Does anyone know how to solve this? And does "initial digit" mean the leftmost or the rightmost digit?

jasperE3 wrote: Find all $5$-tuples of different four-digit integers with the same initial digit such that the sum of the five numbers is divisible by four of them. There are 4 equations we want to solve:-$ka=(b+c+d+e),br=(a+c+e+d),cm=(a+b+e+d),dx=(a+b+c+e)$ Without the loss of generality,we will assume that $a \ge b \ge c \ge d \ge e$ First we will handle the equality cases own by one:- 1)$a=b=c=d=e$:- Over here any pair will work. 2)$a=b=c=d \neq e$:- Over here we want to solve $ak=(3a+e)$,so testing the values for $k=2,3,4,5,6$ none of which work.($k>6$ is impossible by size arguments) 3)$a=b=c \neq d \neq e$:- Again $ak=(2a+d+e)$ so obviously $k>3$ by size arguments and $k>4$ is impossible so $k=3$ implying $2a=d+e$ so just choose $d=a+y,e=a-y$ or $e=a+y,d=a-y$ which contradicts $a \ge b \ge c \ge d \ge e$ so there are no solutions. 4)$a=b \neq c \neq d \neq e$:- So here $ak=c+d+e$ and since $a=b>c>d>e$ so $ak=c+d+e \le a+a+a = 3a$ so $k=2,3$ and no solutions exist. 5)$a=b \neq c=d \neq e$:- Again $ak=a+2c+e \le a+2a+a=4a$ and testing the cases none of which work. The good thing now is that we have discarded the equality case to get a stricter inequality $a>b>c>d>e$ So since $ka=a+c+e+d<4a$ so $k=2,3$ for which no solutions exist other than $(a,b,c,d,e) = (60t,70t,84t,101t,105t), \; t \in \{17,18,19\}$,hence the only solutions are $\boxed{(a,b,c,d,e) = (60t,70t,84t,101t,105t), \; t \in \{17,18,19\} \text{and it's permutations to be the only solutions}}$ Now we will show that this is the only solution,since the solutions to this equation are given by $at,bt,ct,dt,et$ so we want to solve $lcm(a,b,c,d)*t=a+b+c+d+e$ and it is easy to see that $a,b,c,d,e=60,70,84,101,105$ are the only solutions. @below,thanks for reminding me that the other pair is also a solution.

According to the given information there are five four digit integers $(1) \;\; x_1<x_2<x_3<x_4<x_5$ with the same initial digit $n$, i.e. $(2) \;\; 1000n \leq x_i < 1000(n + 1), \;\; i \in \{1,2,3,4,5\} = M$, $(3) \;\; x_i \mid x_1 + x_2 + x_3 + x_4 + x_5, \;\; i \in M \subset \{k\} = M_k$. Assume $i,j \in M$. Set $S = x_1 + x_2 + x_3 + x_4 + x_5$ and ${\textstyle c_i = \frac{S}{x_i}}$. According to condition (1) we have $(4) \;\; c_i>c_j$ when $i<j$. Moreover by condition (2) ${\textstyle \frac{c_i}{c_j} < \frac{1000(n + 1)}{1000n}}$, i.e. ${\textstyle (5) \;\; \frac{c_i}{c_j} < 1 + \frac{1}{n}}$, yielding ${\textstyle (6) \;\; \frac{c_i}{c_j} < 2}$. Hence by condition (6) $(6 - i) \cdot x_i < S < i \cdot x_i + (5 - i) \cdot (2x_i)$, i.e. $(7 - i) \cdot x_i \leq S \leq (9 - i)\cdot x_i$, or equivalently $(7) \;\; 7 - i \leq c_i \leq 9 - i $. According to (3) $c_i \in \mathbb{N}$ when $i \in M_k$. Then, since $|M_k|=4$, there exists two integers $p,q \in M_k$ where $p - q \geq 3$. Hence $c_q - c_p \geq 3$, which combined with (6) result in $3 \leq c_q - c_p < 2c_p - c_p$, i.e. $c_p > 3$. The fact that $c_i \leq 8$ (by condition (7)) and $c_i < 2c_j$ implies $M_k = \{4,5,6,7\}$ or $M_k = \{5,6,7,8\}$. Hence, if $i,j \in M_k$ with $j - i \geq3$, then by condition (5) ${\textstyle 1 + \frac{1}{n} > \frac{c_i}{c_j} \geq \min\{\frac{7}{4},\frac{8}{5}\} = \frac{8}{5}}$, yielding ${\textstyle n < \frac{5}{3}}$, which give us $n=1$. Let us consider the two cases above: Case 1: $M_k = \{4,5,6,7\}$. Hence $LCM(4,5,6,7) \mid S$ by condition (3), i.e. $420 | S$, which means $S = 420t$ and ${\textstyle \{c_i\}_{i \in M_k} = \{\frac{420t}{m} \; | m \in M_k\} = \{60t,70t,84t,105t\}}$, which implies $x_k = S - \sum_{i \in M_k} c_i = 420t - (60 + 70 + 84 + 105)t = 420t - 319t$, i.e. $x_k=101t$, which according to condition (1) yields $k=4$. Hence $(c_1,c_2,c_3,c_5) = (105t,84t,70t,60t)$. Now $1000 \leq x_i < 2000$ by condition (2) (since $n=1$), implying $1000 \leq 60t < 105t < 2000$, yielding $17 \leq t \leq 19$. Therefore $(x_1,x_2,x_3,x_4,x_5) = (60t,70t,84t,101t,105t), \; t \in \{17,18,19\}$ satisfies the conditions (1)-(3). Case 2: $M_k = \{5,6,7,8\}$. Hence $LCM(5,6,7,8) \mid S$ by condition (3), i.e. $840 | S$, which means $S = 840t$ and ${\textstyle \{c_i\}_{i \in M_k} = \{\frac{840t}{m} \; | m \in M_k\} = \{105t,120t,140t,168t\}}$, which implies $x_k = S - \sum_{i \in M_k} c_i = 840 - (105 + 120 + 140 + 168)t = 840t - 533t$, i.e. $x_k=207t$, which according to condition (1) yields $k=5$. Hence $(c_1,c_2,c_3,c_4) = (168t,140t,120t,105t)$. Now $1000 \leq x_i < 2000$ by condition (2) (since $n=1$), implying $1000 \leq 105t < 207t < 2000$, which is impossible. Conclusion:. There are exactly three 5-tuples of integers satisfying conditions (1)-(3), namely $(x_1,x_2,x_3,x_4,x_5) = (60t,70t,84t,101t,105t), \; t \in \{17,18,19\}$.