Question

Fool's problem of the day (Run down memory lane)

After the successful completion of the today's problem, I thought of posting five of the old problems that haven't been (correctly) answered,

1) Prove that \( \tan(20^\circ) \cdot \tan(40^\circ) \cdot \tan(80^\circ)= \tan(60^\circ) \) [Solved by @Mani_Jha]

2) How many eight digit numbers formed using the digit \( \{7,8,9\} \) has exactly three \( 7 \)'s ? [Solved by @PaxPolaris]

3) what is the remainder when we divide \( \large 9^{{100}^{{101}^{102}}}-5^{81^{82^{83}}} \) by \(13\).
[Solved by @KingGeorge] (Note: There is a far more simpler yet analytic way of solving this!) [EDIT: The simpler way is also evinced by @kingGeorge]

4) If \( n \) and \( m \) are natural numbers such that \( 1 + 2 + 3 + … + n = m^2 \), Find the sum of digits of the greatest such \( n \) , smaller than \( 10^3 \).

[PS:This has a very simply non-programming solution]

5) Prove that \( \sum \limits_{k=0}^{n-r} = \binom{n}{k} \times \binom{n}{r+k} = \binom{2n}{n-r} \) [Excellent hint by @satellite73 and solved by @Mani_Jha]

PS: It's safe to assume that none of these are very easy.

Good luck!

Answer 1

3) -11 or 2

Answer 2

FoolForMath, Makes the site more Fun:)

Answer 3

@cinar: I am sorry, it's not.

Answer 4

Thanks @Open2study

Answer 5

Will I ever solve one of your problems FFM?

Answer 6

Let's keep trying ...!

Answer 7

\[9^3=1(\mod13)\]
\[5^4=1(\mod13)\]

Answer 8

2)
number of choosing a 5 digit number with all being 8 or 9 is \(\large 2^5\)
...times number of way to choose 3 positions for the 7s.....\(\large (_8C_3)\)
\[\large2^5\left(\begin{matrix}8 \\ 3\end{matrix}\right)=32\left( 8\cdot7 \right)=\huge1792\]

Answer 9

Oh no! right after I post my cheap rip-off geometry problem!

Answer 10

@TuringTest: You can! I am sure you will :)

Answer 11

4) 8

Answer 12

4) \[\sqrt{\frac{n(n+1)}{2}}=m=\mathbb{N}\]

Answer 13

4) n=49 is too big

Answer 14

2) To find this, we first need to calculate\[\large 100^{101^{102}} \mod \phi(13)\equiv 100^{101^{102}} \mod 12\]For this we need to calculate\[101^{102} \equiv 1^2\equiv 1\mod 4\]Since \(\phi(12)=4\).That means that \[\large 100^{101^{102}} \mod 12\equiv 100^1 \equiv 4 \mod 12\]Finally, we can find\[\Large 9^{100^{101^{102}}} \equiv 9^4 \mod 13\equiv 9\]

Answer 15

Now we have to do the second part of that problem.

Answer 16

To find \[\Large 5^{81^{82^{83}}} \mod 13\]We need to find \[\large {81^{82^{83}}} \mod 12\]So we need to find \[\large 82^{83} \equiv 2^3 \equiv 0\mod 4\]This means that \[81^{82^{83}} \equiv 9^0 \equiv 1 \mod 12\]Finally, \[\Large 5^{81^{82^{83}}} \equiv 5^1\equiv 5 \mod 13\]

Answer 17

4) could be 9
1 and 8

Answer 18

Now that we have both parts, it's simple to see that \[\large 9-5 \equiv 4\mod 13\]

So the answer to 3) should be 4.

Answer 19

4) @Arnab09 is right:
n=8, m=6
sum is 36

Answer 20

@paxpolaris
how about n=1 m=1

Answer 21

largest sum below 1000

Answer 22

http://www.9math.com/book/sum-first-n-natural-numbers

Answer 23

I think the best way to solve #4 is programing

Answer 24

4) 45 not perfect square???

Answer 25

I tried from n=1 to n=50 bit it takes too long to other numbers..

Answer 26

from 1 to 2000 we have 1,2,9,1,2,9,50,289,1682

Answer 27

I think answer will be 9 since the sum is less than 100, next occurance will be at 50 where sum > 1000

Answer 28

@experimentX i think u need to subtract 1 on all nos.

Answer 29

@expermentx
you put n=1000 in your program right?

Answer 30

i put 2000

Answer 31

but why did i get 2 i don't know ..

Answer 32

I am not good at programing :(

Answer 33

@cinar change it into sine and cosines ... i remember it doing.

Answer 34

Second method of doing 3:

We know that \(9^3 \equiv 1 \mod 13\) and that \(5^4\equiv 1\mod13\)So we only need to calculate \[\large 100^{101^{102}} \equiv 1^{2^0} \equiv 1\mod 3 \]and\[81^{82^{83}} \equiv 1^{2^{3}} \equiv 1 \mod 4\]Thus the problem is reduced to \[9^1-5^1 \equiv 9-5\equiv 4\mod 13\]So we're done.

Answer 35

@FoolForMath Was my second solution the simpler way of doing it?

Answer 36

@KingGeorge: You are right.

