-_-

Kuadrat sempurna abab

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Adakah bilangan kuadrat sempurna empat angka berbentuk $abab$? Jelaskan.

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4 minutes ago, -_- said:

Adakah bilangan kuadrat sempurna empat angka berbentuk abababab ? Jelaskan.

Tidaak ._.

 

abab = k^2

1000a + 100b + 10a + b = k^2

1010a + 101b = k^2

101(10a + b) = k^2

101 = bilangan prima

10a + b = 101 TM

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Just now, Agita Phasa said:

Tidaak ._.

 

abab = k^2

1000a + 100b + 10a + b = k^2

1010a + 101b = k^2

101(10a + b) = k^2

101 = bilangan prima

10a + b = 101 TM


Argumen terakhir harusnya 101 habis membagi 10a+b sih (tapi yha tetep aja ga mungkin, soalnya 10a+b kan rangenya dari 11-99)

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12 minutes ago, -_- said:


Argumen terakhir harusnya 101 habis membagi 10a+b sih (tapi yha tetep aja ga mungkin, soalnya 10a+b kan rangenya dari 11-99)

Langsung kepikiran 10a + b maksimal bernilai 99, jadi TM hhe

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$=\overline{abab} \\ =100\overline{ab} + \overline{ab} \\ = 101\overline{ab}$

Karena $101$ bilangan prima, agar menjadi bilangan kuadrat sempurna haris dikalikan $101$. Padahal $9 < \overline{ab} < 100$. Jadi, tidak ada bentuk kuadrat sempurna $\overline{abab}$.

Edited by Wildan Bagus W

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