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0

One answer can be of the form $ma+nb+kab$. So solving for the above three equations one gets $f(a,b) = 2.907692a + 3.138462b - 1.092308ab$. Putting this relation for $[(-3)*1]*(-5)$ one gets the answer $-35.0059$. Since there are three relations for this $*$ operator by adding a known function of $g(a,b)$ we can create a function of the form $ ma+nb+k.g(a,...


0

The last number in $n$-th row is $n^2$. It's justified since $1+3+\ldots+(2n-1)=n^2$ Hence, the middle number in $n$-th row is $(n-1)^2+n=n^2-n+1$. The $k$-th number in $n$-th row is $(n-1)^2+k$ where $1\leq k \leq 2n-1$.


3

Good luck finding any linearity in that - I can't see anything. For visualization:


2

Your approach is quite reasonable. You can improve it by replacing the manual count with the recognition that there are $2x-1$ numbers in row $x$, so $x-1$ to the left of the median and $x-1$ to the right. You can use this to figure out the change from the median for a given location. A simpler approach is to note that row $x$ has $x^2$ as its last number,...


2

When trying to spot a pattern in a problem like this, it’s often helpful not to reduce the fractions. In this case every numerator is initially $3$, and the fractions are: $$\begin{align*} U_1&=\frac33\\ U_2&=\frac3{15}\\ U_3&=\frac3{35}\\ U_4&=\frac3{63}\\ U_5&=\frac3{99}\;. \end{align*}$$ Since all of the numerators are $3$, we can ...


1

Lets say there are $n$ students numbered $1,2,3...,n$. The lockers which are opened an odd number of times will remain open till the end. The student $i$ interacts with locker $j$ if $i|j$. Thus lockers having odd number of divisors will remain open till the end. Since only perfect squares have odd number of divisors. The number of open lockers is $[\sqrt n]$...


2

Let $S_n=\frac{3n}{2n+1}$. Then $U_n=S_n-S_{n-1}$ This is because $$\sum_{r=1}^{n} U_r - \sum_{r=1}^{n-1} U_r=$$ $$\left(U_1+U_2+\ldots+U_{n-1}+U_n\right)-\left(U_1+U_2+\ldots+U_{n-1}\right)=U_n$$ and as you said, $U_1=1$


1

$$x_1=50$$ $$x_{n+1}=n\left(\frac{x_n}{2}+5\right)$$ My python code to verify it.


1

Let $p_n$ be the $n$'th prime, and let $y_n = p_{n + p_{n}} \mod p_n$, i.e. $p_{n+p_{n}} = x_n p_n + y_n$ where $0 \le y_n < p_n$ and $x_n = \lfloor p_{n+p_n}/p_n \rfloor$. Now the point is that $r_n = p_{n+p_n}/p_n$ will tend to grow, but slowly: $p_n \sim n \log n$, $p_{n+p_n} \sim (n + p_n) \log(n + p_n) \sim n (\log n)^2$, so $r_n \sim \log n$. $...


0

As far as your second question goes, I don't believe prime numbers are randomly distributed. For example, for primes greater than 2, a prime number can only be odd. Further, since every 3rd odd number is divisible by 3, the only potential primes are of the form $6k±1$, for $k\in\mathbb N$ (for primes greater than 3). Perhaps prime numbers are randomly ...


1

The answer is $D$. In 1-3, solid squares go small, small, big in the bottom left. In 2-6, empty squares repeat the pattern by going small, small, big (top left instead). In 1-3, triangles at the top go empty right, full left, full left. In 2-6 the pattern is repeated (at the bottom instead): empty right, full left, full left. So square 6 has a big empty ...


0

I would say D because of the symmetry of the situation and because the first square has colours different --> the corresponding 3rd square has equal colours, the 2nd has equal colours --> corresponding 4th square has different colors, therefore the missing square must have different colours.


1

Don't just take differences, take iterated differences! The differences of the cubes are 7, 19, 37, 61, 91, ..., the differences of these numbers are 12, 18, 24, 30, ..., and the differences of these numbers are all 6. (This is still true if you allow 0 as perfect cube.)


1

$$(n+1)^3 - n^3 = 3n(n+1) +1$$ For example, $$10^3 - 9^3 = 3\cdot 9 \cdot 10 + 1 = 271 \\ 1000 - 729 = 271 $$ That pattern looks a little like the pattern for squares...


0

Have you looked at the problem algebraically? $(a*10^2 + b*10 + c) + (c*10^2 + b*10 + a) = ???$ Then consider what conditions must be placed on a,b,c so that the sum is self-similar. On the flip side, what conditions must be true for a self-similar number to be a reversed sum? Can 101 be the result of a reversed sum? Why is it impossible?



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