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Problem #7 · August 3, 2026

EKUB va EKUK

Ikki natural sonning EKUBi 6 ga, yigʻindisi 84 ga teng. Nechta shunday sonlar jufti mavjud? (Juftlikdagi sonlar tartibi hisobga olinmaydi.)
  1. A3
  2. B4
  3. C6
  4. D7
  5. E13

Solution

Sonlarni 6m6m va 6n6n deb yozamiz, bunda EKUB(m;n)=1EKUB(m; n) = 1 boʻlishi shart. U holda 6(m+n)=846(m + n) = 84, yaʼni m+n=14m + n = 14. Oʻzaro tub m<nm < n juftliklar: (1;13),(3;11),(5;9)(1; 13), (3; 11), (5; 9) — (7;7)(7; 7) mos emas, chunki EKUB(7;7)=7EKUB(7; 7) = 7. Demak juftliklar (6;78),(18;66),(30;54)(6; 78), (18; 66), (30; 54) — jami 3 ta.
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