Question

Show that \[{4.6^n} + {5^{n + 1}}\] when divided by \[20\] leaves remainder \[9\].

Answer 1

Hint:We are asked to show that if the number \[{4.6^n} + {5^{n + 1}}\] is divided by \[20\] it leaves \[9\] as a remainder.
To prove this we need to use the method of mathematical induction, first check for \[n = 1\] and then check for the other values and try to show when the number \[{4.6^n} + {5^{n + 1}}\] is divided by \[20\] it leaves \[9\] as a remainder.

Complete step by step solution:
Given the number \[N = {4.6^n} + {5^{n + 1}}\]
We have to show when the number is divided by \[20\] it leaves remainder \[9\]
We know a number can be written as,
\[{\text{dividend}} = {\text{divisor}} \times {\text{quotient}} + {\text{remainder}}\]
\[ \Rightarrow {\text{dividend}} - {\text{remainder}} = {\text{divisor}} \times {\text{quotient}}\]
(i)
Here, \[{\text{dividend}} = N = {4.6^n} + {5^{n + 1}}\], \[{\text{quotient}} = 20\] and \[{\text{remainder}}
= 9\]

Let the divisor be \[\alpha \]
Putting the values of dividend, quotient, divisor and remainder in equation (i), we get
\[{4.6^n} + {5^{n + 1}} - 9 = 20\alpha \]

Therefore the number \[{4.6^n} + {5^{n + 1}} - 9\] is divisible by \[20\] if \[{4.6^n} + {5^{n + 1}}\] when

divided by \[20\] leaves remainder \[9\].

Let \[P(n) = {4.6^n} + {5^{n + 1}} - 9\] (ii)

Now we will use the method of mathematical induction, in which we will first check for \[n = 1\] and then check for \[n = k\] and \[n = k + 1\]

Putting \[n = 1\] in equation (ii), we get
\[P = {4.6^1} + {5^{1 + 1}} - 9\]
\[ = 4 \times {6^1} + {5^2} - 9\]
\[ = 24 + 25 - 9\]
\[ = 49 - 9\]
\[ = 40\]

Therefore for \[n = 1\] the number \[P = {4.6^n} + {5^{n + 1}} - 9\] is divisible by \[20\]
Now we check for \[n = k\], putting \[n = k\] in equation (ii), we get
\[P(k) = {4.6^k} + {5^{k + 1}} - 9\] (iii)

Now we check for \[n = k + 1\], putting \[n = k + 1\] in equation (ii), we get
\[P(k + 1) = {4.6^{k + 1}} + {5^{k + 2}} - 9\] (iv)

Now we subtract equation (iii) from (iv)
\[P(k + 1) - P(k) = \left( {{{4.6}^{k + 1}} + {5^{k + 2}} - 9} \right) - \left( {{{4.6}^k} + {5^{k + 1}} - 9} \right)\]
\[ \Rightarrow P(k + 1) - P(k) = {4.6^k}(6 - 1) + {5^{k + 1}}(5 - 1) - 9 + 9\]

\[ \Rightarrow P(k + 1) - P(k) = 4 \times {6^k} \times 5 + {5^k} \times 5 \times 4\]
\[ \Rightarrow P(k + 1) - P(k) = 20 \times {6^k} + 20 \times {5^k}\]
\[ \Rightarrow P(k + 1) - P(k) = 20\left( {{6^k} + {5^k}} \right)\]

Therefore, from the above equation we observe that \[P(k + 1) - P(k)\] is a multiple of \[20\] from which we can say that \[P(k + 1)\] and \[P(k)\] are also multiples of \[20\] or divisible by \[20\]. The equations (iii) and (ii) are

\[P(k) = {4.6^k} + {5^{k + 1}} - 9\] and \[P(n) = {4.6^n} + {5^{n + 1}} - 9\]
If \[{4.6^k} + {5^{k + 1}} - 9\] is divisible by \[20\] then \[{4.6^n} + {5^{n + 1}} - 9\] is also divisible by
\[20\]. And we can write
\[{4.6^n} + {5^{n + 1}} - 9 = 20\alpha \]

where \[\alpha \] is known as a divisor. Adding \[9\] on both sides on the above equation we get,
\[{4.6^n} + {5^{n + 1}} - 9 + 9 = 20\alpha + 9\]
\[ \Rightarrow {4.6^n} + {5^{n + 1}} = 20\alpha + 9\]

Therefore, comparing the above equation with \[{\text{dividend}} = {\text{divisor}} \times
{\text{quotient}} + {\text{remainder}}\]

We can say if \[{4.6^n} + {5^{n + 1}}\] is divided by \[20\] it leaves \[9\] as remainder.

Note:Mathematical induction is a special way to prove a given statement. It is used to prove that a statement holds true for all the natural numbers that is \[n = 0,1,2,3......\]. For such types of questions where we need to prove a statement for \[{n^{th}}\] term, we can use the method of mathematical induction.