Because 5 students collect both

=> There are: 14 - 5 = 9 students collect stamps.

                       16 - 5 = 11 students collect postcards.

So in the class there are: \(9+11+5+4=29\) students.

Because x,y are positive integers and \(x\times y=8\)

=> \(\left\{{}\begin{matrix}x=1;y=8\\x=2;y=4\\x=4;y=2\\x=8;y=1\end{matrix}\right.\)

Because \(x^y=y^x\)

=> \(\left\{{}\begin{matrix}x=2;y=4\\x=4;y=2\end{matrix}\right.\)

Because x,y are positive integers and \(x\times y=8\)

=> \(\left(x,y\right)=\left(1,8\right),\left(2,4\right),\left(4,2\right),\left(8,1\right)\)

Because \(x^y=y^x\) 

=> \(\left(x,y\right)=\left(2,4\right)and\left(4,2\right)\)

Form the four numbers a,b,c,d. We have: \(a,b,c,d\in N\) 

=> \(a+b+c+d+1>0+1>1\) (because a,b,c,d are consecutive number so their sum isn't 0)

=> PROVED

sorry, I'd made a mistake. This question is already answered.

Sorry Dao Trong Luan for the wrong title. Can you do it again?

\(\left|36^m-5^n\right|\).

If it's the last title, we have m = 1 and n big [unsolved]

AUTO ANSWER:
Let’s try to count all the possibilities in an organized manner. Alexander can get 12 of one type of cookie in 3 ways. He can get 11 of one type and 1 of another type in 3 ways. If there are 12 glazed, then there is only 1 way to complete the order with chocolate and cherry. If there are 11 glazed, then there are 2 ways (1 chocolate or 0 chocolate). If there are 10 glazed, then there are 3 ways. And so on, until if there are 0 glazed, then there are 13 ways (anywhere from 12 to 0 chocolate). So the answer is 1 + 2 + … + 13 = 13 * 14 / 2 = 91 assortments. Alternative solution: Let’s, instead, use the counting technique known as “stars and bars.” The idea is that we arrange 12 stars to represent the cookies and two bars to separate the cookies into the three different categories. For example, the arrangement ****|**|****** represents the possibility that Alexander buys 4 of the first type of cookie, 2 of the second type, and 6 of the third type. Our question now becomes: How many ways can we arrange the 12 stars and 2 bars? Thus, the number of assortments of a dozen cookies he can buy is 14C2 = 14!/(12! × 2!) = (14 × 13)/(2 × 1) = 7 × 13 = 91 assortments.

AUTO ANSWER:

The sum of any three consecutive integers is divisible by 3. Since 206 is not, we know it is either the sum of the first, second and fourth terms or of the first, third and fourth terms. If we call the first term in Pamela’s sequence x, then the four consecutive terms are x, x + 1, x + 2 and x + 3. Based on the two options above, we can have either x + x + 1 + x + 3 = 206 → 3x = 206 – 4 → 3x = 202 or x + x + 2 + x + 3 = 206 → 3x = 206 – 5 → 3x = 201. The latter is the only one that yields an integer value for x. Solving, we find x = 67. The other integer is x + 1 = 68. 

AUTO ANSWER:

Since Annette can pick 4 baskets in one hour, she must have picked 2 baskets in the half hour the three girls worked together. Similarly, Mary must have picked 2.5 baskets in the half hour. That means Lynn picked the remaining 6 – 2 – 2.5 = 1.5 baskets in the half hour. By herself, Lynn must be able to pick 2 × 1.5 = 3 baskets per hour, so she can pick 3 × 3 = 9 baskets in 3 hours. 

Can you explain why?

You can change your avatar by clicking in your last avatar then take a photo/picture that you .

We have: \(\dfrac{\dfrac{1}{2}-\dfrac{1}{3}}{\dfrac{1}{3}-\dfrac{1}{4}}\times\dfrac{\dfrac{1}{4}-\dfrac{1}{5}}{\dfrac{1}{5}-\dfrac{1}{6}}\times...\times\dfrac{\dfrac{1}{2006}-\dfrac{1}{2007}}{\dfrac{1}{2007}-\dfrac{1}{2008}}=\dfrac{\dfrac{1}{2\cdot3}}{\dfrac{1}{3\cdot4}}\times\dfrac{\dfrac{1}{4\cdot5}}{\dfrac{1}{5\cdot6}}\times...\times\dfrac{\dfrac{1}{2006\cdot2007}}{\dfrac{1}{2007\cdot2008}}\)

\(=\dfrac{3\cdot4}{2\cdot3}\times\dfrac{5\cdot6}{4\cdot5}\times...\times\dfrac{2007\cdot2008}{2006\cdot2007}=\dfrac{2008}{2}=1004\)

There are: \(900.25\%=225\left(students\right)\)are weak students

So there are: \(900-225=675\left(students\right)\) are good students

We have: \(a^2+b^2+c^2\ge ab+bc+ca\)

[\(a^2+b^2+c^2\ge ab+bc+ca\Leftrightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+ac+ca\right)\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2\)

\(\ge0\) (proved)]

=> \(a^2+b^2+c^2+2ab+2bc+2ca\ge ab+bc+ca\)

=> \(\left(a+b+c\right)^2\ge3\left(ab+ac+ca\right)\)

=> \(\left(a+b+c\right)^3\ge3\left(ab+ac+ca\right)\) (proved)

The title must be: a + b = 2

(1st condition): a \(\le\) 0, b \(\ge\) 0

=> \(a^4\ge0;b^4\ge0\)

We have: a + b = 2 with \(a^4;b^4\in Z\)+

=> \(a^4+b^4\ge a+b\ge2\) (proved)

(2nd condition): a \(\ge\) 0; b \(\le\) 2: Do exactly as the 1st condition (proved)

So ............

The answer is very simple: I'm born in 2005, so I'm 13 years old :)

You can change your avatar in your profile. This page has that function to do that. If there're any problems, go to https://www.facebook.com/mathu and send a message to us. 

The area of the square is: \(38\cdot38=38^2=1444\left(m^2\right)\)

AUTO ANSWER (AFTER SEVERAL DAYS): 

This is not a hard question, but many can't solve it. Because they thought: "The question is wrong!". Exactly, 99 = 72 + 27; 45 = 27 + 18; 39 = 18 + 21; So the ?? must be 15. But why does 21 = 13 + 7? We have gone into the wrong MINDWAY.  Let's think of another way. We have 9 + 9 + 7 + 2 = 27 (Sum of the digits in line 1, row 1 + sum of the digits in line 2, row 1 = line 2, row 2). Because 2 + 8 + 1 + 2 = 13 and 3 + 6 + 2 + 1 = 12 so the answer correct is 12. 

(#560 - 1000 best IQ questions in history)

(Solution 1)We have: \(\sqrt{1+1\cdot2\cdot3\cdot4}=1+1\cdot4=5\)

\(\sqrt{1+2\cdot3\cdot4\cdot5}=1+2\cdot5=11\)

.........

\(\sqrt{1+204\cdot205\cdot206\cdot207}=1+204\cdot207=42229\)

(Solution 2) We have: \(\sqrt{1+1\cdot2\cdot3\cdot4}=2\cdot3-1=5\)

\(\sqrt{1+2\cdot3\cdot4\cdot5}=3\cdot4-1=11\)

..............

\(\sqrt{1+204\cdot205\cdot206\cdot207}=205\cdot206-1=42229\)

Alone has done solutions 3+4.