Menu Close

N-lt-aabb-gt-N-amp-N-is-perfect-square-find-N-




Question Number 192786 by MM42 last updated on 27/May/23
N=<aabb>∈N  &  N  is  perfect square  find  N  ?
$${N}=<{aabb}>\in\mathbb{N}\:\:\&\:\:{N}\:\:{is}\:\:{perfect}\:{square} \\ $$$${find}\:\:{N}\:\:? \\ $$$$ \\ $$
Answered by AST last updated on 27/May/23
1100a+11b=x^2 ⇒11(100a+b)=x^2   11∣100a+b⇒100a+b=11p^2 ⇒11∣a+b  (a,b)≠(6,5),b≠2,3,6,7,8(a perfect square cannot  end in 2,3,7 or 8,power of 2 in aabb^(____)  when b = 6 is 1)  ⇒Possible values of (a,b)=(2,9),(7,4)  Of these,only (7,4) gives a perfect square  ⇒N=7744
$$\mathrm{1100}{a}+\mathrm{11}{b}={x}^{\mathrm{2}} \Rightarrow\mathrm{11}\left(\mathrm{100}{a}+{b}\right)={x}^{\mathrm{2}} \\ $$$$\mathrm{11}\mid\mathrm{100}{a}+{b}\Rightarrow\mathrm{100}{a}+{b}=\mathrm{11}{p}^{\mathrm{2}} \Rightarrow\mathrm{11}\mid{a}+{b} \\ $$$$\left({a},{b}\right)\neq\left(\mathrm{6},\mathrm{5}\right),{b}\neq\mathrm{2},\mathrm{3},\mathrm{6},\mathrm{7},\mathrm{8}\left({a}\:{perfect}\:{square}\:{cannot}\right. \\ $$$$\left.{end}\:{in}\:\mathrm{2},\mathrm{3},\mathrm{7}\:{or}\:\mathrm{8},{power}\:{of}\:\mathrm{2}\:{in}\:\overset{\_\_\_\_} {{aabb}}\:{when}\:{b}\:=\:\mathrm{6}\:{is}\:\mathrm{1}\right) \\ $$$$\Rightarrow{Possible}\:{values}\:{of}\:\left({a},{b}\right)=\left(\mathrm{2},\mathrm{9}\right),\left(\mathrm{7},\mathrm{4}\right) \\ $$$${Of}\:{these},{only}\:\left(\mathrm{7},\mathrm{4}\right)\:{gives}\:{a}\:{perfect}\:{square} \\ $$$$\Rightarrow{N}=\mathrm{7744} \\ $$
Commented by BaliramKumar last updated on 27/May/23
perfect square number last two same digit only  00 or 44  ∵ 11∣a+b  ∴ bb ≠ 00             (if bb = 00 then 11= a>9, it′s not possible)
$$\mathrm{perfect}\:\mathrm{square}\:\mathrm{number}\:\mathrm{last}\:\mathrm{two}\:\mathrm{same}\:\mathrm{digit}\:\mathrm{only} \\ $$$$\mathrm{00}\:\mathrm{or}\:\mathrm{44} \\ $$$$\because\:\mathrm{11}\mid\mathrm{a}+\mathrm{b} \\ $$$$\therefore\:\mathrm{bb}\:\neq\:\mathrm{00}\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\left(\mathrm{if}\:\mathrm{bb}\:=\:\mathrm{00}\:\mathrm{then}\:\mathrm{11}=\:\mathrm{a}>\mathrm{9},\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{possible}\right) \\ $$
Commented by MM42 last updated on 28/May/23
bravo
$${bravo} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *