Find the remainder and quotient when dividing $10^{k}+1$ by $11$,where $k=1, 2, 3, 4, 5$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
$(1)$ For $k=1$,$10^{1}+1=11$. By Euclid's Division Lemma,$11 = 11 \times 1 + 0$. Thus,the quotient is $1$ and the remainder is $0$.
$(2)$ For $k=2$,$10^{2}+1=101$. By Euclid's Division Lemma,$101 = 11 \times 9 + 2$. Thus,the quotient is $9$ and the remainder is $2$.
$(3)$ For $k=3$,$10^{3}+1=1001$. By Euclid's Division Lemma,$1001 = 11 \times 91 + 0$. Thus,the quotient is $91$ and the remainder is $0$.
$(4)$ For $k=4$,$10^{4}+1=10001$. By Euclid's Division Lemma,$10001 = 11 \times 909 + 2$. Thus,the quotient is $909$ and the remainder is $2$.
$(5)$ For $k=5$,$10^{5}+1=100001$. By Euclid's Division Lemma,$100001 = 11 \times 9091 + 0$. Thus,the quotient is $9091$ and the remainder is $0$.

Explore More

Similar Questions

Without actually performing the long division,state whether the following rational number will have a terminating decimal expansion or a non-terminating repeating decimal expansion: $\frac{6}{15}$

Every composite number greater than $1$ can be expressed (factorized) as a product of primes,and this factorization is . . . . . . apart from the order in which the prime factors occur. ($0$,$1$,unique)

Write down the decimal expansion of the rational number $\frac{23}{2^{3} 5^{2}}$.

Use Euclid's division lemma to show that the cube of any positive integer is of the form $9m, 9m+1$ or $9m+8$.

Difficult
View Solution

Find the $g.c.d.$ (Greatest Common Divisor) of $81$ and $237$ using Euclid's division algorithm.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo