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Question Number 213482 by efronzo1 last updated on 06/Nov/24

Answered by A5T last updated on 06/Nov/24

3∣Y⇒8+c+2+3+d≡0(mod 3)⇒c+d≡2(mod 3) 3∣X⇒9∣Y : 3+1+a+5+b≡0(mod 3)⇒8+c+2+3+d≡0(mod 9) ⇒a+b≡0(mod 3)⇒c+d≡5(mod 9) 3∣X⇒c+d=5⇒a+b=0,3,6,9 ∧ c+d=5 3∤X⇒a+b∈{1,2,4,5,7,8} ∧ c+d=2,5,8 And the possible values follow from those.

3∣Y⇒8+c+2+3+d≡0(mod3)⇒c+d≡2(mod3)3∣X⇒9∣Y:3+1+a+5+b≡0(mod3)⇒8+c+2+3+d≡0(mod9)⇒a+b≡0(mod3)⇒c+d≡5(mod9)3∣X⇒c+d=5⇒a+b=0,3,6,9∧c+d=53∤X⇒a+b∈{1,2,4,5,7,8}∧c+d=2,5,8Andthepossiblevaluesfollowfromthose.

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