Menu Close

n-N-b-a-N-a-0-In-base-10-n-aabb-1-show-that-n-is-not-prime-2-Give-conditions-on-b-such-that-n-is-perfect-square-3-Determinate-n-such-that-n-is-a-perfect-square-




Question Number 142388 by mathocean1 last updated on 30/May/21
n ∈ N, b, a ∈ N ; a≠0.  In base 10; n=aabb^(−)    1. show that n is not prime.  2. Give conditions on b such that  n is perfect square.  3. Determinate n such that n is a   perfect square.
$$\mathrm{n}\:\in\:\mathbb{N},\:\mathrm{b},\:\mathrm{a}\:\in\:\mathbb{N}\:;\:\mathrm{a}\neq\mathrm{0}. \\ $$$$\mathrm{In}\:\mathrm{base}\:\mathrm{10};\:\mathrm{n}=\overline {\mathrm{aabb}}\: \\ $$$$\mathrm{1}.\:\mathrm{show}\:\mathrm{that}\:\mathrm{n}\:\mathrm{is}\:\mathrm{not}\:\mathrm{prime}. \\ $$$$\mathrm{2}.\:\mathrm{Give}\:\mathrm{conditions}\:\mathrm{on}\:\mathrm{b}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{n}\:\mathrm{is}\:\mathrm{perfect}\:\mathrm{square}. \\ $$$$\mathrm{3}.\:\mathrm{Determinate}\:\mathrm{n}\:\mathrm{such}\:\mathrm{that}\:\mathrm{n}\:\mathrm{is}\:\mathrm{a}\: \\ $$$$\mathrm{perfect}\:\mathrm{square}. \\ $$
Answered by mr W last updated on 31/May/21
1.  n=1000a+100a+10b+b  =1100a+11b  =11(100a+b)≡0(mod 11)  2.  100a+b=11m^2 ≥101 ⇒m≥4  100a+b=11m^2 ≤909 ⇒m≤9  b=11m^2 −100a  3.  n=11^2 m^2 =(11m)^2  with 4≤m≤9
$$\mathrm{1}. \\ $$$${n}=\mathrm{1000}{a}+\mathrm{100}{a}+\mathrm{10}{b}+{b} \\ $$$$=\mathrm{1100}{a}+\mathrm{11}{b} \\ $$$$=\mathrm{11}\left(\mathrm{100}{a}+{b}\right)\equiv\mathrm{0}\left({mod}\:\mathrm{11}\right) \\ $$$$\mathrm{2}. \\ $$$$\mathrm{100}{a}+{b}=\mathrm{11}{m}^{\mathrm{2}} \geqslant\mathrm{101}\:\Rightarrow{m}\geqslant\mathrm{4} \\ $$$$\mathrm{100}{a}+{b}=\mathrm{11}{m}^{\mathrm{2}} \leqslant\mathrm{909}\:\Rightarrow{m}\leqslant\mathrm{9} \\ $$$${b}=\mathrm{11}{m}^{\mathrm{2}} −\mathrm{100}{a} \\ $$$$\mathrm{3}. \\ $$$${n}=\mathrm{11}^{\mathrm{2}} {m}^{\mathrm{2}} =\left(\mathrm{11}{m}\right)^{\mathrm{2}} \:{with}\:\mathrm{4}\leqslant{m}\leqslant\mathrm{9} \\ $$
Commented by mathocean1 last updated on 31/May/21
Sorry...please sir; according to the   second question why have you   100a+b≤909 and ≥101 ?
$$\mathscr{S}{orry}…{please}\:{sir};\:{according}\:{to}\:{the}\: \\ $$$${second}\:{question}\:{why}\:{have}\:{you}\: \\ $$$$\mathrm{100}{a}+{b}\leqslant\mathrm{909}\:{and}\:\geqslant\mathrm{101}\:? \\ $$
Commented by mr W last updated on 31/May/21
1≤a,b≤9  100a+b≤100×9+9=909  100a+b≥100×1+1=101
$$\mathrm{1}\leqslant{a},{b}\leqslant\mathrm{9} \\ $$$$\mathrm{100}{a}+{b}\leqslant\mathrm{100}×\mathrm{9}+\mathrm{9}=\mathrm{909} \\ $$$$\mathrm{100}{a}+{b}\geqslant\mathrm{100}×\mathrm{1}+\mathrm{1}=\mathrm{101} \\ $$
Commented by mathocean1 last updated on 31/May/21
thanks
$${thanks} \\ $$

Leave a Reply

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