Question

Find the form of all positive integers with number of divisors = 10.
What is the smallest positive integer ?

Answer 1

u mean
\(\large \tau(n) =10\)

Answer 2

the smallest :-
well i would consider all divisors are multible of 2
so
k+1=10
k=9

the smallest would be 2^9

Answer 3

now hmm idk what do you mean of the FORM :o
@rational

Answer 4

\[\large (k_1+1)(k_2+1)\cdots = 10\]

Answer 5

\[\large (k_1+1)(k_2+1)\cdots = 1\times 2 \times 5\]

Answer 6

so we can have 3 prime factors max, right ?

Answer 7

\[\large (k_1+1)(k_2+1)(k_3+1) = 1\times 2 \times 5\]

Answer 8

sorry i think we can have two prime factors as "1" makes no sense

Answer 9

make sense , ok keep going

Answer 10

\[\large (k_1+1)(k_2+1) = 2 \times 5\]

Answer 11

wait . ok

Answer 12

\(k_1 = 1\) and \(k_2 = 4\)

or

\(k_1 = 4\) and \(k_2 = 1\)

Answer 13

wait, ur case should be considered as well :
\[k_1 + 1 = 10\]

\(k_1 = 9\)

Answer 14

so the minimum number would be :

min{2^9, 2*3^4, 2^4*3}

Answer 15

my case only gives min

Answer 16

it doesnt

Answer 17

oh :o

Answer 18

min is 2^4*3 = 48

Answer 19

oh

Answer 20

ok what about the form ?

Answer 21

try another easy proof ? :)

http://prntscr.com/4fludo

Answer 22

form is simply \(\large n = p^9\) or \(\large n = p_1p_2^4\)

Answer 23

ok keep going with part b

Answer 24

part b >.<
not easy to me

Answer 25

since \( n \lt \sigma(n)\)
we only need to test \(n \lt 10\)

Answer 26

\(\sigma(2) = 1+2 = 3\)

Answer 27

\(\sigma(3) = 1+3 = 4\)
\(\sigma(4) = 1+2 + 4 = 7\)
\(\sigma(5) = 1+5 = 6\)

Answer 28

then ?

Answer 29

\(\sigma(6) = 1+2+3+6 = 12\)
\(\sigma(7) = 1+7=8\)
\(\sigma(8) = 1+2+4+8 = 15\)
\(\sigma(9) = 1+3+9= 13\)

for \(n \gt 9\), since the \(\sigma(n) \gt 10\), there will not be any numbers with \(\sigma(n)=10\)

Answer 30

ahaa ! cool >.<

Answer 31

one more
http://prntscr.com/4flwjz

Answer 32

shoul i use
\(\sigma (m^2)=\sigma (m)\sigma (m)\)

Answer 33

thats not true as \(\large \gcd(m,m) \ne 1\)

Answer 34

>.<

Answer 35

\(\large \sigma (4 . 2) = 1+2+4+8 = 15 \color{red}{\ne} 21 = \sigma(4)*\sigma(2)\)

Answer 36

multiplicative works only when gcd is 1

Answer 37

roger !

Answer 38

ok so k could be any number , or i can chose w.e i want ?

Answer 39

yeah say \(k = 3\) ?

Answer 40

got it !

Answer 41

how ?

Answer 42

there is infinitly many number combination of 4k, 5k

\(\sigma(m^2)=\sigma(4^2 k^2)= \sigma(4^2) \sigma (k^2)=\sigma(5^2)\sigma(k^2)=\sigma (5^2k^2)=\sigma(n^2)\)

Answer 43

got it :)

Answer 44

thats the reason we need gcd(5,4,k) = 1 is it

Answer 45

so that we can do \(\large \sigma(4^2k^2) = \sigma(4^2)*\sigma(k^2)\)

Answer 46

yeah , we can chose other forms like
bk,ak
but we need \(\sigma( a^2) =\sigma (b^2))

Answer 47

\(\sigma( a^2) =\sigma (b^2)\)

Answer 48

and \(\large \sigma(4^2) = 31 = \sigma(5^2)\)

Answer 49

understood :)

Answer 50

yosh