Algebra Exam Paper

Junior Exam J3

Each problem is worth 7 marks.

Time: 4 hours

No books, notes or calculators permitted

Note: You must answer with proof.


Let aa, bb, xx, yy and zz be integers such that

0xaybz0 \leq x \leq a \leq y \leq b \leq z

a+b+x+y+z=2016a + b + x + y + z = 2016

(a) Find the maximum value of x+y+zx + y + z. Find all xx, yy and zz which satisfy this.

(b) Find the minimum value of x+y+zx + y + z. Find all xx, yy and zz which satisfy this.


The polynomial

P(x)=x3+3x2+ax+bP(x) = x^3 + 3x^2 + ax + b

has 3 distinct integer roots which can be arranged to form an arithmetic progression. Find all aa and bb which satisfy this.


Let AA denote the set containing all non-zero real numbers.

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

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


Assuming aa, bb, cc and dd are positive reals:

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

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

(b) Prove that

1a+1b+4c+16d64a+b+c+d\dfrac {1}{a} + \dfrac {1}{b} + \dfrac {4}{c} + \dfrac {16}{d} \geq \dfrac {64}{a + b + c + d}


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

f(p)+f(pn)=f(n)f(p) + f(pn) = f(n)

where nn is a number and pp is a prime number. It is also known that

f(22014)+f(32015)+f(52016)=2013f(2^{2014}) + f(3^{2015}) + f(5^{2016}) = 2013.

Find the value of

f(20142)+f(20153)+f(20165)f(2014^2) + f(2015^3) + f(2016^5).

Note by Sharky Kesa
6 years ago

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Let the roots are kd,k,k+dk-d,k,k+d such that d>0d > 0.

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

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

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

1d2=b1-d^{2} = b and 1+d+1d+1d2=a1+d+1-d+1-d^{2} = a.

1d2=b1-d^{2} = b and 3d2=a3-d^{2} = a.

Therefore, (a,b)=(3d2,1d2)(a,b) = (3-d^{2},1-d^{2}) for all positive integers dd.

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

Therefore, ab=2a-b = 2.

Substitute back to equation we get


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


If x2+2x+b=0x^{2}+2x+b = 0 has 2 other solutions, we get

Δ=224×1×b>0\Delta = 2^{2} - 4\times 1\times b > 0.

b<1b < 1

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

Nihar Mahajan - 6 years ago

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Subtract 3d2=a3-d^{2} = a and 1d2=b1-d^{2} = b.

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@Samuraiwarm Tsunayoshi thanks!!

Nihar Mahajan - 6 years ago

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Sorry, forgot to state that the roots were distinct.

Sharky Kesa - 6 years ago

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

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Doesn't b<1 b < 1 come directly from the fact that b=1d2 b = 1 - d^2 , since d2 d^2 can take any value in the interval (0,) (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 p1p_1 and p2p_2. Then, f(p1)+f(p1p2)=f(p2)f(p_1)+f(p_1p_2) = f(p_2) and f(p2)+f(p1p2)=f(p1)f(p_2)+f(p_1p_2) = f(p_1). This gives that f(p1)=f(p2)f(p_1) = f(p_2) and f(p1p2)=0f(p_1p_2) = 0. Let f(p)=cf(p) = c for arbitrary prime pp.

Let nn be an arbitrary positive composite number with prime factorization p1p2pkp_1p_2\cdots p_k, where the pip_i are not necessarily unique. Then, we propose that f(n)=(2k)cf(n) = (2-k)c. We will show this by induction. From the formula given, f(n)=f(p2pk)f(p1)f(n) = f(p_2\cdots p_k)-f(p_1). First, note that if k=2k = 2, f(n)=0f(n) = 0 and if k=3k = 3, f(n)=f(p1)=cf(n) = -f(p_1) = -c. Let nin_i have ii non-unique prime factors. Then, if f(nk)=(2k)cf(n_k) = (2-k)c, we have from our equation that f(nk+1)=(2k)cc=(2(k+1))cf(n_{k+1}) = (2-k)c-c = (2-(k+1))c. Our induction is complete.

