1234 Followers. A problem for "1234"

Since $1234$ must exist contiguously, we can consider them as a single digit $A$ .

Since the original number is $8$ digit, the number now reduces to $5$ digit numbers $(A_4A_3A_2A_1A_0)$ that can be formed using digits $\{0, A, 5, 6, 7, 8, 9\}$ .

Total ways to place these $7$ digits in a $5$ digit number, $N(T)$ $=7 \times 6 \times 5 \times 4 \times 3 = 2520$

Total $5$ digit numbers that had $A_4$ as $0$ , $N(Z)$ $= 1 \times 6 \times 5 \times 4 \times 3 = 360$

Total $5$ digit numbers that did not have $A$ , $N(\overline{A})$ $= 6 \times 5 \times 4 \times 3 \times 2 = 720$

Total $5$ digit numbers that did not have $A$ and had $A_4$ as $0$ , $N(\overline{A} \cap Z)$ $= 1 \times 5 \times 4 \times 3 \times 2 = 120$

$\therefore$ Total valid $5$ digit numbers that have $A$ , $P$

$= N(T) - N(Z) - N(\overline{A}) + N(\overline{A} \cap Z)$

$= 2520 - 360 - 720 + 120$

$= 1560$

Prime factorization of $P = 2^3 \times 3 \times 5 \times 13$

Sum of divisors of $P = \frac{2^4-1}{2-1} \times (3 + 1) \times (5+1) \times (13 + 1) = 5040$

Sum of cute divisors of $P = 5040 - 1 - 1560 = 3479$

Hey Sandeep! I checked on the internet and it says that 1 is also considered a proper divisor. I thought you meant the sum of proper divisors excluding one and I subtracted 1 when I entered 3480 and it said that it was wrong. The solution is simple enough: 1234 can be considered a block occupying one space and the other digits occupying one space each. This reduces it to a sort of 5 digit problem. 0 should not occupy the first place and we don't have to worry about this when the 1234 block occupies the 1st place. We then get: (5!×5×4+6!)/2=1560=8×3×5×13. Sum of divisors of 1560=5040. Sum of proper divisors (excluding 1) = 3479.