**Junior Exam J3**

Each problem is worth 7 marks.

Time: **4 hours**

No books, notes or calculators permitted

**Note**: You must answer with proof.

**Q1**

Let \(a\), \(b\), \(x\), \(y\) and \(z\) be integers such that

\[0 \leq x \leq a \leq y \leq b \leq z\]

\[a + b + x + y + z = 2016\]

*(a)* Find the maximum value of \(x + y + z\). Find all \(x\), \(y\) and \(z\) which satisfy this.

*(b)* Find the minimum value of \(x + y + z\). Find all \(x\), \(y\) and \(z\) which satisfy this.

**Q2**

The polynomial

\[P(x) = x^3 + 3x^2 + ax + b\]

has 3 distinct integer roots which can be arranged to form an arithmetic progression. Find all \(a\) and \(b\) which satisfy this.

**Q3**

Let \(A\) denote the set containing all non-zero real numbers.

*(a)* Find all functions \(f:\mathbb {R} \rightarrow \mathbb {R}\) such that \(x f(xy) + f(-y) = x f(x)\).

*(b)* Find all functions \(f:A \rightarrow \mathbb {R}\) such that \(x f(xy) + f(-y) = x f(x)\).

**Q4**

Assuming \(a\), \(b\), \(c\) and \(d\) are positive reals:

*(a)* Find the value of \(a + b + c + d\) if

\[\dfrac {1}{a} + \dfrac {1}{b} + \dfrac {4}{c} + \dfrac {16}{d} = \dfrac {64}{a + b + c + d}\]

*(b)* Prove that

\[\dfrac {1}{a} + \dfrac {1}{b} + \dfrac {4}{c} + \dfrac {16}{d} \geq \dfrac {64}{a + b + c + d}\]

**Q5**

\(f:\mathbb {N}^{+} \rightarrow \mathbb {R}\) is a function such that

\[f(p) + f(pn) = f(n)\]

where \(n\) is a number and \(p\) is a prime number. It is also known that

\[f(2^{2014}) + f(3^{2015}) + f(5^{2016}) = 2013\].

Find the value of

\[f(2014^2) + f(2015^3) + f(2016^5)\].

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## Comments

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TopNewestQ2

Let the roots are \(k-d,k,k+d\) such that \(d > 0\).

By Vieta's formula, we get \((k-d)+(k)+(k+d) = -3\), which gives \(k = -1\).

Therefore, the roots are \((-1-d, -1, -1+d\).

Also, we get \((-1-d)(-1)(-1+d) = -b\) and \((-1-d)(-1) + (-1)(-1+d) + (-1+d)(-1-d) = a\).

\(1-d^{2} = b\) and \(1+d+1-d+1-d^{2} = a\).

\(1-d^{2} = b\) and \(3-d^{2} = a\).

Therefore, \((a,b) = (3-d^{2},1-d^{2})\) for all positive integers \(d\).

My life is complicated. This is the original solution if I didn't state \(d > 0\).

Therefore, \(a-b = 2\).

Substitute back to equation we get

\(x^{3}+3x^{2}+(b+2)x+b\).

We know that \(-1\) is the root of equation. We can factor \((x+1)\) out of the polynomial.

\((x+1)(x^{2}+2x+b)\).

If \(x^{2}+2x+b = 0\) has 2 other solutions, we get

\(\Delta = 2^{2} - 4\times 1\times b > 0\).

\(b < 1\)

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could u explain how u got a - b = 2?

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Subtract \(3-d^{2} = a\) and \(1-d^{2} = b\).

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Sorry, forgot to state that the roots were distinct.

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I let the roots are \(k-d,k,k+d\) such that \(d > 0\). This will definitely give you distinct roots.

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Doesn't \( b < 1\) come directly from the fact that \( b = 1 - d^2 \), since \( d^2 \) can take any value in the interval \( (0,\infty) \)

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Yeah I can be dumdum with IQ of room temperature at any time. =w=

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5) Choose arbitrary primes \(p_1\) and \(p_2\). Then, \(f(p_1)+f(p_1p_2) = f(p_2)\) and \(f(p_2)+f(p_1p_2) = f(p_1)\). This gives that \(f(p_1) = f(p_2)\) and \(f(p_1p_2) = 0\). Let \(f(p) = c\) for arbitrary prime \(p\).

