Video transcript
Tarcísio Gomes de Freitas - Não existe atalho fácil para passar no vestibular. Mas existe um bom ...
29 statements · 1 politicians · July 11, 2026 · 4 min
Statements by Tarcisio. 29 transcribed statements, with topic and stance on the ones the analysis classified.
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29 statementsFull transcript of the video. Statements the analysis classified carry topic and stance.
The governor's challenge, and today's question is from ITA.
And just to remind you, ITA has already opened registrations and has already published the announcement.
The first-phase exam is in September; the second-phase exam is in October.
And the best way to prepare now is to do questions from previous years.
It's practice, practice, practice, practice.
And here is a little math question: let A be the summation from k = 0 to n of the binomial coefficient (n choose k) times 3^k, and let B be the summation from k = 0 to n-1 of (n-1 choose k) times 11^k. If log B minus log A equals log(6561/4), then n equals what? Shall we solve it?
Well, the first thing I need to know is that A equals (n choose 0)·3^0 + (n choose 1)·3^1 + ... + (n choose n)·3^n. And what does that look like?
It looks like a binomial of the form (1 + 3)^n. Could that be true?
Observe: this binomial, (1 + 3)^n, can be expanded as (n choose 0)·1^n·3^0 + (n choose 1)·1^{n-1}·3^1 + ... + (n choose n)·1^0·3^n. Since 1 to any power is 1, this is exactly equal to the sum above, and therefore this equals (1 + 3)^n, which means A = 4^n. An analogous reasoning can be done for B. B will be equal to (n-1 choose 0)·11^0 + (n-1 choose 1)·11^1 + ... + (n-1 choose n-1)·11^{n-1}.
This is equal to (1 + 11)^{n-1}, which is equal to 12^{n-1}.
Having found A and B, I can apply the following property.
Logarithm property.
I know that log B minus log A, assuming they're in the same base, equals log(B/A). And if I know that log(B/A) equals log(6561/4), then B/A = 6561/4.
What is B?
12^(n-1).
What is 4^n?
This will equal 6561/4.
Then I can do some simplifications, right?
I can manipulate this and say, look, what is 12?
Isn't 12 just 3 times 4?
So I can say that this is 3^(n-1) times 4^(n-1).
And here in the denominator, what is 4^n?
I can say it's 4^(n-1) times 4.
This will be equal to 6561/4, which, in fact, is 3^8/4.
I can cancel this 4 with that 4; I can cancel this 4^(n-1) with that 4^(n-1).
What remains?
3^(n-1) will be equal to 3^8.
Therefore n - 1 = 8, and thus n = 9.
That's the answer.
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