$a_1$, $a_2$, $a_3$, ..., $a_n$, ... are positive integers such that for all $n\ge 1$, $a_{n+1}>a_n$ and $a_{a}n=3n$

i) Find a₁ to a₉

ii) Find a₁₀₀

I have got a solution but I am not happy with it (though it is correct), I may post it later but now I have to go to lessons!

Hello again.

Here is my solution:

Try a few a₁ s:

If a₁ =n where n> 3 then a_a 1=a_n =3 while a₁ =n> 3 hence a₁ > a_n , but a₁ should be always less than a_n , so n£ 3

If a₁ =3 then a_a 1=a₃ =3 x 1=3, and you get a₁ =3 and a₃ =3, since a₁ should be less than a₃ , so n is not 3

If a₁ =1 then a_a 1=a₁ =3 x 1=3, so you have a₁ =1 and a₁ =3 which is impossible, so n is not 1

It turns out that a₁ =2 and you can then work out a₁ to a₉ except a₄ and a₅ . But since a₃ =6 and a₆ =8 and a_n are all integers, so a₄ and a₅ must equal to 7 and 8 respectively.

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For the second part,

I worked out:

a₁₁ =20, then a_a 11=a₂₀ =11*3=33, a₃₃ =20*3=60, a₆₀ =99, therefore a₉₉ =180;

a₁₂ =21, then a₂₁ =36, a₃₆ =63, a₆₃ =108, a₁₀₈ =189

there are 8 integers between 180 and 189 (exclusive) and 8 terms of a in this range, so a₁₀₀ =181


Is there a better way of doing it? (It took me ages to get there!)

Thanks.

This question is very similar to 1992 BMO1 question 5. First, let me write a_n as a_(n). You can establish the value of a_(3ⁿ ) and a_(2*3ⁿ ) for all n (inductively), which is not difficult. Now, apply a simple argument to show a_(3ⁿ +k)=a_(3ⁿ )+k and correspondingly a_(2*3ⁿ +k)=3ⁿ⁺¹ +3k. Hence you can work out the value of a_n with ease. [the BMO question asked for a_(1992)]

Kerwin