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USA Math Olympiad

Number of ordered triples – Duke Math Meet 2009 Team Round Problem 7 solution

How many ordered triples of integers (a, b, c) are there such that 1 \le a, b, c \le 70 and a^2 + b^2 + c^2 is divisible by 28.

Any square quantity is 0 or 1 modulo 4

Now sum of three square quantities are 0 or 1 mod 4. Since 28 is 0 mod 4, there fore none of the three squares can be 1 mod 4.

Thus a, b, c are even.

Again a square quantity is always 0, 1, 4, 2 mod 7

Since we want the three squares to add up to 0 mod 7 (28 is 0 mod 7), the possibilities are (0, 0, 0) mod 7 and (1, 4, 2) mod 7

So we have following cases:

Case 1

a, b, c are all even multiples of 7. There are 5 of them. Since we are looking for ordered triplet there are 5^3 = 125 of these$

Case 2

a, b, c are all even numbers. with 1, 6 mod 7, 2 , 5 mod 7 and 3, 4 mod 7

1 mod 7 – 10 elements
6 mod 7 – 10 elements

Hence there are 20 numbers which are 1 or 6 mod 7. 10 of these are even.

Similarly there are 20 numbers which are 2 or 5 mod 7 and 20 numbers which are 3 or 4 mod 7. Half of each of which we will take.

So 10 choices for each set. 10^3 = 1000 choices. Now since we are taking ordered pairs, we must consider all the 3! = 6 permutations. So there are 6000 cases.

Hence answer is 6000 + 125 = 6125

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USA Math Olympiad

Remainder Problem – Duke Math Meet 2009 Team Round Problem 9 solution

What is the remainder when 5^{5^{5^5}} is divided by 13 ?

By Fermat’s Little Theorem

5^{12} = 1 \mod 13

Now if we can find out 5^{5^5} \mod 12 we can find the answer.

By Euler’s theorem 5^{\phi (12) } = 5^4 = 1 mod 12

Finally if we can find 5^5 mod 4 we are done.

Since 5 \equiv 1 mod 4 \Rightarrow 5^5 \equiv 1 mod 4 \implies 5^5 = 4Q + 1

Thus 5^{5^5} = 5^{4Q +1} = 5^{4Q} \times 5 \equiv 1 \times 5 mod 12 (as we have previously computed 5^4 \equiv 1 mod 12 \Rightarrow 5^{4Q} \equiv 1 mod 12 )

Thus 5^{5^5} = 12Q' + 5 \Rightarrow 5^{5^{5^5}} = 5^{12Q' + 5 } = 5^{12Q'} \times 5^5 \equiv 5^5 mod 13 . (since we have previously computed 5^{12} \equiv 1 \mod 13 \implies 5^{12Q'} \equiv 1 mod 13 )

Thus 5^{5^{5^5}} \equiv 5^5 mod 13 . But 5^5 = 3125 \equiv 5 mod 13 . Thus answer is 5.

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USA Math Olympiad

Duke Math Meet 2008 Problem 8 Solution (Individual Round)

Find the last two digits of \sum_{k=1}^{2008} k {{2008}\choose{k}}

Discussion:

(1+x)^n = \sum_{k=0}^{n} {{n}\choose{k}}x^k

We differentiate both sides to have n(1+x)^{n-1} = \sum_{k=1}^{n} k {{n}\choose{k}}x^{k-1}

Put n=2008 and x=1in the above equation to have

2008(1+1)^{2007} = \sum_{k=1}^{2008} k {{2008}\choose{k}}1^{k-1}

Thus \sum_{k=1}^{2008} k {{2008}\choose{k}} = 2^{2007} \times 2008 \equiv 8 \times 2^{2007} \equiv 2^{2010} mod 100

Now 2^{12} = 4096 \equiv -4 mod 100 . Now raising both sides to the power of 6, we have 2^{72} \equiv (-4)^6 \equiv 2^{12} \equiv -4 mod 100 . Again raising both sides to the power 6 we have 2^{432} \equiv -4 \Rightarrow 2^{1728} \equiv 2^8 \equiv 56 mod 100.

Earlier we had 2^{72} \equiv -4 \Rightarrow 2^{216} \equiv -64 \equiv 36 mod 100

Hence 2^{1728} \times 2^{216} \equiv 2^{1944} \equiv 56 \times 36 \equiv 16 mod 100

Finally we know 2^{12} \equiv -4 \Rightarrow 2^{60} \equiv -1024 \equiv 76 mod 100 . Hence 2^{1944} \times 2^{60} \equiv 2^{2004} \equiv 16 \times 76 \equiv 1216 \equiv 16 mod 100

Thus 2^{2004} \times 2^{6} \equiv 2^{2010} \equiv 16 \times 64 \equiv 1024 \equiv 24 mod 100.

The two digits are 24.

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Physics Olympiad USA Math Olympiad

Magnetic Field at the Centre of a Ring

A ring of radius \(R\) carries a linear charge density \(\lambda\). It is rotating with angular speed \(\omega\). What is the magnetic field at the centre?

