N heart

We have to find a $n$ such that $100 \mid n♡$ . Note that $100 \mid n♡ \iff 4 \mid n♡ \text{ and } 25 \mid n♡$ Let's first try to find the set of positive integers $n$ such that $n♡ \equiv 0 \pmod{4}$ . To do this, let's write down the first few residues of $n♡ \pmod{4}$ :
Here's the link in case the image doesn't load:http://s8.postimg.org/iob62zp6d/Untitled.png
In the table above, note that $4♡ \equiv 8♡ \pmod{4}$ . We shall now prove by induction that for all $n \geq 4$ , $n♡ \equiv (n+4)♡ \pmod{4}$ . The base case $n=4$ has already been shown. Assume our assertion is true for some $k \geq 4$ , so $k♡ \equiv (k+4)♡ \pmod{4}$ . Then, note that $(k+1)♡ \equiv (k+1) \times k♡ + 1 \pmod{4}$ $\equiv (k+1) \times (k+4)♡ + 4 \times (k+4)♡ + 1 \pmod{4}$ $\equiv (k+5)\times ( k+4)♡ + 1 \equiv (k+5)♡ \pmod{4}$ This completes our induction step, and hence our claim is true for all $n \geq 4$ . Thus, the residues of $n♡ \pmod{4}$ become periodic with periodicity $4$ . From the table above, we have $7♡ \equiv 0 \pmod{4}$ , so for all $n \geq 4$ , $n♡ \equiv 0 \pmod{4} \iff n \equiv 7 \equiv 3 \pmod{4}$ .

Similarly, we calculate the first $40$ residues of $n♡ \pmod{25}$ :
Here's the link in case the image doesn't load:http://s11.postimg.org/vb7ee4ur7/Untitled.png
Here's the link in case the image doesn't load:http://s15.postimg.org/3pedf06aj/Untitled.png
Note that $15♡ \equiv 45♡ \pmod{25}$ . Similarly, we shall prove by induction that for all $n \geq 15$ , $n♡ \equiv (n+25)♡ \pmod{25}$ . The base case $n=15$ has already been shown. Assume our assertion is true for some $k \geq 15$ , so $k♡ \equiv (k+25)♡ \pmod{25}$ . Then, note that $(k+1)♡ \equiv (k+1) \times k♡ + 1 \pmod{25}$ $\equiv (k+1) \times (k+25)♡ + 25 \times (k+25)♡ + 1 \pmod{25}$ $\equiv (k+26) \times (k+25)♡ + 1 \equiv (k+26)♡ \pmod{25}$ This completes the induction step, and hence our claim is true for all $n \geq 15$ . Thus, the residues of $n♡ \pmod{25}$ become periodic with periodicity $25$ . From the table above, note that $n♡$ is zero for $n=32$ and $n=44$ . Hence, for all $n \geq 15$ , $25 \mid n♡ \iff n \equiv 32/44 \pmod{25} \implies n \equiv 7/19 \pmod{25}$ .

Solving the congruencies $n \equiv 7 \pmod{25}$ , $n \equiv 3 \pmod{4}$ by CRT, we get $n \equiv 7 \pmod{100}$ .
Again, solving the congruencies $n \equiv 7 \pmod{25}$ , $n \equiv 3 \pmod{4}$ by CRT, we get $n \equiv 19 \pmod{100}$ .
To conclude, $100 \mid n♡ \iff n \equiv 7/19 \pmod{100}$ . The largest number $\equiv 7 \pmod{100}$ in the range $[1,1000]$ is $907$ , and the largest number $\equiv 19 \pmod{100}$ in the range $[1,1000]$ is $919$ . Our desired answer will be $\boxed{919}$ .

I'm going to use $f_n$ instead of n-heart , not sure how to write the heart :)

Firstly let's have a non-recursive definition for $f_n$ : $f_n = \sum_{i=1}^{n}{n! \over i!}$

We can readily check that this satisfies the definition of n-heart.

Lemma: $f_{n+k} \equiv (f_n + n!) \mod k$

Proof: We have $f_{n+k} = \sum_{i=1}^{n+k}{(n+k)! \over i!}$ All terms with $i \lt k$ are clearly $\equiv 0 \mod k$ . So, letting $j = i - k$ , we get $\begin{aligned} f_{n+k} &\equiv \sum_{j=0}^{n}{n! \over j!} \mod k\\ &\equiv \sum_{j=1}^{n}{n! \over j!} + n! \\ &\equiv f_n + n! \end{aligned}$

Now, we need to solve $f_n \equiv 0 \mod 100$ . By the Chinese Remainder Theorem we can solve this in two parts: $f_n \equiv 0 \mod 4$ and $f_n \equiv 0 \mod 25$ , since $4$ and $25$ are coprime.

