**Junior Exam J1**

Each question is worth 7 marks.

Time: **4 hours**

No books, notes or calculators allowed.

**Note**: You must prove your answer.

**Q1**

The sequence

\[1, 10, 19, 28, 37, \ldots\]

is defined by the rule that a term is the average of its neighbours (excluding the first term).

*(a)* Prove that \(10^{1000}\) is a term in the sequence.

*(b)* Find the number of times the digit \(5\) occurs in the sum of all the terms in the sequence from \(1\) to \(10^{1000}\).

**Q2**

\(P(n)\) is a function defined as the product of all the factors of \(n\). e.g. \(P(10) = 1 \cdot 2 \cdot 5 \cdot 10\).

*(a)* Find all \(n\) such that \(P(n) = 15n\).

*(b)* Find all \(n\) such that \(P(n) = 15n^2\).

**Q3**

Find all positive integral values of \(a\), \(b\) and \(c\) such that

\[a + b = c^2\]

\[a^2 + b^2 = c^3\]

**Q4**

Find all primes \(p\) and \(q\) such that

\[p^{q + 1} + q^{p + 1}\]

is a perfect square. Also state the perfect square.

**Q5**

Sets \(A\) and \(B\) contain positive integers such that the sum of any 2 elements in set \(A\) are in set \(B\) and the quotient (larger element divided by the smaller element) of any 2 elements in set \(B\) are in set \(A\).

Find the maximum number of elements in \(A \cup B\).

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## Comments

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TopNewestQ2

Let \(d_{i}\) be factors of \(n\) such that \(1 = d_{1} < d_{i} < d_{\sigma_{0}(n)} = n\) for all \(1 < i < \sigma_{0}(n)\). (\(\sigma_{0}(n)\) is the number of divisors \(n\))

The function \(P(n)\) becomes

\(P(n) = \prod\limits_{i=1}^{\sigma_{0}(n)} d_{i}\)

If \(n\) is even, we get

\(\displaystyle P(n) = d_{1}d_{\sigma_{0}(n)}\times d_{2}d_{\sigma_{0}(n)-1}\times\dots\times d_{\sigma_{0}(n)/2}d_{ \sigma_{0}(n)/2 +1} = n^{\Large \frac{\sigma_{0}(n)}{2}}\)

If \(n\) is odd, we get

\(\displaystyle P(n) = d_{1}d_{\sigma_{0}(n)}\times d_{2}d_{\sigma_{0}(n)-1}\times\dots\times d_{(\sigma_{0}(n)+1)/2 -1}d_{(\sigma_{0}(n)+1)/2 +1}\times (d_{(\sigma_{0}(n)+1)/2}^{2})^{1/2} = n^{\Large \frac{\sigma_{0}(n)-1}{2}}\times n^{\Large \frac{1}{2}} = n^{\Large \frac{\sigma_{0}(n)}{2}}\)

For these 2 cases, we get \(P(n) = n^{\Large \frac{\sigma_{0}(n)}{2}}\).

(a): \(P(n) = 15n\)

\(n^{\Large \frac{\sigma_{0}(n)}{2}} = 15n\)

\(n^{\Large \frac{\sigma_{0}(n)-2}{2}} = 15\).

\(n^{\sigma_{0}(n)-2} = 225 = 3^{2}\times5^{2}\).

Take \(\log\) base \(n\) on both sides we get

\(\sigma_{0}(n)-2 = 2\times\log_{n}(15)\)

Since \(\sigma_{0}(n)-2\) is an integer, \(2\times\log_{n}(15)\) is also integer.

Which means \(n = 15\) or \(n = 225\). Check the answers and we get \(\boxed{n = 15}\) is the only solution. ~~~!

(b): \(P(n) = 15n^{2}\)

Similar to (a); we get \(\sigma_{0}(n)-4 = 2\times\log_{n}(15)\). But there are no solutions exists in positive integers. ~~~!

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You know there is a much easier solution in terms of appearance.

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What I did is proving the general formula. =w=

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Prob 3: Squaring the first and dividing we get \(\dfrac{a^2+2ab+b^2}{a^2+b^2}=1+\dfrac{2ab}{a^2+b^2}=\dfrac{c^4}{c^3}=c\) and so we must have \(a^2+b^2\mid 2ab\implies a^2+b^2\le 2ab\). By AM-GM we always have \(a^2+b^2\ge 2ab\) so we must have equality, that is \(a=b\). Thus \(c=2\) giving the only solution \((a,b,c)=(2,2,2)\).Prob 4: If \(p,q > 3\) then \(p^{q+1}\equiv q^{p+1}\equiv 1\pmod{3}\) since they are even powers of prime base. So their sum satisfies \(p^{q+1}+q^{p+1}\equiv 2\pmod 3\) which is not a quadratic residue mod \(3\). So we must have at least one of \(p,q\in\{2,3\}\). WLOG we let \(p\in\{2,3\}\). If \(p=2\) and \(q>2\) then \(2^{q+1}+q^3=t^2\implies q^3=\left(t+2^{(q+1)/2}\right)\left(t-2^{(q+1)/2}\right)\). So \(t+2^{(q+1)/2}=q^i\) and \(t-2^{(q+1)/2}=q^j\) so \(2^{(q+3)/2}=q^j\left(q^{i-j}-1\right)\) and then \(q=2\), a contradiction. On the other hand \((p,q)=(2,2)\) works with square \(16\). If \(p=3\) then \(3^{q+1}+q^4=t^2\) implies \(3^{q+1}=\left(t+q^2\right)\left(t-q^2\right)\) so \(t+q^2=3^i\) and \(t-q^2=3^j\) giving \(2q^2=3^j\left(3^{i-j}-1\right)\). Since \(q\) is prime, we can easily observe that we must have \(j=2\) and \(i=3\) so that \(q=3\); or \(q=2\) and \((i,j)=(2,0)\). We check that none works.Log in to reply

For problem \(4\), you showed that both \(p\) and \(q\) can't be greater than \(3\). Then you concluded that both \(p\) and \(q\) have to be in \(\{2, 3\}\). Do you see what's wrong with that argument?

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Edited.

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For \(q=5, 2^{q+1}+q^3\equiv 0\pmod 3\).

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@Jubayer Nirjhor 's assumption that \(2^{q=1} + q^3 \equiv 2 \pmod 3\).

I didn't say that \(2^{q+1} + q^3 \equiv 0 \pmod 3\) for all primes \(q\). I just showed a counter exampleLog in to reply

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Isn't 4 hours a little too generous for this set?

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To tell you the truth, it was hard enough getting 7 marks in most questions. A couple of them, I made screw ups, even with 4 hours.

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We're talking about juniors.

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Q1

(a)Let each term be \(a_{n}\). The sequence is defined as \(a_{n} = \frac{a_{n-1}+a_{n+1}}{2}\); from this, we know that \(a_{n+1} = 2a_{n}-a_{n-1}\).We can see that \(a_{n} = 9n-8\) for base cases \(n = 1 = 2\).

\[a_{n+1} = 2a_{n}-a_{n-1} = 2(9n-8)-9(n-1)+8\]

\[= 9(n+1)-8\]

Since \(10^{1000} \equiv (1+9)^{1000} \equiv 1^{1000} \equiv -8 \mod 9\) and our sequence contains all positive integers of the form \(9n-8\), \(10^{1000}\) is in our sequence.

(b)Honestly I can't care to do this. I'd be surprised if you had a really elegant solution, but as far as I can tell, it's dealing with nasty repunits.Log in to reply

I'll tell you that it's very easy to do with arithmetic sum formula.

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