Note that 2014=2×19×532014 = 2\times 19\times 53, giving f(20142)=4cf(2014^2) = -4c; 2015=5×13×312015 = 5\times 13\times 31, giving f(20153)=7cf(2015^3) = -7c; and 2016=253272016 = 2^5\cdot 3^2\cdot 7, giving f(20165)=38cf(2016^5) = -38c. Thus, f(20142)+f(20153)+f(20165)=49cf(2014^2)+f(2015^3)+f(2016^5) = -49c.

We have f(22014)+f(32015)+f(52016)=2012c2013c2014c=2013f(2^{2014})+f(3^{2015})+f(5^{2016}) = -2012c-2013c-2014c = 2013, which gives c=1/3c = -1/3. Thus, f(20142)+f(20153)+f(20165)=49/3f(2014^2)+f(2015^3)+f(2016^5) = 49/3.

Michael Lee - 6 years ago

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What age group is this aimed at?

Curtis Clement - 6 years ago

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12 years to 15 years.

Sharky Kesa - 6 years ago

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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.

John Ashley Capellan - 6 years ago

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4a) does not have a unique answer. The equality is satisfied by any aa, bb, cc, and dd that satisfy a=b=d/4a = b = d/4 and c=d/2c = d/2 as long as d0d \neq 0, which gives a+b+c+d=2da+b+c+d = 2d, which can obviously take on any positive value within the constraints of the problem.

Michael Lee - 6 years ago

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2) Let P(x)=P(x+c)=P(x+2c)=0P(x) = P(x+c) = P(x+2c) = 0. Then, expand to find that P(x+2c)P(x)2=3cx2+(6c2+6c)x+(4c3+6c2+ac)=0\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)=3cx2+(3c2+6c)x+(c3+3c2+ac)=0P(x+c)-P(x) = 3cx^2+(3c^2+6c)x+(c^3+3c^2+ac) = 0. Then, [P(x+c)P(x)][P(x+2c)P(x)2]=3c2x+(3c3+3c2)=0[P(x+c)-P(x)]-\left[\frac{P(x+2c)-P(x)}{2}\right] = 3c^2x+(3c^3+3c^2) = 0, which gives x=1cx = -1-c. Thus, the three roots in arithmetic sequence are x=1cx = -1-c, x+c=1x+c = -1, and x+2c=1+cx+2c = -1+c. From Vieta's formulas, this gives a=(1c)(1)+(1c)(1+c)+(1)(1+c)=3c2a = (-1-c)(-1)+(-1-c)(-1+c)+(-1)(-1+c) = 3-c^2 and b=(1c)(1)(1+c)=1c2b = -(-1-c)(-1)(-1+c) = 1-c^2 for integer cc.

Michael Lee - 6 years ago

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3a) If we let y=0y = 0, we have f(0)+xf(0)=xf(x)f(0)+x\cdot f(0) = x\cdot f(x), or f(x)=f(0)+xf(0)xf(x) = \frac{f(0)+x\cdot f(0)}{x} for x0x \neq 0. If we go back to our original equation and also let x=0x = 0, we find that f(0)=0f(0) = 0. Thus, f(x)=0f(x) = 0 for all xRx \in \mathbb{R}.

3b) Since the value of f(0)f(0) is undefined, we take our previous solution of f(x)=f(0)+xf(0)xf(x) = \frac{f(0)+x\cdot f(0)}{x} and substitute cc for f(0)f(0) to get f(x)=cx+1xf(x) = c\frac{x+1}{x}. In 3a), we must consider that this ff does not map 00 to a single value in R\mathbb{R} except in the case that f(0)=0f(0) = 0, but as 0A0 \notin A, we avoid that problem here.

Michael Lee - 6 years ago

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