Consider the prime factorization of all the positive integers written on the board.
All integers on the board are of form $p_1^{a_1}p_2^{a_2}\dots$ Write each integer on the board in the form $(p_1^{b_1}\cdot p_2^{b_2}\dots)^n$ where $\gcd(b_1,b_2,\dots)=1$ and label each $(p_1^{b_1}\cdot p_2^{b_...
I don't really understand in the first example how can 86 divide all of Anna's terms and in the second example how did you get 36 in the sequence if 6 is a divisor of all three numbers
Sure. In the second example there are $2$ powers of $2$, $2$ powers of $3$, $2$ powers of $6$, $1$ power of $18$ and $1$ power of $12$ on the board. Specifically, this means that in Anna's list, the highest power of each $2,3,6,\dots$ etc. that acted as divisors for any of the terms in Anna's sequence were $2^2, 3^2, 6^2\dots$, which follows from the first claim. Now the largest divisor is also the largest number in Anna's list, so it follows that $36$ was that largest number by considering all divisors.
The essence of the first claim is that you first write each number on the board in the form of $(p_1^{a_1}p_2^{a_2}\dots)^n$, where $\gcd(a_1,a_2,\dots)=1$. Then each of the distinct "base parts" are put into different set, and we count how many terms on the blackboard there are which have that same base part.
$86$ would be guaranteed to be one of Anna's number, if it's the highest divisor we have. The point of making all these sets and counting their elements is simple-- imagine having a couple of numbers and writing down all their divisors, but when you encounter a divisor that has already been repeated before, you don't write it again
Give me a moment. I'll try to rephrase myself to make it clearer.
Let's go via the first example: $$ 1,49,49,32,64,8,14,28,56,27,216,81,144,324,36,43,86$$. First goal is to write each number as a power of some integer, which itself can't be written as a power of some other integer.
Doing that, we are left with $$1,7^2,7^2,2^5,2^6,2^3,7*2,7*2*2,7*2*2*2,3^3,6^3,3^4,12^2,18^2,6^2,43,86$$
Now, the number of time each base appears tells us a lot: it tells us what the divisors are which are powers of that base. EG: $7$ appears exactly twice (not that there are other terms which are multiples,not powers of $7$), which means that only $7,7^2$ divide any of our numbers. No power of $7$ higher than that can divide any of the terms of our sequence. We don't know which terms, but that's really irrelevant
With this, we can make a list of all divisors of all the terms, which don't include duplicates. From those divisors, we figure out the maximum divisor, which is the greatest term in Anna's sequence.
Ok. Now everything is much more clearer. Thank you so much. This problem is a 10th grade problem from a famous gazette in Romania but people in the US solve problems form it as well
Ooooh. That makes sense. Until you make an account there we’ll be in touch ok this platform:) I also use this platform more than Art of problem solving
Discussion between math.enthusiast9 a…
Discussion between math.enthusiast9 a…