Question

$\forall nϵN,P\left(n\right):{2.7}^n+{3.5}^n-5$ is divisible by

• A. 64
• B. 676
• C. 17
• D. 24

Answer 1

The correct option is D

24

$P\left(n\right):{2.7}^n+{3.5}^n-5P\left(1\right):=14+15-5=24$

Suppose 24 divides $P\left(n\right):{2.7}^n+{3.5}^n-5$
P(1) is true.
Suppose P(k) is true.
${2.7}^k+{3.5}^k-5$ is divisible by 24.
$\therefore {2.7}^k+{3.5}^k-5=24m{2.7}^{k+1}+{3.5}^{k+1}-5=7.\left({2.7}^k\right)+5.\left({3.5}^k\right)-5=5({2.7}^k+{3.5}^k-5)+{\mathrm{2.2.7}}^k+20=5({2.7}^k+{3.5}^k-5)+{4.7}^k+20=5\left(24m\right)+{4.7}^k+20$

Now, prove that $P_1\left(n\right):{4.7}^n+20isdivisibleby24$
$P_1\left(1\right):28+20=48\to divisibleby24P_1\left(k\right):{4.7}^k+20isdivisibleby24\to assumetrue\Rightarrow {4.7}^k+20=24qNow,{4.7}^{k+1}+20=7.\left({4.7}^k\right)+20={\mathrm{6.4.7}}^k+({4.7}^k+20)={24.7}^k+24q{P}_1(k+1)istrue$

$\therefore {4.7}^n+20isdivisibleby24$

$\therefore 5\left(24m\right)+{4.7}^k+20$ is divisible by 24
P(k) is true $\Rightarrow$ P(k+1) is true
Hence, P(n) is true.