Show-that-19-93-13-99-is-a-positive-integer-divisible-by-162-

Question Number 13606 by Tinkutara last updated on 21/May/17

Show that 19^{93} - 13^{99} is a positive

integer divisible by 162 .

Answered by mrW1 last updated on 21/May/17

\log (19^{93}) = 93 \times \log 19 \approx 119

\log (13^{99}) = 99 \times \log 13 \approx 110

\Rightarrow 19^{93} > 13^{99}

19^{93} = {(1 + 18)}^{93} = 1 + 93 \times 18 + \frac{93 \times 92}{2!} \times 18^{2} + \frac{93 \times 92 \times 91}{3!} \times 18^{3} + \dots

= A_{0} + A_{1} + A_{2} + A_{3} + \dots + A_{93}

s i n c e 18^{2} = 324 = 2 \times 162

\Rightarrow A_{k ⩾ 2} m o d 162 = 0

A_{0} + A_{1} = 1 + 93 \times 18 = 1675 = 10 \times 162 + 55

\Rightarrow 19^{93} m o d 162 = 55

13^{99} = {(1 + 12)}^{99} = 1 + 99 \times 12 + \frac{99 \times 98}{2!} \times 12^{2} + \frac{99 \times 98 \times 97}{3!} \times 12^{3} + \dots

= B_{0} + B_{1} + B_{3} + B_{4} + \dots + B_{99}

s i n c e 12^{4} = 20736 = 128 \times 162

\Rightarrow B_{k ⩾ 4} m o d 162 = 0

3 \times 12^{3} = 5184 = 32 \times 162

\Rightarrow B_{3} m o d 162 = 0

9 \times 12^{2} = 1296 = 8 \times 162

\Rightarrow B_{2} m o d 162 = 0

B_{0} + B_{1} = 1 + 99 \times 12 = 1189 = 7 \times 162 + 55

\Rightarrow 13^{99} m o d 162 = 55

\Rightarrow (19^{93} - 13^{99}) m o d 162 = 55 - 55 = 0

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 21/May/17

s o b e a u t i f u l .

Commented by RasheedSindhi last updated on 22/May/17

Great!

Commented by Tinkutara last updated on 22/May/17

Thanks but how do you calculated

93 \log 19 and 99 \log 13 ? In the textbook

it was mentioned that use of

calculators is not allowed .

Second, can we use in general this

property ?

{(1 + x)}^{n} (\mod y) = (1 + n x) (\mod y)

I found it from internet from Quora

by typing the same question .

Commented by RasheedSindhi last updated on 22/May/17

Mentioned logs can be calculated

by table of logs instead of calculator .

Commented by Tinkutara last updated on 22/May/17

But it was to be proved without tables .

I have done it without tables as follows :

{(\frac{19}{13})}^{2} = \frac{361}{169} > 2 \Rightarrow {(\frac{19}{13})}^{8} > 2^{4} > 13

\Rightarrow 19^{8} > 13^{9}

\Rightarrow 19^{88} > 13^{99}

\Rightarrow 19^{93} > 13^{99}

The only thing : Can we use

{(1 + x)}^{n} (\mod y) = (1 + n x) (\mod y) ?

What are the conditions for x, y and n ?

Commented by prakash jain last updated on 22/May/17

The first step is not essential for

proof .

Rest of the proof remains valid

even if the first step is omitted .

Commented by prakash jain last updated on 22/May/17

The expression need not be + ve

for calculating \mod .

- 11 \equiv 0 \equiv - 22 (\mod 11)