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Question Number 215227 by Ghisom last updated on 01/Jan/25

(20+25)^2 =2025 only 3 other 4 digit numbers: (00+01)^2 =0001 (30+25)^2 =3025 (98+01)^2 =9801

$$\left(\mathrm{20}+\mathrm{25}\right)^{\mathrm{2}} =\mathrm{2025} \\ $$$$\mathrm{only}\:\mathrm{3}\:\mathrm{other}\:\mathrm{4}\:\mathrm{digit}\:\mathrm{numbers}: \\ $$$$\left(\mathrm{00}+\mathrm{01}\right)^{\mathrm{2}} =\mathrm{0001} \\ $$$$\left(\mathrm{30}+\mathrm{25}\right)^{\mathrm{2}} =\mathrm{3025} \\ $$$$\left(\mathrm{98}+\mathrm{01}\right)^{\mathrm{2}} =\mathrm{9801} \\ $$

Answered by Rasheed.Sindhi last updated on 01/Jan/25

determinant (((How to determine such numbers))) (a+b)^2 =100a+b ; a,b are 2-digit numbers a^2 +2ab+b^2 −100a−b=0 a^2 +(2b−100)a+b^2 −b=0 a=((100−2b±(√((2b−100)^2 −4(b^2 −b))))/(2a)) a=((100−2b±2(√((b−50)^2 −(b^2 −b))))/2) =50−b±(√(b^2 −100b+2500−b^2 +b)) =50−b±(√(2500−99b)) ; 00≤b≤25 2500−99b=p^2 ⇒b=00,01,25 b=00 a=50−b±(√(2500−99b)) a=50−0±50=100^(×) , 00 0000 b=01 a=50−1±49=98 , 00 0001 , 9801 b=25 a=25±25=50,00 =50−25±5 30 or 20 3025 ,2025

$$\begin{array}{|c|}{\mathcal{H}{ow}\:{to}\:{determine}\:{such}\:{numbers}}\\\hline\end{array} \\ $$$$\left({a}+{b}\right)^{\mathrm{2}} =\mathrm{100}{a}+{b}\:;\:{a},{b}\:{are}\:\mathrm{2}-{digit}\:{numbers} \\ $$$${a}^{\mathrm{2}} +\mathrm{2}{ab}+{b}^{\mathrm{2}} −\mathrm{100}{a}−{b}=\mathrm{0} \\ $$$${a}^{\mathrm{2}} +\left(\mathrm{2}{b}−\mathrm{100}\right){a}+{b}^{\mathrm{2}} −{b}=\mathrm{0} \\ $$$${a}=\frac{\mathrm{100}−\mathrm{2}{b}\pm\sqrt{\left(\mathrm{2}{b}−\mathrm{100}\right)^{\mathrm{2}} −\mathrm{4}\left({b}^{\mathrm{2}} −{b}\right)}}{\mathrm{2}{a}} \\ $$$${a}=\frac{\mathrm{100}−\mathrm{2}{b}\pm\mathrm{2}\sqrt{\left({b}−\mathrm{50}\right)^{\mathrm{2}} −\left({b}^{\mathrm{2}} −{b}\right)}}{\mathrm{2}} \\ $$$$\:\:\:=\mathrm{50}−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{100}{b}+\mathrm{2500}−{b}^{\mathrm{2}} +{b}} \\ $$$$\:\:\:=\mathrm{50}−{b}\pm\sqrt{\mathrm{2500}−\mathrm{99}{b}}\:;\:\mathrm{00}\leqslant{b}\leqslant\mathrm{25} \\ $$$$\:\:\:\:\:\:\mathrm{2500}−\mathrm{99}{b}={p}^{\mathrm{2}} \Rightarrow{b}=\mathrm{00},\mathrm{01},\mathrm{25} \\ $$$${b}=\mathrm{00} \\ $$$$\:{a}=\mathrm{50}−{b}\pm\sqrt{\mathrm{2500}−\mathrm{99}{b}}\: \\ $$$${a}=\mathrm{50}−\mathrm{0}\pm\mathrm{50}=\overset{×} {\mathrm{100}}\:,\:\mathrm{00} \\ $$$$\mathrm{0000} \\ $$$${b}=\mathrm{01} \\ $$$${a}=\mathrm{50}−\mathrm{1}\pm\mathrm{49}=\mathrm{98}\:,\:\mathrm{00} \\ $$$$\mathrm{0001}\:,\:\mathrm{9801} \\ $$$${b}=\mathrm{25} \\ $$$${a}=\mathrm{25}\pm\mathrm{25}=\mathrm{50},\mathrm{00} \\ $$$$\:\:\:=\mathrm{50}−\mathrm{25}\pm\mathrm{5} \\ $$$$\:\:\:\:\:\mathrm{30}\:{or}\:\mathrm{20} \\ $$$$\mathrm{3025}\:,\mathrm{2025} \\ $$

Commented by MathematicalUser2357 last updated on 08/Jan/25

Where did you get that “p”?

$${Where}\:{did}\:{you}\:{get}\:{that}\:``{p}''? \\ $$

Answered by MrGaster last updated on 01/Jan/25

(20+25)^2 =2025 (45)^2 =2025

$$\cancel{\left(\mathrm{20}+\mathrm{25}\right)^{\mathrm{2}} =\mathrm{2025}} \\ $$$$\left(\mathrm{45}\right)^{\mathrm{2}} =\mathrm{2025} \\ $$

Commented by Frix last updated on 01/Jan/25

You seriously claim that (20+25)^2 ≠(45)^2 ?

$$\mathrm{You}\:\mathrm{seriously}\:\mathrm{claim}\:\mathrm{that}\:\left(\mathrm{20}+\mathrm{25}\right)^{\mathrm{2}} \neq\left(\mathrm{45}\right)^{\mathrm{2}} ? \\ $$

Commented by MrGaster last updated on 01/Jan/25

Sorry, my fault.

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