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


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

${(20 + 25)}^{2} = 2025$ $only 3 other 4 digit numbers :$ ${(00 + 01)}^{2} = 0001$ ${(30 + 25)}^{2} = 3025$ ${(98 + 01)}^{2} = 9801$
	Answered by Rasheed.Sindhi last updated on 01/Jan/25

	$\begin{matrix} H o w t o d e t e r m i n e s u c h n u m b e r s \end{matrix}$ ${(a + b)}^{2} = 100 a + b; a, b a r e 2 - d i g i t n u m b e r s$ $a^{2} + 2 a b + b^{2} - 100 a - b = 0$ $a^{2} + (2 b - 100) a + b^{2} - b = 0$ $a = \frac{100 - 2 b \pm \sqrt{{(2 b - 100)}^{2} - 4 (b^{2} - b)}}{2 a}$ $a = \frac{100 - 2 b \pm 2 \sqrt{{(b - 50)}^{2} - (b^{2} - b)}}{2}$ $= 50 - b \pm \sqrt{b^{2} - 100 b + 2500 - b^{2} + b}$ $= 50 - b \pm \sqrt{2500 - 99 b}; 00 ⩽ b ⩽ 25$ $2500 - 99 b = p^{2} \Rightarrow b = 00, 01, 25$ $b = 00$ $a = 50 - b \pm \sqrt{2500 - 99 b}$ $a = 50 - 0 \pm 50 = \overset{\times}{100}, 00$ $0000$ $b = 01$ $a = 50 - 1 \pm 49 = 98, 00$ $0001, 9801$ $b = 25$ $a = 25 \pm 25 = 50, 00$ $= 50 - 25 \pm 5$ $30 o r 20$ $3025, 2025$
	Commented by MathematicalUser2357 last updated on 08/Jan/25

	$W h e r e d i d y o u g e t t h a t ‘ ‘ p^{″} ?$
	Answered by MrGaster last updated on 01/Jan/25

	${(20 + 25)}^{2} = 2025$ ${(45)}^{2} = 2025$
	Commented by Frix last updated on 01/Jan/25

	$You seriously claim that {(20 + 25)}^{2} \neq {(45)}^{2} ?$
	Commented by MrGaster last updated on 01/Jan/25
	Sorry, my fault.
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