GAUSS

[Phaeacian]

I can't figure out how to do the following problems...

Prove that [54 x (4^6n)] + [(11^n+1) x 6^n] is divisible by 65, though I know it has something to do with modular arithmetic

Also, I can't figure out if the answer to this next question is 0.7, as I don't know whether they walk at constant speeds.Please tell me if you can get another solution without assuming that they following walk at constant speeds.

A and B decide to visit each other and set off at the same time to each others houses. This distance between their houses in 7 km. They met on the road joining their houses, and forgot that they wanted to see each other. After they met, it took B nine times as long to reach A's house as it took A to reach B's house. What is the distance between B's house and where he met A?

I would be most grateful if anyone could provide me with any hints or solutions.

 

[TchrWill]

A & B walking

If their walking speeds were the same, they would meet half way between the two houses and B's time from where they met to A's house could not be 9 times A's time from where they met to B's house.

Let Va = A's walking speed and Vb = B's walking speed.
Let the distance from A's huse to where they met be "x".
T1A= time for A to meet B = x/Va.
T1B = time for B to meet A = (7 - x)/Vb.
Since t1A = T1B, Va/Vb = x/(7 - x).
T2A = time for A to reach B's house from where they met = (7 - x)/Va.
T2B = time for B to reach B's house from where they met = x/Vb.
T2B = 9(7 - x)/Va = x/Vb making Va/Vb = (63 - 9x)/x.
Therefore, (63 - 9x)/x = x/(7 - x).
Multiply out to get a quadratic eqaution which you can solve for x by means of the quadratic formula.

 

[Denis]

A...@ a.......................................... X ..............@ b...B : AB = 17
A's speed = a, B's speed = b, X is where they meet; let meeting time = k

Since speeds are irrelevant, let a = XB (A travels XB in 1 hour):
then B travels XA in 9 hours; let m = 9.

17 = k(a + b) : AB = k hours at combined speed
17 = a + bm : AB = XB + XA = a + bm
Therefore:
a + bm = k(a + b) ; simplify to get:
a = (bm - bk) / (k - 1) [1]

But a = bk [2] : XB = a and XB = bk

[1][2]: bk = (bm - bk) / (k - 1) ; simplify to get:
m = k^2

So, as general case, meeting time = sqrt(m) : m is a given
in this case, k = sqrt(9) = 3
Substitute k=3 back in above equations to get:
XA = 51/4 and XB = 17/4 : 51/4 + 17/4 = 17

 

[Denis]

An observation only on your 1st one:
[54 x (4^6n)] + [(11^n+1) x 6^n]

It's evident that this divides 65 if n = 0:
54 * 4^0 + 11^1 * 6^0
= 54 * 1 + 11 * 1
= 54 + 11
= 65

 

[soroban]

Hello, Phaeacian!

Prove that (54 x 4⁶ⁿ) + (11ⁿ⁺¹ x 6ⁿ) is divisible by 65.

I know it has something to do with modular arithmetic

I hope you're familiar with Modular Arithmetic . . .


The first term is: . 54 x (4³)²ⁿ .= .54 x (64)²ⁿ

. . Since 64 = -1 (mod 65), the first term becomes: . 54 x (-1)²ⁿ .= .54 (mod 65)


The second term is: . 11·11ⁿ x 6ⁿ .= .11 x (11·6)ⁿ .= .11 x (66ⁿ)

. . Since 66 = 1 (mod 65), the second term becomes: . 11 x (1)ⁿ .= .11 (mod 65)


Hence, the expression becomes: . 54 + 11 .= .65 .= .0 (mod 65)


Therefore, the expression is divisible by 65.