How hard does this test look to you?

[gb7nash]

I did number 2 a little differently. Without much rigor, supposing that f(x) = a_nxⁿ + ... + a₀:

f(0): a₀ = odd
f(1): a_n + ... + a₀ = odd

Case 1: the root (call it b) is even

So a_nbⁿ + ... a₁b is even since b is even, causing arbitrary a_kb^k (k>0) to be even. a₀ is odd, so we have an odd number. -> <-

Case 2: the root is odd

Since for any a_k, if a_k is even, a_kb^k is still even. If a_k is odd, a_kb^k is still odd. So since by (2) , a_nbⁿ + ... a₁b is odd. -> <-

Too much work. Think about the parity of the problem - i.e. what it looks like mod 2. In Z2, if P(1) = 1 (odd), and P(0) = 1 (odd), then P(n) = 1 for all n. Therefore P(n) is never 0. QED.

That's certainly much cleaner than my approach.

 

[Shackleford]

Interesting. I'm curious to know if 100 years ago high schoolers had to take as many courses as now. If they had less courses and could specialize themselves say in mathematics as early as they wanted, this might explain the level of difficulty of their book.

That's a good question. One of my university friends from Germany said in high school they chose two subjects on which to focus. He said he had already covered multi-variable calculus in high school. I certainly did not. I took the entire Cal I-III sequence in college.

 

[Ryker]

Too much work. Think about the parity of the problem - i.e. what it looks like mod 2. In Z2, if P(1) = 1 (odd), and P(0) = 1 (odd), then P(n) = 1 for all n. Therefore P(n) is never 0. QED.

I assume you go to "then P(n) = 1 for all n" by induction, since one of n or n + 1 is 0 and the other 1, right? But why would you be able to generalize this from Z2 to R?

 

[mathwonk]

The difference apparently was the selection of students. 100-120 years ago only about 10% of all students even went to high school, so they were what would be called gifted or at least privileged today. My father had only a high school diploma, graduating in 1906, but read Archimedes, Aristotle, Shakespeare, Dante,... In his father's home they had the complete works of Washington Irving, Charles Dickens, ... I still have some of them.

I also have his, or his brother's, copy of Alexander Dumas' unabridged (1400 pages) Count of Monte Cristo, apparently read when he was a middle school student, inscribed in pencil "this the best book I ever read".

All my childhood I recall he read both the New York Times and the Chicago Tribune every Sunday, to get the major news. And we lived in Nashville, Tennessee. Recall that high school students used to learn Euclid, not the bowdlerized trivia we call geometry today. Get a hold of a copy of Euclid and compare it to the contents of even a good geometry book of today.

The best high school geometry book of today I know of is that of Harold Jacobs and it is nowhere near the quality of Euclid. Euclid was written as if the reader could be assumed to be intelligent, whereas todays books are written as if the reader is a moron, or at best a child of limited attention span.

Take a look at Euler's elementary algebra book, "Elements of algebra", written the publishers says for the benefit of a young servant knowing only arithmetic, and "not above mediocrity" in mathematical ability.

http://web.mat.bham.ac.uk/C.J.Sangwin/euler/ElementsAlgebra.html

This only seems to be the first few parts...

 

[jbunniii]

I assume you go to "then P(n) = 1 for all n" by induction, since one of n or n + 1 is 0 and the other 1, right? But why would you be able to generalize this from Z2 to R?

If P(n) were zero for some n in R, then the same would have to be true mod 2. But that possibility has been excluded.

 

[Begoner]

I thought that the exam was somewhat easier than high-school level math competitions in the US (i.e., AIME-level competitions). I was never much good at those, thought; the times in parentheses indicate how long it took me to arrive at the solution, but without writing a neat proof.

1. I didn't do this using any clever technique; I just substituted (e^ix+e^-ix)/2 for cos(x) and solved the resulting equation. (~10 minutes)

2. One of the conditions tells you that if f(x)=0, then x is odd; the other tells you that x is even. (~3 minutes)

3. The first one can be easily done by L'Hopital's rule; the second can be done by looking at the limit of 1/x as x goes to infinity (you get xe^x) rather than the limit of x as x goes to 0. (~5 minutes)

4. This is obvious; given a particular gap between the numbers, there are only a finite amount of squares which are less than that distance apart. (~1 minute)

5. I probably did it the messy way, but I applied Fermat's little theorem and then did some computations. (~10 minutes)

6. This is just crying out for induction to be used, but I got bogged down in calculations until I saw a simpler way. So, by induction, suppose that the statement holds true up to n. Then for n+1, you have that 1-1/n!+(n+1)/[(n-1)!+n!+(n+1)] should equal 1-1/(n+1)!. The proof becomes easier if you notice that (n-1)!+n!+(n+1)! = (n-1)!(n+1)^2, so that the numerator of the second fraction cancels out. Then just take -1/n! and 1/(n-1)!(n+1) to the same denominator to get the result. (~15 minutes)

7. Ugh, I hate problems involving logs, whether physical or mathematical. I didn't even try this one. (~infinity minutes)

