How Cheenta works to ensure student success?

Explore the Back-StoryTest 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)} $.

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)} $.

Cheenta is a knowledge partner of Aditya Birla Education Academy

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

JOIN TRIALAcademic Programs

Free Resources

Why Cheenta?