Let an and bn be two sequences such that an=13+6(n−1) and bn=15+7(n−1) for all natural numbers n. Then, the largest three digit integer that is common to both these sequences, is
Started 3 months ago by Shashank in
Explanatory Answer
an=13+6(n−1)=7+6n
bm=15+7(m−1)=8+7m
For some m, n, let 7+6n=8+7m
6n−7m=1
n=6,m=5
a6=43 and b5=43
Since, L.C.M(6,7)=42, the common terms of the series are of the form 43+42k
Let’s find the smallest 4-digit number of this form…
1000/42≅23.80100042≅23.80
43+42(23)=100943+42(23)=1009
∴43+42(22)=967∴43+42(22)=967 is the largest 3-digit such value.
-
No one is replied to this question yet. Be first to reply!