Matematické Fórum

stuart clark · 25. 02. 2013 10:01

(1) Total number of positive diviser of $10!$ which are is in the form of $5n+2\;,$ where $n\in \mathbb{N}$

(2) Total number of positive diviser of $10!$ which are is in the form of $5n+3\;,$ where $n\in \mathbb{N}$

(3) Total number of positive diviser of $7!$ which are is in the form of $3n+1\;,$ where $n\in \mathbb{N}$

(4) Total number of positive diviser of $7!$ which are is in the form of $3n+2\;,$ where $n\in \mathbb{N}$

Plz explain me in detail.

Thanks

Brano · 25. 02. 2013 15:46

(1)

$10!=2^8 3^4 5^2 7$ so its divisors are of the form $d=2^x3^y7^z5^w$ , where $x\in\{0,1,..,8\};y\in\{0,..,4\};z\in\{0,1\};w\in\{0,1,2\}$ . We need $d\equiv 2\mod 5$ so $w=0$ and $2\equiv 2^x3^y7^z\equiv 2^{x+3y+z}\mod 5$ . Since $2^4=1\mod 5$ we have that $1\equiv x+3y+z\equiv x-y+z\mod 4$ .

Now we want the number of solutions of this equation within the aforementioned ranges. Suppose that $x\in\{0..7\}$ so we have exactly two solutions for $x$ for every choice of parameters $y,z$ what yields $2\times 5\times 2=20$ solutions. Next we need solutions for $x=8$ i.e. $1\equiv z-y\mod 4$ . Similarly suppose $y\in\{0,..,3\}$ and we have one solution for every $z$ i.e. $2$ solutions. And finally we need solutions for $y=4$ so $1\equiv z\mod 4$ and there is exactly one. Altogether we have $20+2+1=23$ solutions i.e. $23$ divisors with the desired form. You can verify it on W|A.

(2)-(4) should be simmilar.

stuart clark · 25. 02. 2013 17:44

Thanks Brano

But would You like to explain me that part

and $2\equiv 2^x3^y7^z\equiv 2^{x+3y+z}\mod 5$ . Since $2^4=1\mod 5$ we have that $1\equiv x+3y+z\equiv x-y+z\mod 4$ .

Brano · 25. 02. 2013 18:18

$d=5n+2$ iff $d\equiv 2\mod 5$

$7\equiv 2\mod 5$ so $7^z\equiv 2^z\mod 5$ .
$2^3\equiv 3\mod 5$ so $2^{3y}\equiv 3^y\mod 5$ .
Since $2^4\equiv 1\mod 5$ then $2^{V-4k}\equiv 2^V\mod 5$ or in other words if $V\equiv W\mod 4$ then $2^V\equiv 2^W\mod 5$ .
Actually you also need that $2^p\not\equiv 1\mod 5$ for $p\in\{1,2,3\}$ to obtain
$2^V\equiv 2^1\mod 5$ iff $V\equiv 1\mod 4$

Is it clear now?

stuart clark

Thanks ↑ Brano: Got it.

plz apolozise me for again asking for explanation of that part

Suppose that $x\in\{0..7\}$ so we have exactly two solutions for $x$ for every choice of parameters $y,z$ what yields $2\times 5\times 2=20$ solutions. Next we need solutions for $x=8$ i.e. $1\equiv z-y\mod 4$ . Similarly suppose $y\in\{0,..,3\}$ and we have one solution for every $z$ i.e. $2$ solutions. And finally we need solutions for $y=4$ so $1\equiv z\mod 4$ and there is exactly one

Brano · 04. 03. 2013 20:12

Sorry, I havent noticed that you've editted your last post into question.

So,

the point is that we are about to count number of solutions of $x-y+z\mod 4\equiv1$ . Suppose that $y,z$ are fixed numbers and $x\in\{0,1,2,3\}$ . You have exactly one solution $x = (1+y-z)\text{ modulo }4$ . Now if $x\in\{0,..,7\}$ we have two solutions $x_1 = (1+y-z)\text{ modulo }4$ and $x_2 = (1+y-z)\text{ modulo }4+4$ . So we only count the number of pairs $y,z$ and multiply it by 2. Doing this we have omitted solutions where $x=8$ so we substitute this into the equation and proceed analogically.