Math CHALLENGE...Post the hardest math SAT I questions you can find..

Test Preparation
81

<p>Yes that is what I meant. I knew id mess up the square root signs online some how. And btw, -6 doesnt work. Only 5.</p>

82

<p>Oops :p I can't believe I did that...I'm so careless...I'm glad that wasn't a test question lol</p>

83

<p>math on SAT I is just that - SAT I math. Let's keep our noses to the SAT grindstone - not in the high altitudes of IMOuntains.
SAT I questions can get only so hard and only in so many (few) ways.
Not all the Putnam people scored well on SAT I. It tests something different then an intellect.
Learn the strategies and be done with the SAT - there is life beyond. </p>

<p>Get bored - do the Mensa books. Or start a new thread: "Hardest math non-SAT I questions".</p>

84

<p>gcf, barely any of the questions on here are SAT I math questions in any way other than that they don't go beyond algebra and geometry. Any AMC question would be fair game on the SAT I Math...doesn't mean they'd put one on there.</p>

85

<p>OK, knowing the trick for solving
"Find the number of factors of
n = 2^2 <em>3^3 * 4^4 * 5^5 * 6^6 * 7^7"
would give you an edge on this Blue Book question
"Find the number of factors of n = p</em>q*r, where p, q, r are different primes",
but you could do fine without it.
I think, 99% of SAT I hard math questions do not require a knowledge of special math technics or facts, as opposed to AMC questions.
SAT I technics - yes.</p>

86

<p>Answer to that is (8!)/2...</p>

87

<p>(Hyper2400 has lost his sanity); He is singing:</p>

<p>Mia-hiiii
Mia-huuu
Mia-haaa
Mia-haha!!!!!</p>

<p>:D :D :D :D</p>

88

<p>"Any AMC question would be fair game on the SAT I Math...doesn't mean they'd put one on there."</p>

<p>AMC12B takers may recall a bizarre question that involved complex numbers...I don'th think that's SAT I-testable...</p>

89

<p>You're right. But most of them are. I took the AMC10 last year so I didn't see that one.</p>

90

<p>First, I just want to agree with towerpumpkin on his answer in #86 and with Hyper on his choice of a war song for the Wholly 2.4 grand quest.
++++++++++++++++++++++++
How many different equations XXX + XXX = 46X exist,
if each X can be any digit between 1 an 9,
and each digit in the equation appears only once?</p>

91

<p>Just a clarification:
123 + 456 = 579 is the same as
456 + 123 = 579</p>

92

<p>Random Guess: 900</p>

93

<p>Damn . . . "each digit appears only once"</p>

<p>Is it possible, only 4 possibilities?</p>

94

<p>173 + 295 = 468
175 + 293 = 468
193 + 275 = 468
195 + 273 = 468</p>

<p>Here's how to whittle down the possibilities:</p>

<p>Let's label ABC + DEF = 46G
In order to get the 4, A + D = 3 or A + D = 4
Assume without loss of generality that A < D
So, if A + D = 4, we have A = 1 and D = 3
1BC + 3EF = 46G
BUT, with the same logic, B + E = 5 or B + E = 6, and 1, 3, and 4 are taken, so no pair of B and E exists.
Therefore, A + D = 3, which means A = 1 and D = 2
1BC + 2EF = 46G
B + E = 15 or B + E = 16; 1, 2, 4, 6 taken
Assume B < E (we can switch them later), so if B + E = 15, B = 7 and E = 8
That means 3, 5, and 9 are left, but no pair of those can generate the third in the ones column.
So, B + E = 16 and B = 7 and E = 9 (for now)
17C + 29F = G
3, 5, 8 left, and 3 + 5 = 8, so C = 3, F = 5, G = 8.</p>

<p>Finally, switch the tens and ones columns to get all the combinations...</p>

<p>IMO, too long for an SAT question...</p>

95

<p>I just realized that there were nine digits in the entire equation, so each digit must appear once. Only four combinations yielded such results.</p>

96

<p>Right answer!</p>

<h1>As it's often a case with an SAT, hard quesrion has a shortcut, which makes the question admissable.</h1>

<p>Lets label digits of the first number a1, a2, a3,
and digits of the second number b1, b2, b3.</p>

<p>Using a cool tanonev expression, without loss of generality assume that a1<b1, a2<b2, a3<c3.</p>

<p>a1<em>a2</em>a3 +</p>

<h2>b1<em>b2</em>b3</h2>

<p>4<strong>6</strong>X</p>

<p>To get 4 in 46X, a1 + b1 could be only 1+2 ot 1+3, which means that "1" is taken.
To get 6 in 46X we can't have 1+5 or 2+4 ("1" and "4" are used), which leaves us with 7+9.
Thus, a1 + b1 = 1+2, and now we have only 3 ,5 and 8 left:
173 +</p>

<h2>295</h2>

<p>468</p>

<p>To find all possible solutions, we should switch 1 and 2, 7 and 9, 3 and 5 independently:
two possibilities for hundreds, two for tens, and two for units,
2<em>2</em>2=8.
Each solution has a symmetrical twin (see calrification in #91) which we should not count, therefore, the final answer is
8/2 = 4.</p>

<p>The question did not require to find all possible solutions - just their number.</p>

97

<p>
[quote]
Math CHALLENGE...Post the hardest math SAT I questions you can find.. </p>

<hr>

<p>i could use the practice ....let's see who can solve it the most efficiently...we can all learn from these efficient methods...

[/quote]
</p>

<p>2 + 2 = _ Fill in the blank</p>

98

<p>Well.... Here's a very difficult question from SAT 2020: </p>

<ol>
<li>S is a set of positive integers containing 1 and 2002. No elements are larger than 2002. For every n in S, the arithmetic mean of the other elements of S is an integer. What is the largest possible number of elements of S?</li>
</ol>

<p>Nothing beyond Precalculus Math...</p>

99

<p>When I said "right answer!", I addressed tanonev (post 94).</p>

<p>Assuming that just one "root" solution exists, there would be 4 combinations, you are correct.
(173 + 295 = 468 is the rooot foor three other combinations).
What if there is more than one "root", or no solutions?</p>

100

<p>(a1 + a2 + a(i-1) + a(i+1) + an) mod (n - 1) = 0 for all i on [1, n]</p>

<p>so, a1 mod (n - 1) = a2 mod (n - 1) = a3 mod (n - 1) = ... = an mod (n - 1)</p>

<p>1 mod (n - 1) = 2002 mod (n - 1)
(n - 1) = some factor of 2001
The largest factor of 2001 is, well, 2001
so, n - 1 = 2001 <=> n = 2002</p>