if-AB-BA-and-BB-three-tow-digits-natural-numbers-if-AB-BA-BB-4-then-find-the-maximum-volue-of-A-B-

Question Number 186864 by mathlove last updated on 11/Feb/23

$${if}\:\:{AB},{BA}\:{and}\:{BB}\:\:{three}\:\:{tow}\:{digits}\:{natural} \\ $$$${numbers}\:{if}\:\:\frac{\left({AB}+{BA}\right)}{{BB}}=\mathrm{4} \\ $$$${then}\:{find}\:{the}\:{maximum}\:{volue}\:{of} \\ $$$$\left({A}+{B}\right)=? \\ $$

Answered by mr W last updated on 11/Feb/23

(((AB+BA))/(BB))=((11(a+b))/(11b))=((a+b)/b)=4 a=3b b_(max) =3 a_(max) =9 ⇒(a+b)_(max) =12

$$\frac{\left({AB}+{BA}\right)}{{BB}}=\frac{\mathrm{11}\left({a}+{b}\right)}{\mathrm{11}{b}}=\frac{{a}+{b}}{{b}}=\mathrm{4} \\ $$$${a}=\mathrm{3}{b} \\ $$$${b}_{{max}} =\mathrm{3} \\ $$$${a}_{{max}} =\mathrm{9} \\ $$$$\Rightarrow\left({a}+{b}\right)_{{max}} =\mathrm{12} \\ $$

Commented by mathlove last updated on 11/Feb/23

$${how}\:{is}\:{BA}={BB}\:\:? \\ $$

Commented by mr W last updated on 11/Feb/23

$${who}\:{said}\:{BA}={BB}? \\ $$

Commented by mathlove last updated on 11/Feb/23

$${you}\:{said}\:\:{BA}={b}\:\:{and}\:\:{BB}={b} \\ $$

Commented by mathlove last updated on 11/Feb/23

$${pleasi}\:{alittelmore}\:{describ}\:{you}\:{solution}\: \\ $$

Commented by mr W last updated on 11/Feb/23

i didn′t say BA=b, BB=b. say the two digits are a and b. (i prefer using lower case, a=A, b=B) AB=ab=10a+b BA=ba=10b+a AB+BA=10a+b+10b+a=11(a+b) BB=bb=10b+b=11b ((AB+BA)/(BB))=((11(a+b))/(11b))=((a+b)/b)=4 ⇒a=3b since a is a digit, it can be at most 9, therefore b is at most 3, a at most 9. b_(max) =3, a_(max) =9.

$${i}\:{didn}'{t}\:{say}\:{BA}={b},\:{BB}={b}. \\ $$$${say}\:{the}\:{two}\:{digits}\:{are}\:{a}\:{and}\:{b}. \\ $$$$\left({i}\:{prefer}\:{using}\:{lower}\:{case},\:{a}={A},\right. \\ $$$$\left.{b}={B}\right) \\ $$$${AB}=\underline{{ab}}=\mathrm{10}{a}+{b} \\ $$$${BA}=\underline{{ba}}=\mathrm{10}{b}+{a} \\ $$$${AB}+{BA}=\mathrm{10}{a}+{b}+\mathrm{10}{b}+{a}=\mathrm{11}\left({a}+{b}\right) \\ $$$${BB}=\underline{{bb}}=\mathrm{10}{b}+{b}=\mathrm{11}{b} \\ $$$$\frac{{AB}+{BA}}{{BB}}=\frac{\mathrm{11}\left({a}+{b}\right)}{\mathrm{11}{b}}=\frac{{a}+{b}}{{b}}=\mathrm{4} \\ $$$$\Rightarrow{a}=\mathrm{3}{b} \\ $$$${since}\:{a}\:{is}\:{a}\:{digit},\:{it}\:{can}\:{be}\:{at}\:{most}\:\mathrm{9}, \\ $$$${therefore}\:{b}\:{is}\:{at}\:{most}\:\mathrm{3},\:{a}\:{at}\:{most}\:\mathrm{9}. \\ $$$${b}_{{max}} =\mathrm{3},\:{a}_{{max}} =\mathrm{9}. \\ $$

Commented by mathlove last updated on 12/Feb/23

$${thanks}\:{mr}\:{now}\:{understand} \\ $$

Answered by manxsol last updated on 12/Feb/23

AB=10A+B BA=10B+A BB=10B+B=11B ((AB+BA)/(BB))=((10A+B+10B+A)/(11B)) ((11A+11Bm)/(11B))=4 (A/B)+1=4 (A/B)=3 A=9 max B=3 max A+B=12 convention number two cifras ab^(−) =10a+b how script ab [select] [←] [←][□^□ ][−] ab^(−)

$$ \\ $$$${AB}=\mathrm{10}{A}+{B} \\ $$$${BA}=\mathrm{10}{B}+{A} \\ $$$${BB}=\mathrm{10}{B}+{B}=\mathrm{11}{B} \\ $$$$\frac{{AB}+{BA}}{{BB}}=\frac{\mathrm{10}{A}+{B}+\mathrm{10}{B}+{A}}{\mathrm{11}{B}} \\ $$$$\frac{\mathrm{11}{A}+\mathrm{11}{Bm}}{\mathrm{11}{B}}=\mathrm{4} \\ $$$$\frac{{A}}{{B}}+\mathrm{1}=\mathrm{4} \\ $$$$\frac{{A}}{{B}}=\mathrm{3} \\ $$$${A}=\mathrm{9}\:{max} \\ $$$${B}=\mathrm{3}\:{max} \\ $$$${A}+{B}=\mathrm{12} \\ $$$${convention} \\ $$$${number}\:{two}\:{cifras} \\ $$$$\overline {{ab}}=\mathrm{10}{a}+{b} \\ $$$${how}\:{script} \\ $$$${ab}\:\:\:\left[{select}\right]\:\left[\leftarrow\right]\:\left[\leftarrow\right]\left[\overset{\Box} {\Box}\right]\left[−\right] \\ $$$$\overline {{ab}} \\ $$

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