Math Is Fun Forum

That's a really nice clock!!!

But not the PC time plzz.........

Hmmm...
Let (100y+10a+b) = M[sup]2[/sup] ; ∀ y ∈ Z[sup]+[/sup] and {a, b} ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

We want (100y+10b+a) to be a perfect square too.
Let (100y+10b+a) = N[sup]2[/sup]

We can see that...
M[sup]2[/sup] - N[sup]2[/sup] = 9(a-b)
or, M[sup]2[/sup] - N[sup]2[/sup] = 9k ; k ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

The question is, in fact, looking for two numbers M[sup]2[/sup] and N[sup]2[/sup] such that their difference is a multiple of 9 i.e. ∈ {0, 9, 18, 27, 36, 45, 54, 63, 81}

M[sup]2[/sup] - N[sup]2[/sup] = 9k
(M+N)(M-N) = 9k
since {M, N} ∈ Z[sup]+[/sup]
∴ (M+N) and (M-N) must be either both Even or both Odd factors of 9k.
Hence, only possible values of 9k are {0, 9, 27, 36, 45, 63, 81}

for (M+N)(M-N) = 9 = 9×1 ⇒ M=5 and N=4

for (M+N)(M-N) = 27 = 9×3 ⇒ M=6 and N=3
for (M+N)(M-N) = 27 = 27×1 ⇒ M=14 and N=13

for (M+N)(M-N) = 36 = 18×2 ⇒ M=10 and N=8

for (M+N)(M-N) = 45 = 9×5 ⇒ M=10 and N=8
for (M+N)(M-N) = 45 = 45×1 ⇒ M=23 and N=22

for (M+N)(M-N) = 63 = 9×7 ⇒ M=8 and N=1
for (M+N)(M-N) = 63 = 63×1 ⇒ M=32 and N=31

for (M+N)(M-N) = 81 = 9×9 ⇒ M=9 and N=0
for (M+N)(M-N) = 81 = 27×3 ⇒ M=15 and N=12
for (M+N)(M-N) = 81 = 81×1 ⇒ M=41 and N=40

for the Special Case of ZHero

(M+N)(M-N) = 0 ⇒ M=N

I referred WikiPedia.

It can be seen that points 2. and 6. are not possible and for the remaining possibilities, only 00 and 44 (since M=N) are the possibilities for the last two digits!!!

Points 4. and 5. have been covered in the M=14 and N=13

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6604 ← cetchup

6604+604+04+4+13 is a Prime number.

6604+604+04+4+13131313 is a Prime number.

6604+604+04+4+31 is a Prime number.

6604+604+04+4+31313131 is a Prime number.

If you get a score of 59.49 (i.e. maximum C grade) in 80% of the subjects, then you need to get at least 9.59 score in the remaining 20% subjects in order to get a C grade ; else you get a D.

If you score 54.50 (i.e. minimum C grade) in 80% of the subjects, then you need to get at least 29.55 score in the remaining 20% subjects to get a C ; or else you get a D again...

That's really......
I mean why would.........
I'm sorry.
I hope you to recover from the shock soon.

I worked upon a Long Hand approach just trying to refute this.

Consider that there are P[sub]1[/sub], P[sub]2[/sub], P[sub]3[/sub], . . . , P[sub]n+2[/sub] points on a circle in that order and n≥2 is even.

I take a fixed point P[sub]a[/sub] and another point P[sub]a+k[/sub].

The idea is that for P[sub]a[/sub] and P[sub]a+1[/sub], no line segment formed with the remaining n points as their ends will intersect P[sub]a[/sub]P[sub]a+1[/sub].
(coz there is no point in minor arc P[sub]a[/sub]P[sub]a+1[/sub])

Similalry, For P[sub]a[/sub] and P[sub]a+2[/sub], (n-1) line segments formed out of the remaining n points will intersect P[sub]a[/sub]P[sub]a+2[/sub]!!
(there is only one point, P[sub]a+1[/sub], in minor arc P[sub]a[/sub]P[sub]a+2[/sub])

For P[sub]a[/sub] and P[sub]a+3[/sub], 2(n-2) line segments formed out of the remaining n points will intersect P[sub]a[/sub]P[sub]a+3[/sub]!!
(there are two points, P[sub]a+1[/sub] and P[sub]a+2[/sub], in minor arc P[sub]a[/sub]P[sub]a+3[/sub])
and so on.......

In each of the cases, the total number of line segments that can be formed out of the remaining n points is n(n-1)/2.

Now, the probability can be found out as..
(total number of intersecting lines for each of the line segments P[sub]a[/sub]P[sub]a+1[/sub], P[sub]a[/sub]P[sub]a+2[/sub], etc) ÷ (total number of line segments that can be drawn from the remaining n points in every case.)

The numerator part is a little twisted !!!!

The probability for all even values of n is

SACK
SOCK
SOUK
SOUR
YOUR
TOUR
TORR
TORE
CORE
CIRE ← (French Origin)
CITE
CITY

Next WAS FOUR ⇒ ZERO ⇒ FIVE #138

NOTE: Plz don't post new words in future until the previous ones have been done or at least reckoned impossible...