Math Is Fun Forum

Look at any term, say 3(7[sup]2[/sup]), and you'll observe that 7 = (2×3+1)

observe... T[sub]1[/sub] = 1(3)[sup]2[/sup] = 1(2×1+1)[sup]2[/sup]

T[sub]2[/sub] = 2(5)[sup]2[/sup] = 2(2×2+1)[sup]2[/sup] etc

We can say that T[sub]n[/sub]= n(2×n+1)[sup]2[/sup]

NOTE: Its true that you'll have to guess it.
Most of the times, they will always be easy (of the form n(an+b)) in the questions given in your textbooks and exams.
Curve fitting is another idea that you just don't need in your school.

That's really tricky but now i got it (after checking the maximum difference possible between two numbers who's Last Two digits are same, may be in different orders)!!

Your last sentence again holds a Lot of Information! Thx!!

I'll try and see if some sort of Algebraic method can help give those numbers

Earlier i was wondering if big perfect square xxxxxxxxxab exists such that xxxxxxxxxba is also perfect square (a,b≠0). Now i see that such a number can't be any greater than 7744

Thanks dude

that's correct!!

but only to a few decimal places!

that's coz you've taken v = 10.8 ∼ 10.8444

you should get Force ∼ 73.441 ∼ 73.5 with v = 10.84

use 10.8[sup]2[/sup] - 0[sup]2[/sup] = 2a[sub]s[/sub]0.08

this should give you a[sub]s[/sub]=734.41 m/s[sup]2[/sup]

can you tell me what exactly did you do?

6357 <--- cetchup

6370

(1+3=4) digit Prime numbers with 6, 3, 7, 0.

3067 is a Prime number.

3607 is a Prime number.

6037 is a Prime number.

6073 is a Prime number.

6703 is a Prime number.

7603 is a Prime number.

Just a small error...
its ∼ 10.8 (m/s)

plz recheck the decimal point..... 

And why of course YES! You can always post your maths/physics problems here (MIF should agree with me here  ).

Somebody will always be here to help!!

Hi soha...

Part-1: To calculate the Velocity of the stone just before it starts penetrating the sandy bed (v).

Velocity of the stone when it just starts penetrating the sandy bed can be calculated using v[sup]2[/sup] - u[sup]2[/sup] = 2a[sub]g[/sub]s

where v = final velocity of the stone (when it just starts penetrating the sandy bed)

u = initial velocity of the stone = 0 m/s (coz nothing's mentioned except that it just falls)

a[sub]g[/sub] = acceleration due to gravity = 9.8 m/s[sup]2[/sup]

s = distance covered = 6 m (distance covered while falling.. given)

Part-2: To calculate the Acceleration due to sandy bed (a[sub]s[/sub]).

Now use the relation v[sup]2[/sup] - u[sup]2[/sup] = 2a[sub]s[/sub]s

here v = final velocity of the stone = 0 m/s (coz it comes to rest after penetrating 8cm)

u = initial velocity of stone (when it just starts penetrating the sandy bed) = the value of v that you will get from Part-1.

a[sub]s[/sub] = acceleration due to sandy bed

s = 8 cm = 0.08 m (distance penetrated.. given)

Part-3: To calculate the Force exerted by the sandy bed (F).

Now use F = M × a[sub]s[/sub]

where F = force exerted by sandy bed (you have to find out)

M = mass of stone = 0.1 kg (given)

a[sub]s[/sub] = acceleration due to sandy bed = value of a[sub]s[/sub] that you got from Part-2.

I'm not gonna solve it fully for you

Hopefully that helps...

Let know if you still have any problems

Hi Tigeree!

that's cool

the tring tring one can be explained but lets leave that part right now

and chuck wud be chafed if he comes to know of you 

Its a nice search engine.

I tried 169 and it says I forgot something

Doesn't work quite good for Large numbers though
e.g. I tried 16553 and it returns the following searches...
Taj ul-Alam
and
Seth Gilliam
etc which don't prove to be any help, whatsoever, to me!?