Sequential divisors?

Recall how to count the number of divisors . Given that $N$ has 10 positive divisors, there are two possibilities for its prime factorization:

$\begin{array}{rl} \textbf{Case 1:} & N=p^1 \times q^4 \\ \textbf{Case 2:} & N=p^9 \end{array}$

where $p$ and $q$ are primes.

Now examine these cases taking into account the second statement.

Case 1 :

Suppose that $p \ne 2$ and $q \ne 2.$ Then,

$2N=2^1 \times p^1 \times q^4.$

This gives 20 divisors, so this cannot be true. Now suppose that $q = 2.$ Then,

$2N=p^1 \times 2^5.$

This gives 12 divisors, so this also cannot be true. Now suppose that $p=2.$ Then,

$2N=2^2 \times q^4.$

This gives 15 divisors. Then if case 1 is true, it must be the case that $p=2.$

Case 2 :

Suppose that $p \ne 2.$ Then,

$2N=2^1 \times p^9.$

This gives 20 positive divisors, so this cannot be true. Now suppose that $p=2.$ Then,

$2N=2^{10}.$

This gives 11 positive divisors, so this also cannot be true. This means that case 2 is impossible. Now we know that case 1 must be true, and it must be the case that $p=2:$

$N= 2^1 \times q^4$

It is not necessary to examine the third statement. If you do examine it, you will note that there are no conflicts. Regardless of the value of $q,$

$4N=2^3 \times q^4.$

This gives $\boxed{20}$ positive divisors.

It's simple number theory that whenever you double a positive value, that amount of positive divisors it has increases by a constant rate.

$N$ has precisely 10 positive divisors.

$2N$ has precisely 10 + 5 positive divisors.

$4N$ has precisely 15 + 5 positive divisors.

Fact : With the prime factor decomposition $N = p_1^{e_1}\cdot p_2^{e_2} \cdots$ , the number of divisors is $\delta(N) = (e_1 + 1)(e_2 + 1)(\cdots).$

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Now $\delta(N) = 10$ factors as either $(9+1)$ or $(4+1)(1+1)$ , producing two cases.

Case 1 : $N = p^9$ . If $q \not = p$ is a prime number, then $\delta(pN) = \delta(p^{10}) = (10 + 1) = 11;\ \ \ \ \delta(qN) = \delta(p^9\cdot q^1) = (9 + 1)(1 + 1) = 20.$ Thus $\delta(3N) = 20$ would be possible if $p = 3, N = 3^9$ ; however, then $\delta(2N) = 11$ not 15. This rules out case 1.

Case 2 : $N = p_1^4\cdot p_2$ . If $q \not= p_1, p_2$ is prime, then $\delta(p_1N) = \delta(p_1^5p_2) = (5 + 1)(1 + 1) = 12; \\ \delta(p_2N) = \delta(p_1^4p_2^2) = (4 + 1)(2 + 1) = 15; \\ \delta(qN) = \delta(p_1^4p_2q) = (4+1)(1+1)(1+1) = 20.$ We see that $\delta(2N) = 15$ requires $p_2 = 2$ , and $\delta(3N) = 20$ it true if moreover $p_1 \not = 3$ . But then $\delta(4N) = \delta(p_2^2N) = \delta(p_1^4p_2^3) = (4+1)(3+1) = \boxed{20}.$

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The smallest value of $N$ for which these statements are true is $N = 5^4\cdot 2 = 1250.$ The divisors are:

• of $N = 1250$ : 1,2,5,10,25,50,125,250,625,1250.

• of $2N = 2500$ : 1,2,4,5,10,20,25,50,100,125,250,500,625,1250,2500.

• of $3N = 3750$ : 1,2,3,5,6,10,15,25,30,50,75,125,150,250,375,625,750,1250,1875,3750.

• of $4N =5000$ : 1,2,4,5,8,10,20,25,40,50,100,125,200,250,500,625,1000,1250,2500,5000.

Let $\tau(m)$ denote the number of positive divisors of a natural number $m$ . Suppose $a$ is the highest power of $2$ dividing $N$ . Then $a+1$ is the highest power of $2$ dividing $2N$ , so that

$\frac{15}{10} = \frac{\tau(2N)}{\tau(N)} = \frac{(a+1)+1}{a+1}$ , which gives a = 1. So now, $a+2 = 3$ is the highest power of $2$ dividing $4N$ , so $4N$ has $\frac{3+1}{1+1} \cdot \tau(N) = 2 \cdot 10 = \boxed{20}$ .

For completeness, we see that the number $N= 1250= 2 \cdot 5^4$ satisfies all conditions of the problem.

The number N can be written as a product of prime numbers:N=(2^x)(3^y)(5^z)...etc. Assume the possibility to choose is D=xyz...etc. Actually D is the number of the divisors.

In this problem, the D of N,2N,3N are given, so in the 2N case, x becomes (x+1), in the 3N case, y becomes (y+1), in the 4N case, what's x now? (x+2),brilliant!

The all things after(and including) "z" is a constant number in this problem, so we can assume it's "P", since we don't know it for the present.

D(N)=xyP=10

D(2N)=(x+1)yP=15

D(3N)=x(y+1)P=20

D(4N)=(x+2)yP=?

Believe you can solve it now :D

$N=2^x*3^y*k,$ where $k$ is not divisible by $2$ or by $3$ . Suppose that $k$ has $z$ number of divisors. Then:

$(x+1)(y+1)z=(xy+x+y+1)z=10=A$

$(x+2)(y+1)z=(xy+x+2y+2)z=15=B$

$(x+1)(y+2)z=(xy+2x+y+2)z=20=C$

From that

$xy+x+y+1=\frac{10}{z}=\frac{2}{3}*\frac{B}{z}=\frac{2}{3}*(xy+x+2y+2)$

Since $3A=2B$ ,

$3xy+3x+3y+3=2xy+2x+4y+4$

$xy+x-y-1=0=(x-1)(y+1)$

$y$ can't be $-1$ , so $x=1$ .

