What is the NO-SHORTCUT approach for learning great Mathematics?
Learn More

February 15, 2021

Test of Mathematics Solution Subjective 32 | Power of 3

Test of Mathematics at the 10+2 Level

Test of Mathematics Solution Subjective 32 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also see: Cheenta I.S.I. & C.M.I. Entrance Course


Show that the number 11...1 with $ 3^n $ digits is divisible by $ 3^n$


We use induction. For n=1; we check 111 is divisible by 3. Assuming that the result host for n=k, we establish that it holds for n=k+1.

The number 111...111 (with $ 3^{k+1} $ digits) can be written in 3 blocks each having $ 3^{k} $ 1's.  Hence we can write it as $ {111...111} \times 10^{(2 \times 3^k)} + {111...111} \times 10^{(3^k)} + {111...111}$ where {111...111} denotes $ 3^k $ 1's.
Taking {111...111} common we have $ (111...111)(10^{(2 \cdot 3^k)} + 10^{(3^k)} + 1)$. By induction (111...111) is divisible by $ 3^k $ and we also have 3 dividing $ (10^{(2 \cdot 3^k)} + 10^{(3^k)} +1 )$ as it's sum of digits is 3 (it has only three 1's and rest are 0) .
Hence (111...111) having $ 3^{(k+1)} $ 1's is divisible by $ 3^{(k+1)} $.

Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy

Cheenta Academy

Aditya Birla Education Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.