 3, we can say that $n$ has to be an odd perfect square.

5. Since $n < 1000$ , we can say that $n$ can be $1^2, 3^2, 5^2, 7^2...29^2, 31^2$ .

6. We can neglect any $n$ that is the square of a prime number (other than 3). This is because the square of the prime has 1, the prime and itself i.e. 3 factors. $n$ must be divisible by number of its factors, which in this case is 3. But 3 is not a factor (written above). Hence the rule out.

So we can narrow down $n$ to be $1^2, 3^2, 9^2, 15^2, 21^2, 25^2$ and $27^2$ .

Checking these 7 numbers, only $1^2, 3^2, 15^2, 21^2$ and $25^2$ satisfy the condition. Hence number of such $n= \boxed{5}$

We know that the only way that the number of positive divisors of $n$ must be odd in order for it to divide $n$ . This is because if it is even, an even number times another positive integer must be an even number, and it is given that the original integer is odd.

We also know that the only numbers that have an odd number of factors are perfect squares (the link about divisors of an integer in the details portion of the problem provides an explanation). Therefore, $n$ must be a perfect square in order for the number of positive divisors of $n$ to divide $n$ .

The odd perfect squares less than 1000 are

$1^2, 3^2, 5^2,\ldots, 29^2, 31^2$

which are equal to

$1, 9, 25, \ldots, 841, 961$

Since there aren't that many of these (only 16), we can just test each of them to see if they work:

1 divides 1, so 1 works.

3 divides 9, so 9 works.

3 doesn't divide 25, so 25 doesn't work.

3 doesn't divide 49, so 49 doesn't work.

5 doesn't divide 81, so 81 doesn't work.

3 doesn't divide 121, so 121 doesn't work.

3 doesn't divide 169, so 169 doesn't work.

9 divides 225, so 225 works.

3 doesn't divide 289, so 289 doesn't work.

3 doesn't divide 361, so 361 doesn't work.

9 divides 441, so 441 works.

3 doesn't divide 529, so 529 doesn't work.

5 divides 625, so 625 works.

7 doesn't divide 729, so 729 doesn't work.

3 doesn't divide 841, so 841 doesn't work.

3 doesn't divide 961, so 961 doesn't work.

(Read 'Divisors of an integer if you don't understand how I got those numbers of factors)

Counting up the ones that work, we have $\boxed{5}$ possible values of $n$ . $\ \blacksquare$

For every divisor $a$ of $N$ , there is always distinct divisor $b = N / a$ unless $N$ is a perfect square and $a$ is $N$ 's principal root. Hence, a number always has an even number of divisors unless it is a square. We thus seek odd squares under $1000$ .

We only need to consider the squares of the odd numbers up to and including $31$ . These odds fall into five classes:

1. The number $1$ , which has $1$ divisor, so it works.
2. The number is prime. In this case, the divisors of its square are $1$ , the square, and the prime; there are $3$ divisors total. Thus, the only prime that works is $3$ .
3. The number is a nonsquare semiprime ( $pq$ , where $p$ and $q$ are prime and $p \neq q$ ). Here, the divisors of the square $p^2 q^2$ may be found by taking $0$ , $1$ , or $2$ of $p$ and similarly for $q$ , leading to $9$ combinations. Since this must divide $p^2 q^2$ , the semiprime must be a multiple of $3$ . Here, only $3*5 = 15$ and $3*7 = 21$ work.
4. The number is the square of a prime. In this case, the square of the number is a fourth power of a prime, so the divisors are the prime raised to the powers $0$ , $1$ , $2$ , $3$ , and $4$ , leading to $5$ total divisors. The only prime square which is a multiple of $5$ is $25$ .
5. The number is a cube of a prime ( $27$ is the only one $\leq 31$ ). Its square is a sixth power of a prime, so the reasoning as in the part above says that there are $7$ total divisors. Since $7$ is not a divisor of $27$ , there are no such cubes that fit the criteria.

Since the smallest numbers which are of the form $p^2 q$ or $p^4$ where $p$ and $q$ are distinct odd primes are $3^2 * 5 = 45$ and $3^4 = 81$ , respectively, these five cases exhaust all odd numbers $\leq 31$ . Of them, only the squares of $1$ , $3$ , $15$ , $21$ , and $25$ fit the criteria of the problem, leading to an answer of $\boxed{5}$ numbers that fit.

As we know, usually a number has evenly many positive divisors, but perfect squares not, since they are the product of two same numbers, as an example, $9^{2} = 9 \times 9$ . So, this makes us left with odd perfect squares. But, at the same time, the square of any prime, $p$ , usually they have only $3$ divisors, that is $1, p$ and $p^{2}$ . So, this has eliminated almost all the squares of prime except $3$ , since $3^{2}$ is divisible by $3$

Now, by using Divisors of an integer, we can investigate the cases that we left now.

1. Since $729 = 27^{2} = 3^{6}$ , so it has $7$ divisors. From here, it is very obvious that $729$ is not divisible by $7$ , since it is the product of six $3$ 's.

2. Since $625 = 25^{2} = 5^{4}$ , it has $5$ divisors. From here, the condition is satisfied, since $5|625$ .

3. $441$ has $(2+1)(2+1) = 9$ divisors, since $441 = 3^{2}7^{2}$ . Therefore, $441$ satisfy the condition because $9|441$ .

4. $225$ also has $9$ divisors, since $225 = 3^{2}5^{2}$ . So, $225$ also satisfy the condition, since $9|225$ .

