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<p>No 3</p><p>Pd soal Tertulis&nbsp;</p><p>2001^k + 2002^k + 2003^k + 2004^k, tentukan nilai K sehingga pd penjumlahan tsbt tidak habis dibagi 5.</p><p>Dari soal tsbt kita dpt simpulkan</p><p>1^k + 2^k + 3^k + 4^k (mod 5) == 1^k + 2^k + 3^k + 2^2k&nbsp;</p><p>Dengan Fermat Little Theorem kita dpt a⁴ kongruen 1 (mod 5), kita tinggal Uji saja Nilai K yg ad pilihan Ganda</p><p>Jika K = 2001, maka</p><p>1^2001 + 2^2001 + 3^2001 + 2^4002</p><p>dengan adanya Fakta a^4 == 1 mod 5 kita dpt tulis</p><p>1 + (2^4^500 x 2) + (3^4^500 x 3) + (2^4^1000 x 4) == 1 + (1x2) + (1x3) + (1 x 4) (mod 5)&nbsp;</p><p>1 + 2 + 3 + 4 (mod 5) == 10 (mod 5) == 0 (mod 5) berarti k = 2021 tidk menjadi solusi.&nbsp;</p><p>Jika K = 2002&nbsp;</p><p>1^2002 + 2^2002 + 3^2002 + 4^2002</p><p>dengan fakta diatas jg kita dpt tulis menjadi</p><p>1 + (2^4^500 x 2 x 2) + (3^4^500 x 3 x 3) +(4^4^500 x 4 x 4) (mod 5)&nbsp;</p><p>1 + (1 x 4) + (1 x 9) + (1 x 16) (mod 5) == 1 + 4 + 9 + 16 (mod 5) == 0 (mod 5) berarti k =2002 jg bukan solusi</p><p>Jika K=2003</p><p>1^2003 + 2^2003 + 3^2003 + 4^2003 (mod 5)&nbsp;</p><p>1 + (2^4^500 x 2^3) + ( 3^4^500 x 3^3) + (4^4^500 x 4^3) mod 5</p><p>1 + (1 x 8) + (1 x 27) + (1 x 64) (mod 5 &nbsp;== 1 + 3 + 2 + 4 == 10 (mod 5) == 0 (mod 5) , Jadi K = 2003 juga bukan solusi</p><p>jika K = 2004</p><p>1^2004 + 2^2004 ^ 3^2004 ^ 4^2004 (mod 5) == 1 + (2^4^501) + (3^4^501) + (4^4^501) (mod 5) == 1 + 1 + 1 + 1 == 4 (mod 5), Jadi K = 2004 yg memenuhi soal tsbt</p><p>&nbsp;</p><p>&nbsp;</p><p>&nbsp;</p><p>&nbsp;</p><p>&nbsp;</p><p>&nbsp;</p><p>&nbsp;</p><p>&nbsp;</p><p>&nbsp;</p><p>&nbsp;</p>