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Dterminer-le-nombre-total-des-nombres-de-3-chiffres-qui-sont-impair-et-divisibles-par-9-compris-entre-100-et-500-formule-si-c-est-possible-




Question Number 211311 by a.lgnaoui last updated on 05/Sep/24
Dterminer le nombre total  des nombres   de (3 chiffres)qui sont impair( et) divisibles   par 9   compris entre 100 et 500.?  formule si c est possible?
$$\boldsymbol{\mathrm{Dterminer}}\:\boldsymbol{\mathrm{le}}\:\boldsymbol{\mathrm{nombre}}\:\boldsymbol{\mathrm{total}}\:\:\boldsymbol{\mathrm{des}}\:\boldsymbol{\mathrm{nombres}}\: \\ $$$$\boldsymbol{\mathrm{de}}\:\left(\mathrm{3}\:\boldsymbol{\mathrm{chiffres}}\right)\boldsymbol{\mathrm{qui}}\:\boldsymbol{\mathrm{sont}}\:\boldsymbol{\mathrm{impair}}\left(\:\boldsymbol{\mathrm{et}}\right)\:\boldsymbol{\mathrm{divisibles}}\: \\ $$$$\boldsymbol{\mathrm{par}}\:\mathrm{9}\:\:\:\boldsymbol{\mathrm{compris}}\:\boldsymbol{\mathrm{entre}}\:\mathrm{100}\:\boldsymbol{\mathrm{et}}\:\mathrm{500}.? \\ $$$$\boldsymbol{\mathrm{formule}}\:\boldsymbol{\mathrm{si}}\:\boldsymbol{\mathrm{c}}\:\boldsymbol{\mathrm{est}}\:\boldsymbol{\mathrm{possible}}? \\ $$$$ \\ $$
Answered by mr W last updated on 05/Sep/24
the first one is 117, the second one  is 117+18, the third one is 117+18+18,  etc.   generally:  N=117+18k  N<500 ⇒k≤21  ⇒0≤k≤21 ⇒22 such numbers
$${the}\:{first}\:{one}\:{is}\:\mathrm{117},\:{the}\:{second}\:{one} \\ $$$${is}\:\mathrm{117}+\mathrm{18},\:{the}\:{third}\:{one}\:{is}\:\mathrm{117}+\mathrm{18}+\mathrm{18}, \\ $$$${etc}.\: \\ $$$${generally}: \\ $$$${N}=\mathrm{117}+\mathrm{18}{k} \\ $$$${N}<\mathrm{500}\:\Rightarrow{k}\leqslant\mathrm{21} \\ $$$$\Rightarrow\mathrm{0}\leqslant{k}\leqslant\mathrm{21}\:\Rightarrow\mathrm{22}\:{such}\:{numbers} \\ $$
Commented by a.lgnaoui last updated on 05/Sep/24
thank you very much.
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}. \\ $$

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