A correct statement regarding a sequences of multiples of $2$ and $3$.

Question

The question is as such:

Ali arranges a sequence of positive integers that is a collection of multiples of 2 and 3 as follows: $$2, 3, 4, 6, 8, 9, 10, 12, \dots$$ The statement below that is true is…
I. The $2009$th term is $3012$.
II. The $6784$th term is $10174$.
III. The $6848$th term is $10270$.
IV. The $7917$th term is $11876$.
V. The $9610$th term is $14415$.

My first idea was that to separate the sequence, singling out the multiples of six. Then, using the sequence $6, 12, 18, ...$ for every four terms: $U_{4n}=6+(n-1)6$, I found out that statements II and III are wrong.

Then, observing the sequence $2, 3, 4, 6, 8, 9, 10, 12, ...$, the $(n+4)$th term is $6$ more than the $n$th term. So, $U_{n+4}=U_n+6$. I used this to test the statements I, IV, and V.

For statement I, $U_{2009}=U_{2005}+6=U_{2001}+12=\dots=U_{1}+\left(\frac{2008}{4}\right)6=2+(502)(6)=3014.$ Wrong.

For statement IV, $U_{7917}=U_{7913}+6=U_{7909}+12=\dots=U_{1}+\left(\frac{7916}{4}\right)6=2+(1979)(6)=11876$, which is correct.

For statement V, $U_{9610}=U_{9606}+6=U_{9602}+12=\dots=U_{2}+\left(\frac{9608}{4}\right)=3+(2402)6=14415$, which is also correct.

There is supposedly only one answer for this question, and I'm not sure what I did wrong, or if my method is already incorrect from the start. Any suggestion will be appreciated, thank you.

Answer 1

Let $a_{n} = 2, 3, 4, 6, 8, 9, 10, 12, ...$ be our sequence.

First observe that every fourth term is a multiple of 6.

$a_{4} = 6, a_{8} = 12, a_{12} = 18, ..., a_{4n} = 6n$

Thus $a_{4n} = 6n$ for all positive integers n.

Using this formula, we can rule out a lot of possibilities.

1. The 2009th term cannot be 3012. This is because 3012 is a multiple of 6 and the 2009th term is not a multiple of 6. [1]

2. The 6784th term cannot be 10174. This is because 10174 is not a multiple of 6 and the 6784th term has to be a multiple of 6. [2]

3. The 6848th cannot be 10270. This is because 10270 is not a multiple of 6 and the 6848th term has to be a multiple of 6. [3]

Statements IV and V both appear to be correct.

Statement IV. The 7917th term is 11876.
Statement V. The 9610th term is 14415.

We can use our formula to calculate $a_{7916}$.

\begin{align*} a_{7916} &= a_{4(1979)} \\ &= 6(1979) \\ &= 6000 + 5400 + 420 + 54 \\ &= 11874 \end{align*}

The next term $a_{7917}$ has to be 11876. This is because 11875 is not a multiple of 3, so it cannot be the next term. But 11876 is an even number and all even numbers are included in our sequence.

We can also use our formula to calculate $a_{9608}$.

\begin{align*} a_{9608} &= a_{4(2402)} \\ &= 6(2402) \\ &= 14412 \end{align*}

The next term $a_{9609}$ has to be 14414. This is because 14413 is not a multiple of 3. The term after that, $a_{9610}$, has to be 14415, since 14415 is a multiple of 3 and it's one more than the last term.

So it looks like there are two correct statements, statement IV and statement V.

Let's check this with a Python program.

% python
>>> seq = []
>>> for i in range(2, 15000):
...     if i % 2 == 0 or i % 3 == 0:
...             seq.append(i)
... 
>>> def a(n): 
...     return seq[n-1]
... 
>>> a(7917)
11876
>>> a(9610)
14415

The Python program indicates there are two correct answers. Statements IV and V are both correct.

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Footnotes:

[1] How do we know that 3012 is a multiple of 6? It's because 3012 is even and it's also a multiple of 3. How do we know that 3012 is a multiple of 3? It's because the sum of digits is

$3 + 0 + 1 + 2 = 6$

When the sum of a number's digits is divisible by 3 the number is divisible by 3.

[2] 10174 is not a multiple of 6 because it is not divisible by 3. 10174 is not divisible by 3 because the sum of its digits is 13 and 13 is not divisible by 3.

[3] 10270 is not a multiple of 6 because it's not a multiple of 3. The sum of its digits is 10 and 10 is not divisible by 3.