Q1)Find the number of 4 digit numbers with distinct digits chosen from the set {0, 1, 2, 3, 4, 5} in which no two adjacent digits are even.

I'm not sure about the answer. I'm getting 60. There's a high chance I may be wrong.

Q2) Let a, b, c > 0. If \(\dfrac 1 c , \dfrac 1 b \quad and \dfrac 1 a\) are in arithmetic progression, and if \(a^2+b^2,b^2+c^2,c^2+a^2\) are in a geometric progression, prove that a=b=c

Somehow, I managed to get a=b=c. but I'm not really confident of my approach.

Q3) Let n be a positive integer such that 2n + 1 and 3n + 1 are both perfect squares. Show that 5n + 3 is a composite number.

I tried this

Let \(2n+1=k^2\) and \(3n+1=m^2\)

\(5n+3=m^2+k^2+1\)

Then I used modulo to bash it out. am I on the right track? Because this doesn't seem to lead to the answer.

Thanks!

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## Comments

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TopNewestQ1) Let the odd digit be \(O\) and even digit be \(E\).So by product rule we can work out each case. So hereby I give the possible combinations:

\[OEOO = 3\times 3 \times 2 \times 1 = 18 \\ OOEO = 3\times 2 \times 3 \times 1 =18 \\ EOOO = 2\times 3 \times 2 \times 1 = 12 \\ OOOE = 3\times 2 \times 1 \times 3 = 18 \\ EOOE = 2\times 3 \times 2 \times 2 = 24 \\ OEOE = 3\times 3 \times 2 \times 2 = 36 \\ EOEO = 2\times 3 \times 2 \times 2 = 24 \\ \Rightarrow 18+18+12+18+24+36+24 = \boxed{150}\]

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Thanks Nihar! :D

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Saale dost ko thanks bolta hain :P

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Chal nahi bolta thanks xD

Waapas liya thanks xD

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For the record, here are the numbers:

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Comp. Science Gawd is here. _/_

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For the 2nd one, We get that \(b=2ac/a+c \) Substitute this in the second equation, \(( {b}^2+{c}^2)/({a}^2+{b}^2)=({ c}^2+{a}^2)/({b}^2+{c}^2) \) By simplifying we get, a=b=c

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@Vishnu Bhagyanath @Satyajit Mohanty @Nihar Mahajan

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Wait in Q1 when there asking for no two adjacent digits are even, doesn't that mean none of the digits are even ? O.O

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Not necessarily I guess. The placement of the digits could be OEOO

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Oh, so no two adjacent digits are both even?

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550 doesn't make sense because the maximum number of four digit numbers with distinct digits from the given set is \(5 \cdot 5 \cdot 4 \cdot 3 = 300\) , which is less than your answer.

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So you just work from there.

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Oh, I like question 3. The main way to show that a number is composite, is to factorize it and then show that the factors are not \(\pm1\)

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1) 278

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