8. Draw a circle of radius 0.5 cm around the center of the circle; at most one point can lie in this region. Hence at least 7 points must lie in the outer annulus. However, this annulus has area 3pi/4. For each point, draw a circle of radius 0.5 cm around it; then 7 such points have area 7pi/4. However, their centers must also lie within the annulus, which is impossible (although this is not easy to prove; there's probably an easier way). (not completed, ~15 minutes)

9. The derivative of the expression is the same as the expression without the first term. Hence, if there is a repeated root, then both the initial expression and its derivative must be 0, so the first term must be 0. But this implies that x = 0, so the entire expression is 1, a contradiction. (~2 minutes)

10. I have no idea. (~infinity minutes)

11. Square the expression to get that sqrt(21)+3sqrt(7)+7sqrt(3) is rational; subtract the original expression from this to get that sqrt(7)+3sqrt(3) is rational. Square this to get that sqrt(21) is rational, which is impossible. (~3 minutes)

12. This can be solved by a system of equations; in drawing the top of the rhombus, you create a smaller isosceles triangle that is similar to the initial one. If you call the bottom of this triangle (i.e., the top of the rhombus) x, and one of its sides a, then you have that a/[sqrt(5)/2] = x/1 and that a+x=sqrt(5)/2. This can be solved to find x. (~10 minutes)

13. First note that (b+i/b)^4 = b^4+4b^2i-4i/b^2-6+1/b^4 = 4i(b^2-1/b^2). Square this to get -16(b^4-2+1/b^4)=-16(4)=-64. Then square this once more to get 4096 as the answer. (~3 minutes)

1. I read jbuniii's answer before trying this.

3. a. This just requires finding out what 10 000 is in base 6.

b. We have 36a+6b+c=81c+9b+a, so b = 5/3(7a-16c). Thus, 7a-16c=3, so b = 5. Trial-and-error yields that a=5 and b =2, so the number is 552. (~15 minutes)

4 a. Suppose a^2+b^2=c^2. If a or b is divisible by 2, then ab/2 is also divisible by 2. If neither a nor b is divisible by 2, then looking at a^2+b^2 mod 4 shows that c^2 is divisible by 2 but not by 4, which is impossible. So the area is divisible by 2. If neither a nor b is divisible by 3, then a^2+b^2 is equal to 2 mod 3; but c^2 cannot be equal to 2 mod 3. So either a or b is divisible by 3 and the area is divisible by 3. So the area is divisible by 6.

b. Drop some perpendicular bisector, l, which divides some side of the triangle into p and q. Then l(p+q) is rational by assumption, and p+q is rational by assumption, so l must be rational. Also, p = sqrt(a^2-l^2) and q = (b^2-l^2) = r-p, where a and b are the other sides of the triangle. So if p is irrational, then sqrt(b^2-l^2)=k-sqrt(a^2-l^2), so b^2-l^2=k^2-2sqrt(a^2-l^2)+a^2-l^2, which implies that sqrt(a^2-l^2) is rational, a contradiction. Hence p and q are both rational, and we have the desired partition. (~15 minutes)

7. This is obvious; if a and b are the common differences in the arithmetic progressions, then a line is given by (a1, b1)+t(a, b). (~1 minute)

 

[jbunniii]

#10 in the first part is quite easy to see geometrically.

An equivalent question is this: given three unit-magnitude complex numbers z1, z2, z3, such that

z1 + z2 + z3 = 3sqrt(3)/2 + 3i/2 = 3e^(i*pi/6),

what are z1, z2, and z3?

But the magnitude of the right-hand side is 3. Therefore z1, z2, and z3 must all have the same angle. And since the angle of the right-hand side is pi/6, the same must be true of z1, z2, and z3 (modulo 2pi).

 

[Ryker]

If P(n) were zero for some n in R, then the same would have to be true mod 2. But that possibility has been excluded.

Oh OK, so just to see if I'm following this. We know that 0 and 1 are just 0 and 1 mod 2, and that the properties of R can be scaled down to Z2 (but not the other way around). So if P(n) = 0 for some n in R, then by this fact it needs to be zero mod 2. And since every member of R can be "modded out" mod 2, we know every n is either 0 or 1 mod 2 (but we cannot say which of the two it is, for example, we can't determine whether the square root of 2 is 0 or 1), P of both of which is different from zero. Is that what you meant or am I not getting this right?

 

[jbunniii]

Oh OK, so just to see if I'm following this. We know that 0 and 1 are just 0 and 1 mod 2, and that the properties of R can be scaled down to Z2 (but not the other way around). So if P(n) = 0 for some n in R, then by this fact it needs to be zero mod 2. And since every member of R can be "modded out" mod 2, we know every n is either 0 or 1 mod 2 (but we cannot say which of the two it is, for example, we can't determine whether the square root of 2 is 0 or 1), P of both of which is different from zero. Is that what you meant or am I not getting this right?

Yes that's right.

To say that a number n is (or, to put it technically, "is congruent to") 0 mod 2 means exactly that n is even. Similarly if n is 1 mod 2 then it's odd.