Answer 37

last one maybe a verbal description after writing \(\binom{2n}{n-r}=\binom{2n}{n+r}\)

Answer 38

i have made some progress with 1

Answer 39

i have shown cos20cos40cos80 = 1/8

Answer 40

sin20sin40sin80 = sqrt(3)/8

Answer 41

has someone already completed 1?

Answer 42

most likely, for problem 4) either (n and n+1/2) must be a perfect square or ( n/2 and n+1) must be a perfect square.

the possibilities are 1 4 9 16 25 36 49 for n, n/2, n+1, n+1/2 are these ... it would simplify our job a lot easier than programming.

let n=1, then (n+1)/2 = 1 => so n= 1, sum = 1 sum^2 = 1
let n/2=1, then (n+1) => 3 => nps
let n+1=1, n/2=0 => nps
let (n+1)/2=1, n=1, n=1, sum^2 = 1

let n=4, then (n+1)/2 => 5/2 => nps
let n/2 = 4, then (n+1) => 9 => sum = 36
let n+1 = 4, then n=5 => nps
let (n+1)/2 = 4, then n=7 => nps

let n=9, then (n+1)/2=5 => nps
let n/2=9, then (n+1) = 19 => nps
let (n+1)/2 = 9, then n=17 => nps
let n+1 = 9, n/2=4 => sum = 36

let n=16, then (n+1)/2=17/2 => nps
let n/2=16, then (n+1) = 33 => nps
let (n+1)/2 = 16, then n=31 => nps
let n+1 = 16, then n/2=15/2 => nps


let n=25, then (n+1)/2=13 => nps
let n/2=25, then (n+1) = 51 => nps
let (n+1)/2 = 25, then n=49 => sum is greater than 1000
let n+1 = 25, then n/2=12 => nps

similarly we can check for 36

the answer would be 8

Answer 43

i am now trying to show that

Answer 44

how did you do it? im stuck :/

Answer 45

@eigenschmeigen i don't think anyone has ... i saw the proof for sin20sin40sin60 somewhere ... i won't be posting answer

Answer 46

proof for the cos:

\[\sin2\theta = 2\sin\theta \cos\theta\]
\[\frac{\sin2\theta}{ 2\sin\theta}= \cos\theta\]

\[cos\theta cos2\theta cos 4\theta = \frac{sin2\theta sin4\theta sin8\theta}{8sin\theta sin2\theta sin4\theta} \]

let\[\theta = 20\]
\[\sin{160} = \sin20\]
so we get:
\[cos20cos40cos80= \frac{1}{8} \]

Answer 47

Problem 4) solution is attached.

Answer 48

so aswer for #4 is 346

Answer 49

i can prove #1 real quick with my calculator :)

Answer 50

@cinar
according to @robtobey , n=288
sum = (1+288)*288/2

@dumbcow
Good 'method', I've never thought of it though :(

Answer 51

http://www.goiit.com/posts/list/trignometry-prove-that-a-tan-20-tan-40-tan-60-tan-80-3-1066775.htm;jsessionid=02F7750B298F2A1B7497F6EC7CF5A45E.node1#1548740 First step for number 1.

Answer 52

Actually, it probably isn't.

Answer 53

http://en.wikipedia.org/wiki/Square_triangular_number
FOR NUMBER 4.

Answer 54

@FoolForMath ?

Answer 55

i dont think the idea was to google the problems :P

Answer 56

and @dumbcow that wouldn't be proving it, as the calculator would have inaccuracy

Answer 57

i agree .. with eigenschmeigen

Answer 58

i am determined to prove the sine part today

Answer 59

it will happen

Answer 60

i will make it so

Answer 61

best of luck ... i saw it on internet ... beware of google.

Answer 62

i was only half-kidding, yes i realize that is not a mathematical proof

Answer 63

\[\tan(60+A)\times tanA \times \tan(60-A)=\tan3A\]
This identity works for the first one. Taking A =20, we write 80=60+20 and 40=60-20
So the result is tan (3*20)=tan60

Answer 64

@Mani_Jha: Congratz! :) Do you know how the prove that identity itself?

Answer 65

I can prove that identity for the sine and cosine functions.(using the sin(A+-B) formulae) Then I can divide to obtain that identity.

Answer 66

For the last problem, the first term represents the sum of binomial coefficients of :
\[(1+x)^{n}=\left(\begin{matrix}n \\ 0\end{matrix}\right)+\left(\begin{matrix}n \\ 1\end{matrix}\right)\]........
The second term represents the sum of the binomial coefficients of a part of the same series:
\[(1+x)^{n}=....\left(\begin{matrix}n \\ n-r\end{matrix}\right)x ^{n-r}+\left(\begin{matrix}n \\n-r-1\end{matrix}\right)\]...........
When we multiply both all we have to do is to collect the coefficients of x^(n-r) in the multiplication of the series: That will be:
\[nC0\times nCn-r+nC1 timesnCn-r-1...........\]
which leads to the Left hand side(the sum of the subscripts should always be n-r)
The problem is simply collecting the coeffecient of x^n-r in the expansion of:
\[(1+x)^{2n}\]

Answer 67

I give up easily, then I just use google. Most of these problems are beyond me anyways. I haven't actually taken any formal math beyond algebra two.