E.696. Să se arate că dacă a+3b+5c+7d ⋮ 17,a+3b+5c+7d ~\vdots~ 17,a+3b+5c+7d ⋮ 17, atunci și 53a+57b+61c+65d ⋮ 17.53a+57b+61c+65d ~\vdots~ 17.53a+57b+61c+65d ⋮ 17.
53a+57b+61c+65d=(51a+2a)+(51b+6b)+(51c+10c)+(51d+14d)=53a+57b+61c+65d=(51a+2a) + (51b+6b) + (51c+10c)+(51d+14d)=53a+57b+61c+65d=(51a+2a)+(51b+6b)+(51c+10c)+(51d+14d)= =51⏟M17(a+b+c+d)+2(a+3b+5c+7d⏟M17)=M17.=\underbrace{51}_{M_{17}}(a+b+c+d)+2(\underbrace{a+3b+5c+7d}_{M_{17}})=M_{17}.=M1751(a+b+c+d)+2(M17a+3b+5c+7d)=M17.