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Question Number 215999 by lmcp1203 last updated on 25/Jan/25

if the fraction ((m^2 +25m)/(m+1)) is reductible. how many values does m take if is a 2 digit number? thanks

$${if}\:{the}\:{fraction}\:\frac{{m}^{\mathrm{2}} +\mathrm{25}{m}}{{m}+\mathrm{1}}\:\:{is}\:{reductible}.\:{how}\:{many}\:{values}\:{does}\:{m}\:\:{take}\:{if}\:{is}\:{a}\:\mathrm{2}\:{digit}\:\:{number}?\:{thanks} \\ $$

Answered by Rasheed.Sindhi last updated on 25/Jan/25

((m^2 +25m)/(m+1))=m(((m+25)/(m+1))) =m(((m+1+24)/(m+1)))=m(1+((24)/(m+1))) ((m^2 +25m)/(m+1)) is reducible⇒((24)/(m+1)) is reducible. 24=2^3 .3 m+1=2p,3q,6r m=2p−1,3q−1,6r−1 ∧ 10≤m≤99 { ((10≤2p−1≤99⇒6≤p≤50)),((10≤3q−1≤99⇒4≤q≤33)),((10≤6r−1≤99⇒2≤r≤16)) :} m=2p−1: 6≤p≤50 : 45 values m=3q−1: 4≤q≤33 : 30 values m=6r−1: 2≤r≤16 : 15 values common Total 45+30−15=60 values

$$\frac{{m}^{\mathrm{2}} +\mathrm{25}{m}}{{m}+\mathrm{1}}={m}\left(\frac{{m}+\mathrm{25}}{{m}+\mathrm{1}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:={m}\left(\frac{{m}+\mathrm{1}+\mathrm{24}}{{m}+\mathrm{1}}\right)={m}\left(\mathrm{1}+\frac{\mathrm{24}}{{m}+\mathrm{1}}\right) \\ $$$$\frac{{m}^{\mathrm{2}} +\mathrm{25}{m}}{{m}+\mathrm{1}}\:{is}\:{reducible}\Rightarrow\frac{\mathrm{24}}{{m}+\mathrm{1}}\:{is}\:{reducible}. \\ $$$$\mathrm{24}=\mathrm{2}^{\mathrm{3}} .\mathrm{3} \\ $$$${m}+\mathrm{1}=\mathrm{2}{p},\mathrm{3}{q},\mathrm{6}{r} \\ $$$${m}=\mathrm{2}{p}−\mathrm{1},\mathrm{3}{q}−\mathrm{1},\mathrm{6}{r}−\mathrm{1}\:\wedge\:\mathrm{10}\leqslant{m}\leqslant\mathrm{99} \\ $$$$\begin{cases}{\mathrm{10}\leqslant\mathrm{2}{p}−\mathrm{1}\leqslant\mathrm{99}\Rightarrow\mathrm{6}\leqslant{p}\leqslant\mathrm{50}}\\{\mathrm{10}\leqslant\mathrm{3}{q}−\mathrm{1}\leqslant\mathrm{99}\Rightarrow\mathrm{4}\leqslant{q}\leqslant\mathrm{33}}\\{\mathrm{10}\leqslant\mathrm{6}{r}−\mathrm{1}\leqslant\mathrm{99}\Rightarrow\mathrm{2}\leqslant{r}\leqslant\mathrm{16}}\end{cases} \\ $$$${m}=\mathrm{2}{p}−\mathrm{1}:\:\mathrm{6}\leqslant{p}\leqslant\mathrm{50}\::\:\mathrm{45}\:{values} \\ $$$${m}=\mathrm{3}{q}−\mathrm{1}:\:\mathrm{4}\leqslant{q}\leqslant\mathrm{33}\::\:\:\mathrm{30}\:{values} \\ $$$${m}=\mathrm{6}{r}−\mathrm{1}:\:\mathrm{2}\leqslant{r}\leqslant\mathrm{16}\::\:\mathrm{15}\:{values}\:{common} \\ $$$$\mathcal{T}{otal}\:\mathrm{45}+\mathrm{30}−\mathrm{15}=\mathrm{60}\:{values} \\ $$

Commented by lmcp1203 last updated on 25/Jan/25

thank you

$${thank}\:{you}\: \\ $$

Answered by A5T last updated on 25/Jan/25

((m^2 +25m)/(m+1))=m+24−((24)/(m+1)) ⇒gcd(m^2 +25m,m+1)=gcd(24,m+1) ((m^2 +25m)/(m+1)) is reducible when (24,m+1)≠1 That is when m+1 contains either 3 or 2 as a prime factor Cardinality of such numbers: ⌊((99)/3)⌋+⌊((99)/2)⌋−⌊((99)/6)⌋−(⌊(9/3)⌋+⌊(9/2)⌋−⌊(9/6)⌋) =33+49−16−3−4+1=60

$$\frac{\mathrm{m}^{\mathrm{2}} +\mathrm{25m}}{\mathrm{m}+\mathrm{1}}=\mathrm{m}+\mathrm{24}−\frac{\mathrm{24}}{\mathrm{m}+\mathrm{1}} \\ $$$$\Rightarrow\mathrm{gcd}\left(\mathrm{m}^{\mathrm{2}} +\mathrm{25m},\mathrm{m}+\mathrm{1}\right)=\mathrm{gcd}\left(\mathrm{24},\mathrm{m}+\mathrm{1}\right) \\ $$$$\frac{\mathrm{m}^{\mathrm{2}} +\mathrm{25m}}{\mathrm{m}+\mathrm{1}}\:\mathrm{is}\:\mathrm{reducible}\:\mathrm{when}\:\left(\mathrm{24},\mathrm{m}+\mathrm{1}\right)\neq\mathrm{1} \\ $$$$\mathrm{That}\:\mathrm{is}\:\mathrm{when}\:\mathrm{m}+\mathrm{1}\:\mathrm{contains}\:\mathrm{either}\:\mathrm{3}\:\mathrm{or}\:\mathrm{2}\:\mathrm{as}\:\mathrm{a} \\ $$$$\mathrm{prime}\:\mathrm{factor} \\ $$$$\mathrm{Cardinality}\:\mathrm{of}\:\mathrm{such}\:\mathrm{numbers}: \\ $$$$\lfloor\frac{\mathrm{99}}{\mathrm{3}}\rfloor+\lfloor\frac{\mathrm{99}}{\mathrm{2}}\rfloor−\lfloor\frac{\mathrm{99}}{\mathrm{6}}\rfloor−\left(\lfloor\frac{\mathrm{9}}{\mathrm{3}}\rfloor+\lfloor\frac{\mathrm{9}}{\mathrm{2}}\rfloor−\lfloor\frac{\mathrm{9}}{\mathrm{6}}\rfloor\right) \\ $$$$=\mathrm{33}+\mathrm{49}−\mathrm{16}−\mathrm{3}−\mathrm{4}+\mathrm{1}=\mathrm{60} \\ $$

Commented by lmcp1203 last updated on 25/Jan/25

thank you

$${thank}\:{you} \\ $$

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