<p>So I took the SAT in May last year and for the last math grid-in problem I made a program on my calculator that did the problem for me. I couldn't figure it out so I just used my calculator to do it for me and as a result got an 800 on the math. It wasn't a pre-made program, there's no way I could have done that anyway, I made it on the spot and the calculator was a ti-83, something every student can have on the test. I seriously doubt it was cheating and it was, in some ways, harder to do than actually solving the problem.</p>
<p>The Problem
The problem asked for the number of numbers under 53 that gave 3 as a remainder when you divided 53 by those numbers. I had no idea how to do this problem and only had a few minutes left so I turned to my programming skills.</p>
<p>How it worked (Dull read)
I made a for loop that went through all the numbers between 1 and 53 and at each number divided 53 by that number and stored it in a variable. Then I took the fraction part of the number, multiplied it by 53 and the product was 3 printed out the divisor. Then I simply hand checked and counted up the numbers printed.</p>
<p>Background
I get bored in math class pretty often so I spend a lot of time dicking around on my ti-83 calculator. I've learned how to program quite well on it so when I reached the last grid-in problem and couldn't figure it out using classical mathematics and logic so I resorted to boolean logic and mathematics. It was a simple problem from a computer standpoint and the code was less than 10 short lines.</p>
<p>Although this did get me a perfect score I want to make some points clear.
1. This was the last problem and I had extra time
2. I know how to program pretty damn well
3. This method will take longer than basic math 99.9% of the time</p>
<p>I don't know if you'll find this useful but I thought it was a cool trick and I'm actually pretty proud of it.</p>
<p>It’s just 5, 10, 25, and 50; factors of 50, right? I highly doubt that coding on the spot will help without a qwerty keyboard…</p>
<p>That just made me a headache. (sorry, family expression coined by my younger brother when he was 4)</p>
<p>I don’t even care if it’s cheating, I hope you guys get into great schools. Math is not my best class. When MIT was talking to me at a camp this summer I asked my dad what to say to them, and his answer was “I don’t think you have seen my math scores, coach.”</p>
<p>I think what moltenicee was right. I guess you didn’t see it through at the moment.
But anyhow, was the code more or less like this??</p>
<p>int x = 0, numberCount = 0;
while ( x < 53)
{ if( x % 53 == 3)
numberCount++
print
x++
else
x++
}</p>
<p>Just came back from programming class haha. But yeah those kind of problems are what %(modulus) is used for.</p>
<p>@moltenicee
I get your point but I’ve always hated remainders and I couldn’t think of any practical way to solve the problem at the time. Also the stress of the moment was killing me and, since I knew it worked, the calculator method eased my mind.</p>
<p>@calipretty3
There was no modulus on the calculator but there was a function (fPart()) that took the fraction part of a number which I used to find the remainder.</p>
<p>The exact code was this:
0 -> C
FOR(X,1,53)
IF fPart(53/X)*53 = 3
THEN
C+1 -> C
END
END
DISP C</p>
<p>It’s a weird ugly form of basic but I had grown used to typing in it as I had made games like snake and battleship during math class.</p>
<p>An easy way to solve the problem:</p>
<p>We have 53 ≡ 3 (mod n) so we will let 53 = nk + 3, where n > 3 and k is a positive integer (note that n=2 and n=1 technically work but the “remainder” isn’t 3). Therefore nk = 50. The only possible values for n are divisors of 50 greater than 3.</p>
<p>Since 50 = (2^1)(5^2), the number of factors of 50 is (1+1)(1+2) = 6. However we must exclude 1 and 2 since they are less than 3. There are 6-2 = 4 numbers that work.</p>