Discussion:

Linear charge density $$ \lambda=\frac{Q}{2\pi R}
$$
When the ring is rotated about the axis, the motion of the electrons in a circular orbit is equivalent to a current carrying loop.
Current $$ I=\frac{Q}{T}=\frac{Q\omega}{2\pi}$$
since Time period \(T=2\pi/\omega\).
Now, magnetic field around the centre of a current carrying loop is given by $$ B=\mu_0I/2R$$
Putting the value of \(I\) in the above equation, we get
$$ B=\frac{\mu_0\omega}{2}.\frac{Q}{2\pi R}
$$$$ \Rightarrow B=\frac{\mu_0\lambda\omega}{2}
$$

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Physics Olympiad USA Math Olympiad

Specific Heat of a Rigid Triangular Molecule

A rigid triangular molecule consists of three non-collinear atoms joined by rigid rods. The constant pressure molar specific heat \(C_p\) of an ideal gas consisting of such molecules is

(a) \(6R\)

(b) \(5R\)

(c) \(4R\)

(d) \(3R\)

Degrees of freedom are the number of independent parameters that define its configuration
If \(N\) be the number of particles in a system and \(k\) be the number of constraints between the number of degrees of freedom is given by $$ f=3N-k$$ $$f=(3*3)-3$$ $$=6$$
Relation between \(f\) and \(C_p\) $$ C_p=(f/2+1)R$$ $$ \Rightarrow C_p=(6/2+1)R$$ $$C_p=4R$$

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Physics Olympiad USA Math Olympiad

Work Done on Compression of Gas

A cylinder contains \(16g\) of \(O_2\). The work done when the gas is compressed to \(75\%\) of the original volume at constant temperature of \(27^\circ\) is ________.

Discussion:

Given mass of \(O_2\), m=\(16g\)

Number of moles of \(O_2\), $$n=\frac{m}{M}$$ where \(M\)=molecular weight of \(O_2\)=\(32g\)
$$n=\frac{16}{32}=\frac{1}{2}$$

Temperature \(T=27^\circ=300K\)

If \(V_1\) be the original volume and \(V_2\) be the final volume

Work done by the gas in the isothermal process $$ W_0=nRTlog(V_2/V_1)$$$$=0.5*8.31*ln(3/4)$$$$=150*8.31*ln(3/4)$$$$=-358.56J$$

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Opportunity USA Math Olympiad

UCLA full scholarship program – MUMS

Cheenta Opportunity is an initiative for the benefit of Cheenta Olympiad candidates. We dig up opportunities and resources available all around the world for our students. 


University of  California, Los Angles is one of the leading universities in the world. Its mathematics department is held in great esteem by all who matter. UCLA Math department offers a Merit-based scholarship to students with an exceptional background in mathematics.

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Physics Olympiad USA Math Olympiad

A Problem on Doppler Effect

A train passes through a station with constant speed. A stationary observer at the station platform measures the tone of the train whistle as \(484Hz\) when it approaches the station and \(442Hz\) when it leaves the station. If the sound velocity is \(330m/s\), then the tone of the whistle and the speed of the train are
(a) \(462hz, 54km/h\)
(b) \(463Hz, 52Km/h\)
(c) \(463Hz, 56Km/h\)
(d) \(464Hz, 52Knm/h\)

Solution:

When train approaches the station, the frequency heard by the observer
$$ n_1=n\frac{v}{v-v_s}=n(\frac{330}{330-v_s})$$
Here, $$ v=330m/s$$
n is the actual frequency of the whistle
$$ 484 =n(330/330-v_s)$$….. (i)
When the train leaves the station $$ n_2=n\frac{v}{v+v_s}=n(\frac{330}{330+v_s}) $$
$$ 442=n(\frac{330}{330+v_s})$$…. (ii)
Divide Eqs (i) by (ii), we get
$$ \frac{484}{442}=330+v_s/330-v_s$$
$$ 1.09=(330+v_s)/(330-v_s)$$
$$ 330+v_s=1.09(330-v_s)$$
$$v_s=\frac{31.35}{2.09}$$$$=15m/s$$
Substituting \(v_s\) in Eqn (i) gives $$ 484=n(330/330-15)$$ $$=n(330/315)$$ $$n=\frac{484*21}{22}$$
$$=462Hz$$

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College Mathematics Corner USA Math Olympiad

Supremum of function (TIFR 2014 problem 13)

Question:

Let \(S\) be the set of all tuples \((x,y)\) with \(x,y\) non-negative real numbers satisfying \(x+y=2n\) ,for a fixed \(n\in\mathbb{N}\). Then the supremum value of \(x^2y^2(x^2+y^2)\) on the set \(S\) is:

A. \(3n^6\)

B. \(2n^6\)

C. \(4n^6\)

D. \(n^6\)

Discussion:

Write the expression in terms of \(x\) only by substituting \(y=2n-x\).

Let \(f(x)=x^2(2n-x)^2(x^2+(2n-x)^2)\). Here, \(x\in[0,2n]\).

Note that for \(x=0\) or \(x=2n\) the function \(f(x)=0\). Also, \(f\) is positive everywhere else on the interval.