5. $81$ has $5$ divisors, since $81=3^{4}$ . But, $81$ is not a multiple of $5$ , so $81$ is rejected.

6. $1$ has only $1$ divisor. So, $1$ satisfy the condition, since $1|1$ .

From the cases above, we have $\boxed {5}$ odd positive integers that satisfy the condition. Starting from the smallest, they are $1, 3, 225, 441, 729$ .

Solution

The only $n$ that satisfy the conditions are: $1, 9, 225, 441, 625$

Therefore, the answer is $\boxed{5}$ .

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Motivation

At first glance, this question may seem very complicated and full of cases to consider. But notice that the number of positive divisors of any integer (with prime factorization of $p_{1} ^ {k_ {1}} \times p_ {2} ^{k_{2}}\times ... \times p_ {x} ^{k_{x}}$ for primes { $p_{1}, p_{2}, ..., p_{x}$ } and positive integers { $k_{1}, k_{2}, ..., k{n}$ } ) $= (k_{1} +1)(k_{2} +1) ... (k_{x} +1)$ .

Observing that an odd number does not have $2$ as a factor, all powers of the prime factorisation of an odd integer satisfying the question would be even. (This is so that the product of $(k_{1} +1)(k_{2} +1) ... (k_{x} +1)$ will not be even.)

Thus, the possible $n$ satisfying the question would simply be:

$1 , 3^{2} = 9 , 3^{2} \times 5^{2} = 225, 3^{2} \times 7^{2} = 441, 5^{4} = 625$

I found the following numbers 9, 225, 441, & 625. 4 should be the answer

Number of Divisors of number expressed of the form ab11⋅ab22⋅ab33⋯abrr where a1, a2 are primes and b1, b2 are their respective powers..

then no of divisors is 1⋅(b1+1)⋅(b2+1)⋅(b3+1)⋯(br+1) and therefore

Notice that every number has an even number of divisors except the square numbers, since factors occur in pairs. Odd numbers must have only odd factors........I think there Are 5 numbers...........1,9,225,441,625

n is odd =>divisors of n is also odd =>n is a square of an odd number. number of divisors of a prime number square is 3. therefore only odd prime number square possible is 9. other possible values of n are 1,81,225,441,625,729.(n<1000 =>31 is the max value of root of n to be tried) out of the given numbers ,using trial or error method,81&729 can be rejected. therefore the possible values of n are 1,9,225,441,625. hence answer is 5.

n must be a square.other wise no. of divisors will be even which divids odd n.n odd. now,we have choice for n to be,(1square,3square,5square,....,31square). a lot of choice.we need to eliminate some.

supp,n= prime square.then n have 3 divisors.3 divides prime square.3 divides prime.only choice the prime is 3.now the other prime squares other than 3 square are eliminated.

now check for the rest 6 options.(easy calculation).you will get 5 answers.

Because $n$ is odd, any divisor must also be odd. Thus, the number of divisors of $n$ must be odd. Therefore, $n$ must be an odd, perfect square under 1000. Then, $n\in\{1^2, 3^2, ..., 31^2\}$ . Testing, we find that only $1^2, 3^2, 15^2, 21^2, 25^2$ work, giving 5 values. (The testing can be somewhat simplified by the observation that any prime squared other than 3 cannot work, because the number of factors must be 3, but a prime cannot be divisible by anything besides itself and 1).

An odd number will always be divisible only by an odd number. Hence total no. of divisors for any number to satisfy this property must be odd.

This property will hold true for any number of the form $n^{n-1}$ where n is odd . Such numbers are : $9 = 3^{2}$ => total divisors $(2+1)=3$ and 3 divides 9 other is $625 = 5^{4}$ => total divisors = $(4+1) =5$ and 5 divides 625 Another number can be 7^{6} but that is greater than 1000 hence only these two numbers.

As number of divisors should be odd, we can expect following total number of divisors : $3,5,7,9,11,13,15,17,19,21,23...$ Starting from 3 and 5 we have got 2 numbers already. A number 7^{6} won't be included as it is greater than 1000. And similarly 11,13.. and primes excluded because they will be achieved only by (10+1) or (12+1).

Left out option is only for 9 and that can be achieved by $(2+1)*(2+1)$

i.e. odd numbers of pattern $a^{2} * b^{2}$ Also a and b should be odd. Such numbers <1000 are only 2 :

$225 = 3^{2} * 5^{2}$

$441= 3^{2} * 7^{2}$

Other numbers of such pattern are greater than 1000.

That makes a total count of 4. And in the end don't forget 1 Because it has only divisor i.e. 1 and 1 divides itself. :) :) That's why answer is $5$ .

I didn't apply much of formulas here except of the basic counting of divisors, it was just an analytic problem which took me long time to solve. Concentration is the key.

There are few things to note here:

1. Since n is odd, therefore the number of its divisor is also odd => n is an odd square number
2. We observe that if $n$ = $k^{2}$ = $qp^{2}$ = $q^{2} * p^{2}$ = $q^{2} * p^{2} * r^{2} = ...$ , therefore its number of divisor is $(2+1)*(2+1)*(2+1)*... = 3^{m}$ . Thus, n is divisible by 3
3. As n is less than 1000, k is less than 31. Since n is divisible by 3, k is also divisible by 3 . Therefore, k is [3, 9, 15, 21, 27]

Therefore there are only 5 options of n: $­3^{2}$ , $9^{2}$ , $15^{2}$ , $21^{2}$ and $27^{2}$ , correspond to 9, 81, 225, 441 and 729 respectively.