Let \(n\) be an arbitrary positive composite number with prime factorization \(p_1p_2\cdots p_k\), where the \(p_i\) are not necessarily unique. Then, we propose that \(f(n) = (2-k)c\). We will show this by induction. From the formula given, \(f(n) = f(p_2\cdots p_k)-f(p_1)\). First, note that if \(k = 2\), \(f(n) = 0\) and if \(k = 3\), \(f(n) = -f(p_1) = -c\). Let \(n_i\) have \(i\) non-unique prime factors. Then, if \(f(n_k) = (2-k)c\), we have from our equation that \(f(n_{k+1}) = (2-k)c-c = (2-(k+1))c\). Our induction is complete.

Note that \(2014 = 2\times 19\times 53\), giving \(f(2014^2) = -4c\); \(2015 = 5\times 13\times 31\), giving \(f(2015^3) = -7c\); and \(2016 = 2^5\cdot 3^2\cdot 7\), giving \(f(2016^5) = -38c\). Thus, \(f(2014^2)+f(2015^3)+f(2016^5) = -49c\).

We have \(f(2^{2014})+f(3^{2015})+f(5^{2016}) = -2012c-2013c-2014c = 2013\), which gives \(c = -1/3\). Thus, \(f(2014^2)+f(2015^3)+f(2016^5) = 49/3\).

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What age group is this aimed at?

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12 years to 15 years.

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Question 4: Using Cauchy-Schwarz, the inequality is proven trivially.

Using the equality case for C-S, this implies that a^2 = b^2 = (c^2)/4 = (d^2)/16 implying that a = b = c/2 = d/4. Performing the substitutions gives a + b + c + d = 8.

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4a) does not have a unique answer. The equality is satisfied by any \(a\), \(b\), \(c\), and \(d\) that satisfy \(a = b = d/4\) and \(c = d/2\) as long as \(d \neq 0\), which gives \(a+b+c+d = 2d\), which can obviously take on any positive value within the constraints of the problem.

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2) Let \(P(x) = P(x+c) = P(x+2c) = 0\). Then, expand to find that \(\frac{P(x+2c)-P(x)}{2} = 3cx^2+(6c^2+6c)x+(4c^3+6c^2+ac) = 0\) and \(P(x+c)-P(x) = 3cx^2+(3c^2+6c)x+(c^3+3c^2+ac) = 0\). Then, \([P(x+c)-P(x)]-\left[\frac{P(x+2c)-P(x)}{2}\right] = 3c^2x+(3c^3+3c^2) = 0\), which gives \(x = -1-c\). Thus, the three roots in arithmetic sequence are \(x = -1-c\), \(x+c = -1\), and \(x+2c = -1+c\). From Vieta's formulas, this gives \(a = (-1-c)(-1)+(-1-c)(-1+c)+(-1)(-1+c) = 3-c^2\) and \(b = -(-1-c)(-1)(-1+c) = 1-c^2\) for integer \(c\).

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3a) If we let \(y = 0\), we have \(f(0)+x\cdot f(0) = x\cdot f(x)\), or \(f(x) = \frac{f(0)+x\cdot f(0)}{x}\) for \(x \neq 0\). If we go back to our original equation and also let \(x = 0\), we find that \(f(0) = 0\). Thus, \(f(x) = 0\) for all \(x \in \mathbb{R}\).

3b) Since the value of \(f(0)\) is undefined, we take our previous solution of \(f(x) = \frac{f(0)+x\cdot f(0)}{x}\) and substitute \(c\) for \(f(0)\) to get \(f(x) = c\frac{x+1}{x}\). In 3a), we must consider that this \(f\) does not map \(0\) to a single value in \(\mathbb{R}\) except in the case that \(f(0) = 0\), but as \(0 \notin A\), we avoid that problem here.

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