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calculati suma tuturor nr.naturale de 3 cifre care impartite la 31 dau restul 13 ?



Răspuns :

X:31=13rest r
r<31 deci: r∈{0,1,2,3,4,5,.....,30}
X=31x13+r=403+r
X∈{403+0, 403+1,..... ,403+30}
deci suma tuturor nr nat care respecta cerinta este:
403x30+1+2+3+4+5+....+30=12090+30x31:2=12090+30x31=12090+930=13020
Sper ca e corect :))
106,199,292,385;
Sper ca te-am ajutat.