Prove by induction that 11^n - 6 is divisible by 5 for all positive integer n.

Let P(n) be the statement that 11n - 6 is divisible by 5.

BASE CASE: Let n = 1.  This gives 111 - 6 = 5, obviously divisible by 5, therefore we know that P(1) is true.

HYPOTHOSIS STEP: Assume that P(k) is true for some positive integer k.  We can write this a different way: 11k – 6 = 5m where m is also a positive integer.

INDUCTION STEP: We will now show that P(k+1) is true.

P(k+1) states that 11k+1 – 6 is divisible by 5.

11k+1 – 6 = 11 * (11k) – 6  

                = 11* (6 + 5m) – 6  (now we use our hypothesis step, with rearranged expression 11k = 6 + 5m)

                = 55m + 60   (multiplying out the brackets gives)

                = 5 * (11m + 12)  (now factorising again)

Which shows that this is a factor of 5 and that P(k+1) is true.

CONCLUSION: Since P(k+1) is true given P(k), and we know that P(1) is true, we have proved by induction that P(n) is true for all positive integer n.

SH
Answered by Sally H. Further Mathematics tutor

45275 Views

See similar Further Mathematics A Level tutors

Related Further Mathematics A Level answers

All answers ▸

Find the general solution for the determinant of a 3x3 martix. When does the inverse of this matrix not exist?


z = 4 /(1+ i) Find, in the form a + i b where a, b belong to R, (a) z, (b) z^2. Given that z is a complex root of the quadratic equation x^2 + px + q = 0, where p and q are real integers, (c) find the value of p and the value of q.


Express the complex number (1+i)/(1-i) in the form x+iy


Prove ∑r^3 = 1/4 n^2(n+1)^2


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences