There are 13 weights. It is known that any 12 of them could be placed in 2 scale cups with 6 weights in each cup in such a way that balance will be held.
Prove the mass of all the weights is the same, if it is known that:
a) the mass of each weight in grams is an integer;
b) the mass of each weight in grams is a rational number;
c) the mass of each weight could be any real (not negative) number.
For each pair of real numbers \(a\) and \(b\), consider the sequence of numbers \(p_n = \lfloor 2 \{an + b\}\rfloor\). Any \(k\) consecutive terms of this sequence will be called a word. Is it true that any ordered set of zeros and ones of length \(k\) is a word of the sequence given by some \(a\) and \(b\) for \(k = 4\); when \(k = 5\)?
Note: \(\lfloor c\rfloor\) is the integer part, \(\{c\}\) is the fractional part of the number \(c\).
Prove that for any natural number \(a_1> 1\) there exists an increasing sequence of natural numbers \(a_1, a_2, a_3, \dots\), for which \(a_1^2+ a_2^2 +\dots+ a_k^2\) is divisible by \(a_1+ a_2+\dots+ a_k\) for all \(k \geq 1\).
Let \(n\) be some positive number. It is obvious that \[2n-1<2n.\] Take another positive number \(a\), and multiply both sides of the inequality by \((-a)\) \[-2na +a< -2na.\] Now, subtracting \((-2na)\) from both sides of the inequality we get \(a<0\).
Thus, ALL POSITIVE NUMBERS ARE NEGATIVE!
Suppose \(a \neq b\). We can write \[-a = b - (a+b)\] and \[-b = a - (a+b)\] Since \((-a)b = a(-b)\), then \[( b - (a+b))b = a(a - (a+b))\] Removing the brackets, we have \[b^2 - (a+b)b = a^2 - a(a+b)\] Adding \(\left(\frac{a+b}{2}\right)^2\) to each member of the equality we may complete the square of the differences of two numbers \[\left(b - \frac{a+b}{2}\right)^2 = \left(a - \frac{a+b}{2}\right)^2\] From the equality of the squares we conclude the equality of the bases \[b - \frac{a+b}{2} = a - \frac{a+b}{2}.\] Adding \(\frac{a+b}{2}\) to both sides of equality we get \(a=b\). Therefore, WE HAVE SHOWED THAT FROM \(a\neq b\) IT FOLLOWS \(a=b\).
Let \(x\) be equal to 1. Then we can write \(x^2=1\), or putting it differently \(x^2 -1 =0\). By using the difference of two squares formula we get \[(x+1)(x-1)=0\] Dividing both sides of the equality by \(x-1\) we obtain \[x+1=0,\] in other words \(x=-1\). But earlier we assumed \(x=1\). THUS \[-1=1\ !\]
In every right-angled triangle the arm is greater than the hypotenuse. Consider a triangle \(ABC\) with right angle at \(C\).
The difference of the squares of the hypothenuse and one of the arms is \(AB^2 -BC^2\). This expression can be represented in the form of a product \[AB^2 -BC^2 = (AB - BC)(AB+BC)\] or \[AB^2 -BC^2 = -(BC - AB)(AB+BC)\] Dividing the right hand sides by the product \(-(AB-BC)(AB+BC)\), we obtain the proportion \[\frac{AB+BC}{-(AB+BC)} = \frac{BC-AB}{AB-BC}.\] Since the positive quantity is greater than the negative one we have \(AB+BC>-(AB+BC)\). But then also \(BC-AB>AB-BC\), and therefore \(2BC>2AB\), or \(BC>AB\), i.e. THE ARM IS GREATER THAN THE HYPOTENUSE!
In how many ways can you rearrange the numbers 1, 2, ..., 100 so the neighbouring numbers differ by not more than 1?
I have three positive integers. When you add them together, you get \(15\). When you multiply the three numbers together, you get \(120\).
What are the three numbers?