Similarly $2A=C$ ,

$2xy+2x+2y+2=xy+2x+y+2$

$xy+y=0=y(x+1)$

$x=1$ , so $y$ has to be $0$ .

From these $z=5$ .

Then $4N$ has

$(x+3)(y+1)z=4*1*5=\boxed{20}$

divisors.

let N = (p1^n1) (p2^n2) (p3^n3) ... (pi^ni); p are prime numbers, p1<p2<p3<...<pi, n are positive integrals

The number of divisors of N: (n1+1)(n2+1)...(ni+1) = 10

Since n >0, n+1>1, (n1+1)(n2+1)...(ni+1) = 2x5 or 5x2 or 10 => n1=1, n2=4 or n1=4, n2=1 or n1=9

N = p1 (p2^4) or (p1^4) p2 or p1^9

2N = 2 p1 (p2^4) or 2 (p1^4) p2 or 2*(p1^9)

Since 2 is a prime number, 2 = p1 or 2 = p2 or 2=/= p1 or p2

if 2 = p1: 2N = (2^2) (p2^4) or (2^5) p2 or 2^10, Number of divisors: 3x5=15 or 6x2=12 or 11=> N=2*p2^4

if 2 = p2: N= p1 (2^4) or (p1^4) 2 or (p1^9)*2, since p1<p2, 2 is the smallest prime, it is not possible.

if 2 =/= p1 or p2, N = 2 p1 (p2^4) or 2 (p1^4) p2 or 2*(p1^9) => number of divisors: 2x2x5 or 2x5x2 or 2x10, =/= 15

THEREFORE, N = 2*(p2^4)

3N = 2 * 3 * (p2^4)

since 3 is also a prime number, 3 = p2 or 3 =/= p2

if 3 =p2, N = 2 * 3^4, 3N = 2 * 3^5, number of divisors: 2x6=12 =/= 20

so 3 =/= p2, N=2 * (p2^4), 3N=2 * 3 * (p2^4), number of divisor: 2x2x5 = 20

THEREFORE, N = 2 * (p2^4), p2 =/= 3

4N = 4 * 2 * (p2^4) = (2^3) * (p2^4), number of divisors: 4x5 = 20

@Andy Hayes gave a very good general term solution but I did it with making cases like this:

Case 1:

Let Different Divisors of N be : 1,2,3,4,5,6,7,8,9,10 then Divisors of 2N : 1,2,3,4,5,6,7,8,9,10,12,14,16,18,20 ( = 15 divisors) Divisors of 4N = 2(2N) : 1,2,3,4,5,6,7,8,9,10,12,14,16,18,20,24,28,32,36,40 ( = 20 divisors)

Case 2:

As you can see by question that all divisors cannot be ODD because if this happened, 2N will give 20 divisors and 4N will give 40 divisors

Case 3:

this is what i used - Trying to find Smallest Possible Value of N by looking at question and case 1 we can simply tell that upto 10 divisors none can be divisible by 3 or it won't make 3N with 20 divisors So here is what i came off with N = 1 2 4 5 7 8 10 11 13*14 So 2N = 1,2,4,5,7,8,10,11,13,14,16,20,22,26,28 ( =15 divisors) 3N = 1,2,4,5,7,8,10,11,13,14,3,6,12,15,21,24,30,33,39,42 ( = 20 divisors) 4N = 2(2N) = 1,2,4,5,7,8,10,11,13,14,16,20,22,26,28,32,40,44,52,56 ( = 20 divisors)

Please respond if i had done anything wrong :)

I will first try to determine whether $N$ is divisible by 2 and/or 3.

Let $\displaystyle N = 2^a \cdot 3^b \cdot \prod_{i=3}^n p_i ^{r_i}$ , where $a$ and $b$ are non-negative integers to be determined, $p_3, p_4, \ldots, p_n$ are distinct prime numbers larger than 3, and $r_3, r_4,\ldots , r_n$ are positive integers.

We define $\sigma_0 (X)$ as the total number of positive divisors of positive integer $X$ . Then,

\begin{aligned} 10 &=& \sigma_0(N) = (a + 1)(b+ 1) \prod_{i=3}^n (r_i + 1) \qquad\qquad\qquad \tag 1 \\ 15 &=& \sigma_0(2N) = (a + 2)(b+ 1) \prod_{i=3}^n (r_i + 1) \qquad\qquad\qquad \tag 2 \\ 20 &=& \sigma_0(3N) = (a + 1)(b+ 2) \prod_{i=3}^n (r_i + 1) \qquad\qquad\qquad \tag 3 \\ \end{aligned}

$(2) \div (1)$ : $\dfrac{15}{10} = \dfrac{a+2}{a+1} \quad\Leftrightarrow\quad a = 1$ .

$(3) \div (1)$ : $\dfrac{20}{10} = \dfrac{b+2}{b+1} \quad\Leftrightarrow\quad b = 0$ .

From $(1)$ : $\displaystyle 10 = (1 + 1)(0 + 1) \prod_{i=3}^n (r_i + 1)\quad\Leftrightarrow\quad \prod_{i=3}^n (r_i + 1) = 5 \quad\Leftrightarrow\quad n = 3 .$ Hence, $N$ can be expressed as $2p^4$ , where $p$ is a prime larger than 3.

Our answer is $\sigma_0(4N) = \sigma(4\cdot 2p^4) = \sigma(2^3 \cdot p^4) = (3+1)(4+1) = \boxed{20}$ .