#number system

60 messages · Page 1 of 1 (latest)

ember marsh
delicate fieldBOT
potent hill
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lets assume

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$\sqrt{N} = 2^a 3^b 5^c ....$

tender quarryBOT
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ᴰᴱᴬᴰ ★ ᴋɪʟʟᴇʀ ⁰⁴

879304647356067870.png
potent hill
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wait

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so lets take

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sqrt N as a number

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perfect number

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so 1 to sqrt N has 11 factors

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means including 1 has 12

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so there pair also exist

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so total is 24

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but include sqrt N

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so its 25

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@ember marsh

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25?

ember marsh
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Yes 25 is right

potent hill
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ok

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did you get it

ember marsh
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Is it necessary to consider n as a perfect square?

potent hill
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as 24 is not in option

ember marsh
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So what

potent hill
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otherwise there would be two cases

ember marsh
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So*

potent hill
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for sqrt N as perfect

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it will be count

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like

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take 20

ember marsh
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Ok ok

potent hill
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ok

ember marsh
potent hill
ember marsh
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Because double should be of 11 not 12

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If there are 12 factors then that means the power was 11

potent hill
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wont you add 1

ember marsh
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I added one that's why 12 factors

potent hill
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factors are btw 1 and sqrtN

ember marsh
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I know

potent hill
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1 === N

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so 24

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and sqrt N

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so 25

ember marsh
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Ok i was wrong

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But it's just N needs to be a perfect square

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It can't be a non perfect square but that reasoning of 24 was wrong

potent hill
ember marsh
potent hill
potent hill
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not*

ember marsh
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You english is bad brother. You are unable to explain your point

potent hill
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i know

ember marsh
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Ok I got it

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The no. Of factors on both side of √n are equal if the total no. Of factors are odd

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I think this is the trick?

potent hill
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just observations