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Question Number 181275 by Linton last updated on 23/Nov/22

what is the sum of all even factors of 1000?

$${what}\:{is}\:{the}\:{sum}\:{of}\:{all} \\ $$$${even}\:{factors}\:{of}\:\mathrm{1000}? \\ $$

Commented by Rasheed.Sindhi last updated on 24/Nov/22

����

Commented by Linton last updated on 24/Nov/22

well done

$${well}\:{done} \\ $$

Commented by Acem last updated on 24/Nov/22

♥♥

$$\heartsuit\heartsuit \\ $$

Commented by Frix last updated on 23/Nov/22

1000=2^3 5^3 sum of even factors is (1+5+5^2 +5^3 )(2+2^2 +2^3 )=2184

$$\mathrm{1000}=\mathrm{2}^{\mathrm{3}} \mathrm{5}^{\mathrm{3}} \\ $$$$\mathrm{sum}\:\mathrm{of}\:\mathrm{even}\:\mathrm{factors}\:\mathrm{is} \\ $$$$\left(\mathrm{1}+\mathrm{5}+\mathrm{5}^{\mathrm{2}} +\mathrm{5}^{\mathrm{3}} \right)\left(\mathrm{2}+\mathrm{2}^{\mathrm{2}} +\mathrm{2}^{\mathrm{3}} \right)=\mathrm{2184} \\ $$

Commented by Rasheed.Sindhi last updated on 23/Nov/22

Sir this is a complete solution, more than a comment!

$${Sir}\:{this}\:{is}\:{a}\:{complete}\:{solution}, \\ $$$$\boldsymbol{{more}}\:\boldsymbol{{than}}\:\boldsymbol{{a}}\:\boldsymbol{{comment}}! \\ $$

Commented by Frix last updated on 23/Nov/22

I′ve been afraid others might claim it is not a complete solution because I didn′t explain it...

$$\mathrm{I}'\mathrm{ve}\:\mathrm{been}\:\mathrm{afraid}\:\mathrm{others}\:\mathrm{might}\:\mathrm{claim}\:\mathrm{it}\:\mathrm{is} \\ $$$$\boldsymbol{{not}}\:\mathrm{a}\:\mathrm{complete}\:\mathrm{solution}\:\mathrm{because}\:\mathrm{I}\:\mathrm{didn}'\mathrm{t} \\ $$$$\mathrm{explain}\:\mathrm{it}... \\ $$

Commented by Acem last updated on 23/Nov/22

@Sir Rasheed, Have a good evening bro.

$$@{Sir}\:{Rasheed},\:{Have}\:{a}\:{good}\:{evening}\:{bro}. \\ $$

Answered by Acem last updated on 23/Nov/22

even×odd= even the 1000= 2^3 5^3 2^i 5^j ∀ i : 2^i is even , ∀j : 5^j is odd The request is the sum of the even factors of 1000 Sum= Σ_(i=1) ^3 2^i Σ_(j=0) ^3 5^j that means we have 12 numbers Sum= 14 × 156= 2 184 It′s funny appearing the question 1 0 0 0

$$\:{even}×{odd}=\:{even} \\ $$$$\:{the}\:\mathrm{1000}=\:\mathrm{2}^{\mathrm{3}} \:\mathrm{5}^{\mathrm{3}} \: \\ $$$$\:\mathrm{2}^{{i}} \:\mathrm{5}^{{j}} \:\:\forall\:{i}\::\:\mathrm{2}^{{i}} \:{is}\:{even}\:,\:\forall{j}\::\:\mathrm{5}^{{j}} \:{is}\:{odd} \\ $$$$\:{The}\:{request}\:{is}\:{the}\:{sum}\:{of}\:{the}\:{even}\:{factors}\:{of}\:\mathrm{1000} \\ $$$$\:{Sum}=\:\underset{{i}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}\mathrm{2}^{{i}} \:\underset{{j}=\mathrm{0}} {\overset{\mathrm{3}} {\sum}}\mathrm{5}^{{j}} \:{that}\:{means}\:{we}\:{have}\:\mathrm{12}\:{numbers} \\ $$$$\:{Sum}=\:\mathrm{14}\:×\:\mathrm{156}=\:\mathrm{2}\:\mathrm{184} \\ $$$$ \\ $$$$\:{It}'{s}\:{funny}\:{appearing}\:{the}\:{question}\:\mathrm{1}\:\:\mathrm{0}\:\:\mathrm{0}\:\:\mathrm{0} \\ $$$$\: \\ $$

Commented by Rasheed.Sindhi last updated on 24/Nov/22

Is the following correct?: Sum of even factors of 1000 =2×Sum of all factors of 500

$${Is}\:{the}\:{following}\:{correct}?:\: \\ $$$${Sum}\:{of}\:\boldsymbol{{even}}\:{factors}\:{of}\:\mathrm{1000} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{2}×{Sum}\:{of}\:\boldsymbol{{all}}\:{factors}\:{of}\:\mathrm{500} \\ $$

Commented by Linton last updated on 24/Nov/22

well done

$${well}\:{done} \\ $$

Commented by Rasheed.Sindhi last updated on 24/Nov/22

@Acem, I also think so.

$$@{Acem},\:{I}\:{also}\:{think}\:{so}. \\ $$

Commented by Acem last updated on 24/Nov/22

I think yes, because either it factors were even or odd the result after multiplying them by 2 will be even, right?

$${I}\:{think}\:{yes},\:{because}\:{either}\:{it}\:{factors}\:{were}\:{even} \\ $$$$\:{or}\:{odd}\:{the}\:{result}\:{after}\:{multiplying}\:{them}\:{by}\:\mathrm{2} \\ $$$$\:{will}\:{be}\:{even},\:{right}? \\ $$

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