Lacsap's triangle

Lascap's triangle is very closely related to Pascal's triangle: The entry in the $i$ th row and $j$ th column of Lascap's triangle is equal to $2^{P_{(i,j)}}$ , where $P_{(i,j)}$ is the entry in the $i$ th row and $j$ th column of Pascal's triangle. To prove this, just note that the elements on the edges of the triangle are $2^1$ and use the fact that $2^a\cdot2^b = 2^{a+b}$ .

We note that $32768 = 2^{15}$ , and so the problem becomes finding how many times the number $15$ appears in Pascal's Triangle. It is very easy to check that it appears $4$ times, and so our answer is $\fbox{4}$ .

Notice that each term is equal to 2 to the power of the corresponding term in pascal's triangle. Because 32768=2^15, we are basically asking ourselves how many times 15 occurs in pascal's triangle, which is 4 times.

Lacsap's Triangle is very similar to Pascal's triangle; each value $n$ in Lacsap's triangle is just $2^n$ in Pascal's triangle. Since $32768=2^{15}$ , we must find the number of times 15 appears in Pascal's triangle. Obviously, we have $\binom{15}{1}$ and $\binom{15}{14}$ are equal to 15, and we can also see that $\binom{6}{2}$ and $\binom{6}{4}$ are as well. Checking other values (we only have to check a few numbers in the first 16 rows), we see that there are no more, so our final answer is $\boxed{4}$ .

We first notice that all the terms in the triangle will be powers of two. We also notice that $32768$ is $2^{15}$ . We also notice that the second diagonal from the left and the second diagonal from the right are just increasing by a power of two: $2,2^{2},2^{3},...$ so we are sure to get at least two $2^{15}$ . Notice now the sub-triangle enclosed by the two second diagonals. On the third row of this triangle, we can see another two $2^{15}$ but after that all the terms just get bigger and bigger. Thus there are only $4$ entries that are equal to $2^{15}=32768$ .

You will found the answer if you differentiate once (dy/dx will be 2n) then divide each number by 2. We know that $2^{15}$ = $32768$

Lacsap triangle after differentiation, (the answer is $30$ ) $2$ , $2$ $2$ , $2$ $4$ $2$ , $2$ $6$ $6$ $2$ , $2$ $8$ $12$ $8$ $2$ , $2$ $10$ $20$ $20$ $10$ $2$ , $2$ $12$ $30$ $40$ $30$ $10$ $2$ , ...... The other two found at the two end of Row 15, that is either $2$ $30$ ... or ... $30$ $2$ . Hence, the number of solutions is $4$ .

Just by yourself, start to extend this triangle. Note that any given value except for 2 will repeat 4 times. Twice in one of the inner rows, and twice on the diagonal next to the side. You see that the numbers there go 2, 4, 8, 16, etc. Because 32,768 is a power of two, we already know that it will show up there. And, in the 7th row, there are the other 2, for a total of 4 times.

The number 32768.

The last digit is 8. Except the 8 because the last digit after multiplication in above assumptions is 8. Count the digit(3276). which are 4.

32768 = $2^{15}$

Observed that the 2nd diagonals from the top is a geometric sequence of 2 to the power of an integer. While the 3rd diagonals is 2 to the power of arithmetic sum of integers. Each 2nd and 3rd diagonal will give $2^{15}$ so the total number is 4.

Note that from the triangle's pattern we knew there will be no more other answers in other diagonals because we observed that the numbers below the two 32768 in 3rd diagonals must be larger than it.

Note 32758 = 2 ^ 15. As all entries in the triangle will be powers of 2 starting with 2^1, (2^1 2^1), (2^1 2^2 2^1) and we know that when multiplying two numbers of the same base we can add the indices the problem can be simplified to finding the number of times 15 arises in Pascals triangle. This can most easily be found by computing the first few rows.