How Does Modulus Calculation Aid in Solving Exponential Equations?

[chickenguy]

hi, this is me again.i was wondering if you could teach me how to do these 2 questions (?) means to the power of

1)a)find the remainder when 7(100) (I.E 7 TO THE POWER OF 100) is divided to the power of 25( i know that you are supposed to use mod10 or mod25 or something, but i don't know how to use ,modulus)

b)Find the remainder when 2(603) X 3(201) is divided by 25

c)hence show that 29 X 7(100) + 54 X 2(603) X 3(201) is divisible by 25


2. Two cars left towns A and C at the same time to drive to the other town, passing each other at B and both traveling at different constant speeds, The car from A travels 20km/h faster than the car from C. The distance from A to B is 10km more than the distance from C to B. The car from C completed the journey from B to A in 35 minutes. Find the distance from A to C


so, A-----------------------------B----------------------------C
x + 10 km x km

 

[robert Ihnot]

Looking at 1a). Well, sometimes you have to be a little foxy with these and play around with them. Noting 7^2=49==-1 Mod (25), we are about 90% home already. So we look at 7^100==(-1)^50 Mod 25. So its only a short step now...

 

[VietDao29]

1. b can be done like 1.a. Just notice that 603 = 3 x 201.
1. c Notice that:
*If: $$a_1 \equiv b_1 \mbox{ mod } c$$ and $$a_2 \equiv b_2 \mbox{ mod } c$$
$$\Rightarrow (a_1 + a_2) \equiv (b_1 + b_2) \mbox{ mod } c$$
*If:
$$a_1 \equiv b_1 \mbox{ mod } c$$ and $$a_2 \equiv b_2 \mbox{ mod } c$$
$$\Rightarrow (a_1 a_2) \equiv (b_1 b_2) \mbox{ mod } c$$

For question 2, call car starts from A car 1, and car starts from C car 2.
Car 1, and car 2 travels at constant speed, and car 1 travels 20 km / h faster than car 2. That means for every 1 hour, car 1 can travels more than car 2 : 20 km. And for $$n \mbox{, } n \in \mathbb{R}$$ hour(s) car 1 travels more than car 2 : $$20n$$ (km).
So when car 1 reaches B, it has travel a distance AB, when car 2 reaches B, it has travel a distance CB, and AB = BC + 10 km. They reach B at the same time, car 1 travels more than car 2 : 10 km (using the same amount of time), so can you find how much time they have spent to reach B?
Compare the two amount of time car 2 spends on CB, and BA, and there length AB = 10 + BC, can you find car 2's speed? From there, you can solve for AC.
Viet Dao,

 

[chickenguy]

thanks, but i still don't understanf how to use mod and how it works, also, with question 2, could you eexplain a bit more?? i don't understand where the n comes from etc.

 

[VietDao29]

Let call $v_1$ (km / h) the speed of car 1, $v_2$ (km / h) the speed of car 2. We have $v_1 = v_2 + 20$ So for a specific amount of time t (in hours), car 1 travels: $$d_1 = v_1 t = v_2t + 20t$$, car 2 travels $$d_2 = v_2 t$$ For for every t (h), car 1 travels more than car 2: $$\Delta d = d_1 - d_2 = 20t$$ (km). That's where the t comes from.
Is it clear enough?
Viet Dao,

 

[chickenguy]

sorry for being a persistent annoyance, but could you possibly explain how modulus works(question 1) because i have no idea, and explain question 2 a bit more( i still don't understand

 

[VietDao29]

For the first question : You are asked to find the remainder of 7 ^ 100 divided by 25.
You will try to make it into something that's easy to take the remainder. You notice that:
$7 ^ 2 = 49$, so: $7 ^ 2 \equiv -1 \mbox{ mod } 25$.
Using:
---------
$$a_1 \equiv b_1 \mbox{ mod c and } a_2 \equiv b_2 \mbox{ mod } c \Rightarrow (a_1 a_2) \equiv (b_1 b_2) \mbox{ mod } c$$ (1)
---------
$$a_1 \equiv b_1 \mbox{ mod c and } a_2 \equiv b_2 \mbox{ mod } c \Rightarrow (a_1 + a_2) \equiv (b_1 + b_2) \mbox{ mod } c$$ (2)
---------
So: $$7 ^ {100} = {(7 ^ 2)} ^ {50} = 7 ^ 2 \times 7 ^ 2 \times 7 ^ 2 \times ... \times 7 ^ 2$$ 50 copies of 7 ^ 2.
Using (1), you have:
$$7 ^ {100} \equiv 7 ^ 2 \times 7 ^ 2 \times 7 ^ 2 \times ... \times 7 ^ 2 \equiv (-1)(-1)(-1)...(-1) \equiv (-1) ^ {50} \equiv 1 \mbox{ mod } 25$$ (50 copies of -1).
So $$7 ^ 100 \equiv 1 \mbox { mod } 25$$
Then, do the same for b. Remember $$2 ^ {603} \times 3 ^ {201} = (2 ^ 3 \times 3) ^ {201}$$ and $$2 ^ 3 \times 3 \equiv ? \mbox{ mod } 25$$?
For c: Also use (1), and (2) will give you $$29 \times 7 ^ {100} + 54 \times 2 ^ {603} \times 3 ^ {201} \equiv 0 \mbox{ mod } 25$$ Which means it's divisible by 25.
For Q2, please read my post again. I am showing you that for a specific amount of time t, car 1 travels more than car 2 : 20t (km).
So that means, for every 1 hour car 1 will travel more than car 2 : 20 x 1 = 20 km, for every 2 hours car 1 travels more than car 2 : 20 x 2 = 40 km, for every 2.5 hours car 1 travels more than car 2 : 20 x 2.5 = 50 km.
So when they both reach B, car 1 travel more than car 2 : 10 km. So how long does it take car 2 to travels from C to B, and car 1 to travel from A to B?
From there you will know how long it takes car 2 to travel from C to B, you also know it takes car 2 : 35 minutes (7 / 12 hour) to travel from B to A, and AB = BC + 10, so can you find car 2's speed?
Viet Dao,