So we want to find \(sup\{f(x)|x\in (0,2n)\}\). Note that it exists because the interval is compact and \(f\) is continuous.

One can straightaway take derivative and compute, or one can do the following:

Take log. Note that now we are only working on the open interval \((0,2n)\).

\(log(f(x))=2logx+2log(2n-x)+log(x^2+(2n-x)^2)\)

Now take derivative.

\(\frac{f'(x)}{f(x)}=\frac{2}{x}+\frac{-2}{2n-x}+\frac{2x-2(2n-x)}{x^2+(2n-x)^2}\)

\(=\frac{4n-4x}{x(2n-x)}+\frac{4n-4x}{x^2+(2n-x)^2}\)

\(=4(n-x)[\frac{1}{x(2n-x)}+\frac{1}{x^2+(2n-x)^2}]\).

Now, \(f'(x)=0\) if and only if \(4(n-x)[\frac{1}{x(2n-x)}+\frac{1}{x^2+(2n-x)^2}]=0\).

Note that \([\frac{1}{x(2n-x)}+\frac{1}{x^2+(2n-x)^2}]>0\) for all \(x\in (0,2n)\).

Therefore, \(f'(x)=0\) if and only if \(x=n\). If now, \(x\) is slightly bigger than \(n\) then \(\frac{f'(x)}{f(x)}<0\) and since \(f(x)>0\) we have \(f'(x)<0\) in that case. And if \(x\) is slightly smaller than \(n\) then \(f'(x)>0\).

This proves that indeed the point \(x=n\) is a point of maxima.

Therefore, the supremum value is \(f(n)=2n^6\). So the correct answer is option B.

 

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College Mathematics Corner USA Math Olympiad

Mapping properties (TIFR 2014 problem 12)

Question:

There exists a map \(f:\mathbb{Z} \to \mathbb{Q}\) such that \(f\)

A. is bijective and increasing

B. is onto and decreasing

C. is bijective and satisfies \(f(n)\ge 0\) if \(n\le 0\)

D. has uncountable image.

Discussion:

Option D is out of the question right away. Because the co-domain \(\mathbb{Q}\) is countable, and the image set is a subset of the co-domain set, any subset of countable set is countable, so image set has to be countable.

Let’s assume \(f\) is onto and decreasing. Then \(…\ge f(-1) \ge f(0) \ge f(1) \ge f(2) \ge … \).

At this point, we don’t know for sure whether \(f(0)=f(1)\) or not. But, for the map to be onto, we must have strict inequality somewhere in the above chain of inequalities.

Let’s say \(f(a)>f(a+1)\). Then we ask what is the pre-image of the rational number \(\frac{f(a)+f(a+1)}{2}\)?

Let \(f(n)=\frac{f(a)+f(a+1)}{2}\)

Since, \(f(n)=\frac{f(a)+f(a+1)}{2}>f(a)\) and \(f\) is decreasing, \(a>n\).

Also, since \(f(n)=\frac{f(a)+f(a+1)}{2}<f(a+1)\) and \(f\) is decreasing, \(n>a+1\).

The two inequalities \(a>n\) and \(n>a+1\) together give a contradiction.

Therefore, option B is false.

What did we use here, only onto-ness and that f is decreasing. If instead our function \(f\) was bijective and increasing then we will get a similar kind of contradiction:

Suppose \(f\) is bijective and increasing. Then \(f(0)<f(1)\) (One-to-one ness guarantees the strict inequality)

We have \(\frac{f(0)+f(1)}{2}\in\mathbb{Q}\).

Therefore there exists \(m\in\mathbb{Z}\) such that \( f(m)= \frac{f(0)+f(1)}{2} \).

We then have \(f(0)<f(m)<f(1)\) which means \(0<m<1\), a contradiction because there is no natural number in between 0 and 1.

So option A is false.

We know that \(\mathbb{Q}\) does have a bijection with \(\mathbb{Z}\), in other words \(\mathbb{Q}\) is countable.

Now, both the set \(A=\{n\in \mathbb{Z}| n\le 0 \}\) and \(B=\{q\in \mathbb{Q}| q\ge 0\}\) are countably infinite, therefore has a bijection between them. To see this, let \(f_1:A\to \mathbb{Z}\) and \(f_2:B\to \mathbb{Z}\) be bijections. Then \(f_2^{-1}o f_1:A \to B\) is a bijection.

Similarly \(C=\{n\in \mathbb{Z}| n> 0 \}\) has a bijection with \(D=\{q\in \mathbb{Q}| q< 0\}\).

Now, we have \(h:A\to B\) and \(g:C\to D\) two bijections.

Then define \(f(n)=h(n)\) for \(n\in A\) and \(f(n)=g(n)\) for \(n\in C\) to get a bijection from \(\mathbb{Z}\) to \(\mathbb{Q}\). This is true because \(A \cup B= \mathbb{Z}\) and \(C \cup D= \mathbb{Q}\).

And this \(f\) satisfies the condition of option C: \(f\) is bijective and satisfies \(f(n)\ge 0\) if \(n\le 0\)

Therefore, option C is true.