Algebra
Product polynomial from three values — divisors
MJAT_TS6_P2
Grade 12
Question:
$(a+1)(b+1)(c+1)(d+1)=3$, $(a+2)(b+2)(c+2)(d+2)=7$, $(a+3)(b+3)(c+3)(d+3)=13$, $(a+4)(b+4)(c+4)(d+4)=21$. The number of positive integral divisors of $(a+5)(b+5)(c+5)(d+5)(a+6)(b+6)(c+6)(d+6)$ equals:
Step-by-Step Solution
Key Concept: $f(x)=(a+x)(b+x)(c+x)(d+x)=(x-1)(x-2)(x-3)(x-4)+\text{lower terms}$. From $f(1)=3,f(2)=7,f(3)=13,f(4)=21$: $f(n)=n^2+n+1$. So $f(5)=31$ and $f(6)=43$.
$(a+5)(a+6)\cdots(d+5)(d+6)$ product $=55\times 163=8965=5\times 11\times 163$. Divisors $=\mathbf{8}$.
Correct Answer: 8