If $f_n \equiv 0 \mod 4$ then repeated application of the lemma tells us that $f_n \equiv f_{n \mod 4} + n! \mod 4$ . So we just need to compute $f_n + n!$ for $n = 1, 2, 3, 4$ (I choose this range to avoid using $f_0$ ), and we find: $\begin{aligned} f_1 + 1! &= 2 \\ f_2 + 2! &= 5 \\ f_3 + 3! &= 16 \\ f_4 + 4! &= 65 \end{aligned}$

So we must have $n \equiv 3 \mod 4$ .

We can solve $f_n + n! \equiv 0 \mod 25$ in the same way. I did this by brute force, although it is not particularly brute considering that you can compute $f_n \mod 25$ up to $f_{25}$ easily just by only keeping track of the last two digits; and since $10! \equiv 0 \mod 25$ we don't need to worry about $n!$ beyond the step $f_9 + 9!$ .

It turns out that $f_7 + 7! \equiv 0 \mod 25$ , and $f_{19} + 19! \equiv 0 \mod 25$ , and no other solutions. So we have $n \equiv 7 \mod 25$ or $n \equiv 19 \mod 25$ .

So we have $f_n \equiv 0 \mod 100$ if and only if $n \equiv 7\mbox{ or }19 \mod 25 \\ n \equiv 3 \mod 4$

Combining these two conditions gives $n \equiv 7 \mbox{ or } 19 \mod 100$ , so the biggest solution under $1000$ is $\boxed{919}$ .

[Note: You might need to scroll right and left to view the full solution.]

If the last two digits are zeroes, this means that $n\heartsuit$ is congruent to 0 mod 25 and mod 4

Carefully writing out the congruences of $n\heartsuit$ modulo 25,

$\begin{array} { c | c c c c c c c c c c c c c c c c c c c c c c c c c c c c } n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 & 24 & 25 \\ n \pmod{25} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 & 24 & 0\\ n \heartsuit \pmod{25} & 1 & 3 & 10 & 16 & 6 & 12 & 10 & 6 & 5 & 1 & 12 & 20 & 11 & 5 & 1 & 17 & 15 & 21 & 0 & 1 & 22 & 10 & 6 & 20 & 1 \\ \hline n & 26 & 27 & 28 & 29 & 30 & 31 & 32 & 33 & 34 & 35 & 36 & 37 & 38 & 39 & 40 & 41 & 42 & 43 & 44 & 45 \\ n \pmod{25} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20\\ n \heartsuit \pmod{25} & 2 & 5 & 16 & 15 & 1 & 7 & 0 & 1 & 10 & 1 & 12 & 20 & 11 & 5 & 1 & 17 & 15 & 21 & 0 & 1 \\ \end{array}$

First observe that $n\heartsuit \equiv (n+25) \heartsuit \pmod{25}$ for all $n\geq 15$ . We will prove this by induction.

Base case: We have $15 \heartsuit \equiv 1 \equiv 40 \heartsuit \pmod{25}$ .

Induction step : If $k \heartsuit \equiv (k+ 25 ) \heartsuit \pmod{25}$ , then $(k+1) \heartsuit \equiv (k \heartsuit) \times (k+1) + 1 \equiv (k+25) \heartsuit \times (k+1+25) + 1 \equiv (k+26) \heartsuit \pmod{25}$ . Hence done.

Thus, for $n \geq 15$ , we know that $n \heartsuit \equiv 0 \pmod{25}$ if and only if $n \equiv 7, 19 \pmod{25}$ (check the above table).

For modulo 4, a similar observation holds. The congruences are:

$\begin{array} { c | c c c c c c c c c c c c c c c c } n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & \\ n \pmod{4} & 1 & 2 & 3 & 0 & 1 & 2 & 3 & 0 & 1 & 2 & 3 & 0 \\ n \heartsuit \pmod{4} & 1 & 3 & 2 & 1 & 2 & 1 & 0 & 1 & 2 & 1 & 0& 1 \\ \end{array}$

We can similarly show that for $n \geq 4$ , $n \heartsuit \equiv (n+4) \heartsuit \pmod{4}$ , and that $n \heartsuit \equiv 0 \pmod 4$ if and only if $n \equiv 3 \pmod {4}$ .

Combining the above using Chinese Remainder Theorem, we know that for $n \geq 15$ , $n \heartsuit \equiv 0 \pmod {100}$ if and only if $n \equiv 7, 19 \pmod{100}$ . Hence, the largest possible solution is 919.

[Edits for clarity. Latex edits - Calvin]