87 75, 68, 61, 54, 47, 40, 33, 26 63 28 51 16 39, 32, 25, 18, 11, 4
The rule is -12 as a common difference for "down"
The rule is -7 as a common difference for "across"
0+1+2+3+4+5+6+7 =28 base 10
28/8 =3 + 4 as a remainder
so, the answer is 34 base 8.
ant101 what is the answer then?
Incorrect, tertre. we know it has to be 5 and up.
No, that's incorrect.
We can set up variables: 7x=5964, 12x=5964/7*12=10224 total votes.
Hmm, I give a take at this.
Okay, thank you!
This is old, but...
(Derived from AoPS' Alcumus, thank you!)
You are very welcome!
:P
Is this supposed to have a pic with it?
Hi Rom,
Thanks, but I hope you do not think it was a request from me.
This question has NOT captured my imagination.
I was just trying to make the asker and other answerers think a little more.
If you are looking at it because it interests you personally then I appologise for butting in :)
Yes, I guess so. Thanks Chris.
Don't you know the words 'thank you'!
It was very nice of guest to go to this trouble for you !
You can spoon feed babies but sometimes they are still not happy.
Are you sure of your equation? Is it + 5 at the end....or should it be (x+5) multiplier?
Read your question! What do n and m stand for?
No votes are 7 out of 12 yes votes are 5 out of 12
7/12 = 5964 votes
total votes = 5964 * 12/7 = 10224 total votes
\(5(2a-5+b)\)
I'll give you a hint.
\(\text{since }f(z) \text{ has the stated property }\forall z\\ f(1) \text{ is equidistant from }(0,0i), \text{ and }(1,0i)\\ \text{The locus of these points is }\left(\dfrac 1 2 , i y\right),~\forall y \in \mathbb{R}\\ \text{From this you can figure out }a\\ \text{Once you have }a \text{ finding }b \text{ is trivial}\)
2.
7(4-x)=2(y-3) ⇒ 28 - 7x = 2y - 6 ⇒ 2y + 7x = 34 (1)
4(x-y)=7-y ⇒ 4x - 4y = 7 - y ⇒ 3y - 4x = - 7 (2)
Multiply (1) by 3 and (2) by -2
6y + 21x = 102
-6y + 8x = 14 add these
29x = 116 divide both sides by 29
x = 4
And using (1) to find y
2y + 7(4) = 34
2y + 28 = 34
2y = 6
y = 3
So (x, y) = (4, 3)
1. 2(x+y)=y+10
4(x+y)=5y+32
Simplify
2x + 2y = y + 10 ⇒ 2x + y = 10 ⇒ y = 10 - 2x (1)
4x + 4y = 5y + 32 ⇒ 4x - y = 32 (2)
Sub (1) into (2) for y
4x - (10 - 2x) = 32
6x - 10 = 32
6x = 42
x = 7
And using (1) y = 10 - 2(7) = -4
So (x, y) = (7, -4)
\(\text{Using Euler's Theorem}\\ 17^{\varphi(102)}\equiv 1 \pmod{102}\\ \varphi(102) = 32\\ 1507 = 32 \cdot 47 + 3\\ 17^{1507} = 17^{32 \cdot 47 + 3} = 17^{32\cdot 47}17^3 \equiv 17^3 \pmod{102}\\ \text{this we can solve by brute force w/o too much trouble}\\ 17^3 = 4913 = 48\cdot 102 + 17 \equiv 17 \pmod{ 102} \)
I think something like this has more of a number theory type solution.
Chinese remainder theorem or something.
I will tinker with it and see if anything pops out.
thank you !
please help
The last answer is correct....
(Derived from AoPS' Alcumus, thank you!)
