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Start by dividing the islanders into 3 groups of 4. Placing two groups results in two possible outcomes. 

(1) If they balance, they are all real. That means one of the remaining is more or less.

Weigh 3 of the regular weighing guys against 3 of the remaining ones. If they balance, the last guy is more or less. 

If it is lighter, one of the ones on the seesaw must be less. Weigh 2 of the suspicious ones. If they balance, then the third guy is less. If they don't balance, the lighter one is the guy we are looking for.

If it is heavier, one of the ones on the seesaw must be more. Weigh 2 of the suspicious ones. If they balance, then the third guy is more. If they don't balance, the heavier one is the guy we are looking for.

(2) If at first they don't balance, the remaining 4 are real. Replace 3 of the ones on the "heavier" side with 3 of the ones on the "lighter" side and add 3 real guys on the "lighter" side. If the same side is still heavier, it means that either the one that wasn't replaced on the "heavier" side is heavier or the one not replaced on the "lighter" side is lighter. Choose either of them, and weigh them against a real guy.

Also, if they balance at the second use, then one of the three "lighter" ones were fake. Weigh two against each other, If they balance, then the third guy is fake. If they don't balance, the lighter one is the guy we are looking for.

 I do not need to use the see saw at all.

I did a visual inspection and noticed that 11 were small children and one was a huge adult.

So I decided that the adult was the odd one out.   :)

Hi guest. Your method assumes that the odd person out is heavier than the others.

This may not be the case. The odd person out might be the lightest one. 

Assume one is heavier than the rest:

1 - First weighing. Put 6 on one side of the see-saw and the other 6 on the other side. Whichever side sinks or goes down, that side has the heaviest person.

2 - Second weighing. Split those into 3 and 3 and whichever side sinks, that side has the heaviest person.

3- Third weighing. Of those 3, weigh 2 of them. if the see-saw sinks, then that is the heaviest person. If the see-saw balances, then the remaining person is the heaviest.

I can't do it with 3 uses of saw, but I can do it with 4 uses of the see-saw. 

There are 12 people. A,B,C,D,E,F G H I J K, and #. Let's assume # is the odd one. Since we do not know if # is heavier or lighter than the others, this makes it a tad bit harder. 

1) Divide the 12 people into two groups. A - F and G - #. We measure one of the groups, 3 on one side and 3 on the other. Let's just say we measure ABC to DEF. If they are equal, that means the odd duck is in the second group. If they are not even, we go to step 2. For our purpose, let's assume Group 2 has the odd one, the #.

2) Choose 4 of the 6 from the group and measure them. Let's just say we measure GH and IJ. If they are equal, then the odd duck is in the group of 2. If they are not equal, then the odd duck is inside the group of 4.

3) a) If the group has 2 people, in our case K and #, we choose K or # and measure it with one of the normal ones. In our case, let's measure K and G. If they are equal, then the non-measured one (#) is the odd duck, but if they are not equal, then the one that was chosen to be measured with the normal one, in our case K, is the odd duck.

3) b) If the group has 4 people, randomly choose 2 people and take them out. Let's make a new scenario, where the group of 4 is I, J, K and #. We randomly choose out K and #. We measure I and J. If they are equal, the odd one is either K or #, which we can find by using step 2 above. If I and J are not equal, then do as above in step 2.

4) More of step 2. The 4th step is only needed in some scenarios. 

- PM

P.S. please send the link to the solution...I want to know a quicker way. 

For this question, I'm going to assume you are asking the number of sheets of paper (sides) in which the old page # (side) and the new page # (side) have the same units digit. 

This problem might seem difficult, but a little bit of magic (algebra and logic) solves it. 

Page 1 is replaced with 50, Page 2 is replaced with 49...ect. page 25 is replaced with page 26. Notice how all the old and new pages add up to 51. 

51 is an odd number, and the only way to get an odd number through addition with two numbers is to have and odd number + even number. So this is algebraically and logically impossibe. For any integer A, A + A = 2A, which is always even. An odd number and an even number cannot have the same units digit, so the answer is \(\boxed{0}\) pages share the same ones digit.

Hope this helps,

- PM 

If you want to check, plugging in 14 for X, we have \(\dfrac{20-14}{30} = \dfrac{20}{30 + 5(14)} \Rightarrow \dfrac{6}{30} = \dfrac{20}{100} = \dfrac{1}{5}\). 

This is true, so the answer is 14.

- PM

Let's call one pack of gum X. 

The ratio of # of cherry to # of grape is \(\dfrac{20}{30} = \dfrac{2}{3}\). Using this, we have the following equality.  

\(\dfrac{20-X}{30} = \dfrac{20}{30 + 5X}\). The ratio to the left is the ratio of gum if chewman loses 1 pack of cherry, and the ratio to the right is the ratio of gum if chewman finds 5 packs of grape. 

Cross multiplying, we have \((20-X)(30+5X) = (20)(30) \Rightarrow (20-X)(30+5X) = 600\). Factoring, we have \(-5x^2 + 70x + 600 = 600 \Rightarrow -5x^2 + 70x = 0 \Rightarrow 5x^2 = 70x \Rightarrow \boxed{x = 14}\).

There are 14 pieces of gum in one pack of gum.