You can do arithmetic mod 2 and it's consistent with doing the same arithmetic in R and then "scaling it down" to Z2. For example, an even number times any number is even [0 times any number is 0, mod 2]. The sum of two odd numbers is even [1 + 1 is 0, mod 2]. But once you're in Z2 you have lost any information about the original number, except for its even/odd status.

 

[Ryker]

Yeah, I was confused at first because I thought we were going from Z2 to R, not the other way around. Thanks for the clarification, upon which I have to say that this solution is brilliant

 

[doodle_sack]

I can probably answer 2-4 questions but then I'm toast. I'm entering a double major in math and computer science. If I can't do this entrance exam, is it a bad indicator that i'll do bad in math and computer science?

NO! Those questions are problem-solving based (mostly taken by regional mathematical olympiad contestants).

To answer your question, I have heard many many mathematicians say that having the "problem-solving" mindset and skill will help you become a mathematics researcher but it is nevertheless a requirement.

Also, these problems looks foreign to many who have not seen olympiad problems before, they are not as hard as they seem to be (trust me!), but indeed you need above-average mathematical maturity.

 

[Mépris]

Do all of you realize that this is an exam that one has to pass to actually *ENTER* the BS course? It's taken by 17/18 year olds.

 

[qwerty68]

1 very easy:
cos3x+cosx=2cos2x*cosx
<=>
cosx=0 or cos2x=0
=>x

 

[Mépris]

http://web.mat.bham.ac.uk/C.J.Sangwin/euler/ElementsAlgebra.html

This only seems to be the first few parts...

Amazing. I wish I knew of this when I was twelve. I remember our teacher torturing us by having us copy pages from our (bad) textbook into our exercise books. He claimed that it would help us because apparently, the more you progressed in your schooling, the more you'd have to copy. I thought it was bollocks and didn't bother after a while. Not sure how I long I stuck with it though. Ironically enough, those failing to get the "work" done were given lines like "I should always do my homework" to copy a few hundred times! Madness!
Anyway, this looks like an awesome book.

I came back to this thread because some time next month, when I'm done with SATs, I will have to prepare for this maths exam. The syllabus is http://www.cmi.ac.in/admissions/syllabus/ugmath-syllabus.pdf" . They have included a "Suggested Reading" section but I don't think I'll be able to get a hold of these books. I don't know of anyone who's going anywhere near India any time soon (most are by Indian authors). Does anyone know of any good books that will be able to provide me with a rigorous grounding on these topics?

I do know some relatively advanced maths: differential equations, multivariable calc, vectors in 3-d, geometry (co-ordinate, circular measure, etc), some complex numbers, a.p/g.p and a few things here and there. The thing is, I feel I don't know enough of these topics and I would like to learn more. I know how to integrate by parts but I have no clue what's going on as I'm doing it. I have a formula, plug in values into it and voila! I hate doing maths this way and would like to make some sense of things.

I badly need to be able to ace this test because it's the only place I can afford which can assure me a solid education maths and/or physics.

Any suggestions, guys?

 

[EricVT]

As a high-school graduate with no college mathematics experience I would not have been able to do any of those problems. Not a single one.

As a senior college student with 4 years of college mathematics behind me...I could have answered most of those questions if given enough time.

Now that it has been 3 years since I've had a course in mathematics...I feel about like that high school student again.

Rough.

 

[in-a-box]

Amazing. I wish I knew of this when I was twelve. I remember our teacher torturing us by having us copy pages from our (bad) textbook into our exercise books. He claimed that it would help us because apparently, the more you progressed in your schooling, the more you'd have to copy. I thought it was bollocks and didn't bother after a while. Not sure how I long I stuck with it though. Ironically enough, those failing to get the "work" done were given lines like "I should always do my homework" to copy a few hundred times! Madness!
Anyway, this looks like an awesome book.

I came back to this thread because some time next month, when I'm done with SATs, I will have to prepare for this maths exam. The syllabus is http://www.cmi.ac.in/admissions/syllabus/ugmath-syllabus.pdf" . They have included a "Suggested Reading" section but I don't think I'll be able to get a hold of these books. I don't know of anyone who's going anywhere near India any time soon (most are by Indian authors). Does anyone know of any good books that will be able to provide me with a rigorous grounding on these topics?

I do know some relatively advanced maths: differential equations, multivariable calc, vectors in 3-d, geometry (co-ordinate, circular measure, etc), some complex numbers, a.p/g.p and a few things here and there. The thing is, I feel I don't know enough of these topics and I would like to learn more. I know how to integrate by parts but I have no clue what's going on as I'm doing it. I have a formula, plug in values into it and voila! I hate doing maths this way and would like to make some sense of things.

I badly need to be able to ace this test because it's the only place I can afford which can assure me a solid education maths and/or physics.

Any suggestions, guys?

http://www.ncert.nic.in/NCERTS/textbook/textbook.htm

The standard textbooks can be found there for free. They look pretty good, actually

Enjoy!