Interesting Sequence

Let $a_1<a_2<a_3<a_4<a_5<a_6$ be the six integers.If $a_4 \geq12$

Then $a_5 \geq 2a_4 \geq 24$ and $a_6 \geq 2a_5 \geq 48$

$\rightarrow a_4+a_5+a_6 \geq 84$

So $a_4<12$

Thus the first four numbers are $1,2,4,8$

Let $a_5=8m$ and $a_6=8mn$ [where $m,n$ are positive integers]

So $8m+8mn=79-(1+2+4+8)=64$

or, $m(n+1)=8$

This leads to a unique solution $m=2$ & $n=3$

So $a_6=8*2*3=\boxed{48}$

We note that the smallest possible sequence is:

$1+2+4+8+16+32=63$

As the difference of sums of the required and smallest sequences is $79-63=16$ , we can solve the problem by adding $16$ to the last term $32$ making it $48$ , which is a multiple of the term before $16$ . Therefore the sequence is:

$1+2+4+8+16+\color{#3D99F6}{48}=\color{#3D99F6}{79}$

And the largest term is $\boxed{\color{#3D99F6}{48}}$ .

It should be noted that the solution is unique. We note that by replacing the other terms with the next smallest possible number, the sequence sum is larger than 79. See below:

\[\begin{array} {} 1+2+4+8+16+32& = \color{red}{63} \\ 1+2+4+8+16+\color{blue}{48} & = \color{blue}{79} \\ 1+2+4+8+\color{red}{24} +48& = \color{red}{87} \\ 1+2+4+\color{red}{12} +24+48& = \color{red}{91} \\ 1+2+\color{red}{6} +12+24+48& = \color{red}{93} \\ 1+\color{red}{3} +6+12+24+48& = \color{red}{94} \\ \color{red}{2} +4+8+16+32+64& = \color{red}{126} \end{array}\]

Label the terms from right to left: $\langle a_6, a_5, a_4, a_3, a_2, a_1 \rangle$ . Define the tail sum $T_n$ as $T_n = \sum_{i=1}^n a_i.$ Since every term in this sum is a multiple of $a_n$ , we have $a_n | T_n$ . This will be our guiding principle in considering possible terms.

Because the sequence must be strictly increasing, $a_n \geq 2a_{n+1}$ . It is easy to see that $T_n \geq (2^n-1)a_n.$ This greatly reduces the number of cases we must consider: given a desired sum $T_n$ , the only candidates for $a_n$ are $a_n = m\cdot a_{n+1} \ \ \ \text{with} \ \ \ m | \frac{T_n}{a_{n+1}}\ \ \ \text{and}\ \ \ 2 \leq m \leq \frac{T_n/a_{n+1}}{(2^n-1)}.$ In particular, if $T_n/a_n$ is prime, no such value $a_n$ exists; the sequence cannot be completed.

The problem starts with $T_6 = 79$ . Since $n_6 | T_6$ , this implies $n_6 = 79$ (which obviously is too big) or $n_6 = 1$ : $\langle 1, \dots\rangle.$ Moving on, we now have $T_5 = 78$ . Candidates for $a_5$ must satisfy $a_5 | 78$ , and moreover $2 \leq a_5 \leq \tfrac{78}{31} < 3$ . This only leaves $a_5 = 2$ : $\langle 1, 2, \dots\rangle.$ Next, $T_4 = 76$ , and $a_4 = 2m$ with $m | 38$ and $2 \leq m \leq \tfrac{38}{15} < 4.$ This results in $m = 2, a_4 = 4$ : $\langle 1, 2, 4\dots\rangle.$ Then, $T_3 = 72$ , and $a_3 = 4m$ with $m | 18$ and $2 \leq m \leq \tfrac{18}{7} < 3.$ The only solution is $m = 2, a_3 = 8$ . $\langle 1, 2, 4, 8\dots\rangle.$ Next, $T_2 = 64$ , so $a_2 = 8m$ with $m | 8$ and $2 \leq m \leq \tfrac{8}{3} < 3.$ Again, $m = 2, a_2 = 16$ : $\langle 1, 2, 4, 8, 16\dots\rangle.$ Finally, $T_1 = 48$ , which of course gives $a_1 = 48$ : $\langle 1, 2, 4, 8, 16, 48\rangle.$

Note: This method seems longer than the other solutions. However, it is algorithmic and could easily be implemented as a computer program. For other values than 79, there may be multiple solutions: at every step of the algorithm there may be a recursive branching of possible values.