Hope this helps,

- PM 

Factor 343!
2^337×3^169×5^83×7^57×11^33×13^28×17^21×19^18×23^14×29^11×31^11×37^9×41^8×43^7×47^7×53^6×59^5×61^5×67^5×71^4×73^4×79^4×83^4×89^3×97^3×101^3×103^3×107^3×109^3×113^3×127^2×131^2×137^2×139^2×149^2×151^2×157^2×163^2×167^2×173×179×181×191×193×197×199×211×223×227×229×233×239×241×251×257×263×269×271×277×281×283×293×307×311×313×317×331×337 (918 prime factors, 68 distinct). 
It follows from the above that:
[343! / 7^57] is an integer. And:

[343! / 7^57] mod 7 = 6

121, 484, 676

217=216+1 = (36)6 + 1

45 = 42+3 = (7)6 + 3

217 x 45 = (216+1)(42+3) = (216)(42)+(216)(3)+42+3 mod 6 = 3

3 is the units digit of 217d x 45d written in base 6

My solution got deleted, so just to save my time, the answer is 175+6k.

Oh I forgot to add the dimensions

4.5 ft by 7.5 ft by 11.25 feet

I have not thought about it very hard but the HCF of 450, 750 and 1125 is 75

so maybe the answer is 6*9*10=540

Let's use a Stars and Bars approach on this. There are 30 votes(stars) and 4 choices, but to get 4 catergories we only need 3 bars. In order to find where these bars would go, we can do 33C3 which is 5456.

- Daisy

\(\text{I get the following strings} \\ \left( \begin{array}{cccccc} \text{A} & \text{B} & \text{D} & \text{G} & \text{V} & \text{Z} \\ \text{A} & \text{B} & \text{D} & \text{K} & \text{N} & \text{Z} \\ \text{A} & \text{B} & \text{D} & \text{M} & \text{N} & \text{V} \\ \text{A} & \text{B} & \text{G} & \text{H} & \text{K} & \text{Z} \\ \text{A} & \text{B} & \text{G} & \text{H} & \text{M} & \text{V} \\ \text{A} & \text{B} & \text{G} & \text{K} & \text{M} & \text{P} \\ \text{A} & \text{B} & \text{H} & \text{K} & \text{M} & \text{N} \\ \text{A} & \text{D} & \text{G} & \text{H} & \text{K} & \text{M} \\ \end{array} \right)\)

xxxxx

Edited:

The one root is 4  

So the factored quadratic  will look like this:

(2x-8)(2x+?)      the  - 8 and ?   have to multiply to -24     soooo ? = 3

(2x-8)(2x+3)   =   4x^2  -10x  -24        and the two roots are 4 and - 12/8

Thanx , Chris for pointing out that I originally had  the lead coefficient incorrect! 

1. Let the first term be a, and let the common difference be d. Then the four positive integers are a, a+d, a+2d, and a+3d. The sum of these four positive integers is 4a+6d=46, so 2a+3d=23. Solving for d, we find that =(23-2a)/3. The third term is a+2d=a+2(23-2a)/3=(46-a)/3. Thus, to maximize this expression, we minimize a, so a must be 1. Substituting 1 in, we find that the largest possible third term is 15.

2. 94 =F + 26D,  
27 =F + 93D, solve for F, D
F =120 - First term
D = -1 - Common difference

3.  As we begin, we note that no side of the triangle can exceed 29 units in length. Let’s start at the 20–20–20 triangle. We first count the number of triangles that have 2 sides of length less than 20. Let the third side be of length 20+k where 2≤k≤9. This we do by taking away k units in length from the first and the second sides, by subtracting at least one unit from each. For each k, We can do it in ⌊k/2⌋ ways. Thus, counting for each possible k, we have 1+1+2+2+3+3+4+4=201+1+2+2+3+3+4+4=20 cases.

Let’s now go back to the 20–20–20 triangle and count the number of cases where only one side has a length <20<20 units. We can take away 1 to 18 units from this one side and redistribute it to the other two such that no side exceeds 29 units in length. Let us take away k units from this one side, such that 1≤k≤9. In order for the triangles to be distinct, the redistribution can be done in ⌊k/2⌋+1⌊k/2⌋+1 ways. Then, for all possible values of k in this range, we have 1+2+2+3+3+4+4+5+5=29 cases. Now, let us consider the cases where 9
Finally, we add up all our cases:

1+20+29+25=751+20+29+25=75 distinct triangles. Also, subtract one because of the equilateral one.

4. I really can't think of another way other than listing.

The x-intercept refers to when y=0. 

Also, since the x-intercept is -5, then that means one of the legs of the right triangle is 5. 

Using the formula for the area of a triangle, \(\frac{5h}{2}=10 \rightarrow 5h=20 \rightarrow h=4\)

Now you have 2 points, (-5, 0) and (0, 4) (using (0, -4) also works i think)

Find the slope to be \(\boxed{\frac{4}{5}}\)

Work backwards. 

To get five on the last f(x), then x could be -3 for the first f(x). 

Or, on the f(x)=x+3, it would be impossible. 

So you need -3 after the first f(x).

To do that, you could start off with -6, get to -3 and get 5.

Or you could start off with x=1, get to -3, and get 5. 

There are two possible values for x: \(\boxed{\{-6, 1\}}\) .

4,125 = 3 * 5^3 * 11 

Since the smallest m must have 5^3 and 11 as factors, then the smallest m = 15, because:

15! / 4,125 =317,011,968 -- The quotient and m! is multiple of 4125.

Similarly,  281,600 = 2^10 * 5^2 * 11, then the smallest n must have these 3 factors as well. And 12 is the smallest n, because: 12! / 281,600 =1,701 -- The quotient and n! is multiple of 281,600

n - m =12 - 15 = - 3