If $N = \frac{{n_1}(n_1+1)}{2}\cdot{d_1}$, is it possible to have $N = \frac{{n_2}(n_2+1)}{2}\cdot{d_2}$? If $$N = \frac{{n_1}(n_1+1)}{2}\cdot{d_1},$$
is it possible to have $$N = \frac{{n_2}(n_2+1)}{2}\cdot{d_2}?$$
Here, $d_i, n_i \in \mathbb{N}$, $d_i > 1$ and $n_1 \neq n_2$.
That is,
Question 1

If a number can be represented as a nontrivial multiple of a triangular number, can it then be represented as a nontrivial multiple of another triangular number?

Added January 27 2017
and
Question 2

If such a number can be so represented (as in Question 1), must it necessarily be (a nontrivial multiple of or) the least common multiple of the triangular numbers
  $$T(n_1)=\dfrac{{n_1}(n_1 + 1)}{2}$$
  and
  $$T(n_2)=\dfrac{{n_2}(n_2 + 1)}{2}?$$

Thanks to users Alex Macedo, dxiv, and Stahl for their initial comments!
 A: Let $\dfrac{n_1(n_1+1)}{n_2(n_2+1)} = \dfrac{d_2}{d_1}$
It follows that 
$\dfrac{{n_1}(n_1+1)}{2}\cdot{d_1} =  \dfrac{{n_2}(n_2+1)}{2}\cdot{d_2}$
Example $T(6) = 21$ and $T(8)=36$. So $\dfrac{T(6)}{T(8)} = \dfrac{7}{12}$
Hence 
$\dfrac{{6}(6+1)}{2}\cdot{12} =  \dfrac{{8}(8+1)}{2}\cdot{7}$
If $n_1$ and $d_1$ are given {It would have been nice if you were a bit more explicit}, then you want to find $n$ and $d$ that solve
$\dfrac{{n}(n+1)}{2}\cdot{d} = \dfrac{n_1(n_1+1)}{2}\cdot{d_1}$
Which can be expressed as
$n^2 + n - \dfrac{n_1(n_1+1) d_1}{d} = 0$
and can be solved by inspection.
Example
\begin{align}
   \dfrac{4(4+1)}{2}\cdot{3} &=  \dfrac{{n}(n+1)}{2}\cdot{d} \\
   n^2 + n - \dfrac{60}{d} &= 0 \\
   n &= \dfrac{-1 + \sqrt{1+\dfrac{240}{d}}}{2} \\
\end{align}
The possible integer values of $\dfrac{240}{d}$ are 
$$\{1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 40, 48, 60, 80, 120, 240\}$$
We note that $1+\{3, 8, 15, 24, 48, 80, 120\}$ are perfect squares.
Which leads to $d \in \{80, 30, 16, 10, 5, 3, 2\}$
and $n \in \left\{\dfrac 12, 1, \dfrac 32, 2, 3, 4, 5\right\}$.
Removing the fractions, we end up with
$(n,d) = \{(1,30), (2,10), (3,5), (4,3), (5,2) \}$
