Intermediate Competition Maths thread

[alchemy]

Question attached.

[brightsky]

Let l be length of long and s be length of short. l - 4l/7 = s - 4s/10.
3l/7 = 3s/5
s/l = 5/7

[alchemy]

Find the last two digits of 2²²².

I know it's got something to modulo 100 in the end, as that should return the last two digits but I'm sure there must be some preliminary steps before doing that.

[Orb]

Find the last two digits of 2²²².

I know it's got something to modulo 100 in the end, as that should return the last two digits but I'm sure there must be some preliminary steps before doing that.

Well you don't technically have to use modulo if you're unfamiliar with the concept.

I'll try expressing this in a student-friendly way:

Let's express everything starting from 2^1 till we see a pattern. Try to focus on the LAST TWO digits as stated by question!

2^1=2
2^2=4
2^3=8
2^4=16
2^5=32
2^6=64
2^7=28 (notice how i ignored the 1, it's unnecessary because the last two digits won't be affected by anything in the hundreds' or further)
2^8=28x2=56 (you can just ignore the 2)
2^9=12
2^10=24

and so the pattern goes 48,96,92,84,68,36,72,44,88,76,52,04 (REPEATS HERE)

So it repeats once every 20 times from 2^2 onwards.

Hence 2^222= 2^20x11+2= last two digits of 04


P.S This is essentially the mod concept, except rephrased into understandable terms for people who are unfamiliar!

 

[brightsky]

Most arguments of this irk are a little artificial. We observe that 2^2 = 4 mod 100. If we add multiples of 20 to the power, the modulo stays the same. Now, 222 = 2 + 20*11. This means that 2^222 = 4 mod 100. So the last two digits are 04.

[jammin]

Most arguments of this irk are a little artificial. We observe that 2^2 = 4 mod 100. If we add multiples of 20 to the power, the modulo stays the same. Now, 222 = 2 + 20*11. This means that 2^222 = 4 mod 100. So the last two digits are 04.

How would you know without prior knowledge that by adding multiples of 20 that the power would stay the same?

[alchemy]

I'm bored
Question is attached.
Thanks (:

[Conic]

From the diagram:











The area of the square is , and the area of the circle is , so the ratio is

[Orb]

I'm bored
Question is attached.
Thanks (:

bored so you decided to do AMC questions? wowwww nice man

when i'm bored I play games >.<

[alchemy]

bored so you decided to do AMC questions? wowwww nice man

when i'm bored I play games >.<

.XD I was feeling mathsy

[alchemy]

If x and y are positive integers which satisfy x^2-8x-1001y^2=0, what's the lowest possible value of x+y.

Also, would I be able to use Vieta's somehow to solve this one?

[brightsky]

I didn't manage to incorporate Vieta's theorem in my solution, but...

x^2 - 8x - 1001y^2 = 0

Let x + y = k for the sake of convenience. Then x = k - y. Sub this into the equation above.

(k - y)^2 - 8(k - y) - 1001y^2 = 0

Now solve for k.

k^2 - 2yk + y^2 - 8k + 8y - 1001y^2 = 0
k^2 - (2y + 8)k + (8y - 1000y^2) = 0
k = (y + 4) +- sqrt[(y+4)^2 - (8y - 1000y^2)], where +- represents the 'plus/minus' sign
k = y + 4 +- sqrt(1001y^2 + 16)

Now, since x and y are both positive integers, it follows that k = x + y > 0. Let us first investigate the negative square root:

y + 4 - sqrt(1001y^2 + 16) > 0
0 < y < 1/125

This is no good, because there are no positive integers between 0 and 1/125. Hence, we need now only to consider the positive square root, which is always greater than 0.

We require the smallest value for k. To obtain the smallest value for k, we need 1001y^2 + 16 to be as small as possible. But since it is under a square root, and k is a positive integer, we require 10001y^2 + 16 to be a perfect square as well. We start from y = 1 (the smallest possible value for y) and continue to test increasing values of y until we find one that works.

If y = 1, k = 3sqrt(113)+5, which is not an integer.
If y = 2, k = 2sqrt(1005) + 6, which is not an integer.
If y = 3, k = 102, which *is* an integer!

Hence, the smallest possible value of x + y is 102.