# Junction Combination Theorem

Standard

Suppose that, $a_{k_1} + s_d(a_{k_1}) = a_{k_2} + s_d(a_{k_2}) = \dots = a_{k_r} + s_d(a_{k_r}) = N$ for $k_1, k_2, \ldots k_r > 1$.$s_d(n)$ denotes the sum of the digits of n. Thus all of $a_{k_1}, a_{k_2}, \ldots , a_{k_r}$ yield $N$. That is, $N$ has $r$ “generators”. Numbers with more than one generator (i.e. $r>1$) are called “Junction Numbers” by Kaprekar.

Let $N-1$ and $N+1$ be two “Junction Numbers”, such that they have equal number of digits, say, $d$ digits. If $P_1, P_2, \ldots$ be the generators of $N-1$ and $R_1, R_2, \ldots$ be the generators of $N+1$.
Then,

$1(0)_{(1)_d} P_1 \quad, \quad 1(0)_{(1)_d} P_2 \quad, \quad \ldots$

$(9)_{(1)_d} R_1 \quad, \quad (9)_{(1)_d} R_2 \quad, \quad \ldots$

give the generators of $1(0)_{(1)_d} N$.

Let us consider an example to understand this theorem.

Consider two Junction Numbers 519 (generated by both 498 and 507) and 521 (generated by both 499 and 508).

Here $N-1 = 519$ and $N+1 = 521$

$\therefore P_1 = 498 \quad, \quad P_2 = 507$
$\therefore R_1 =499 \quad, \quad R_2 = 508$

$d = 3 =$ Number of digits in $N-1$ or $N+1$ or $N$.

$\therefore (1)_d = (1)_3 = 111$
Then according to the theorem,
$1(0)_{111} 498 \quad, \quad 1(0)_{111} 507$

$(9)_{111} 499 \quad, \quad (9)_{111} 508$

give the four generators of $1(0)_{111} 520$.

$1(0)_{111} 520$ is a number with 115 digits

Kaprekar named this theorem as Kaprekar’s Last Theorem in 1962, when he was seriously ill and feared that his death was nearing. He miraculously recovered and named it as Junction Combination Theorem.

Smallest “Junction Numbers”

• Smallest “Junction Number” with two generators is $\mathbf{101}$
• Smallest “Junction Number” with three generators is $\mathbf{10000000000001}$ or $\mathbf{1(0)_{12}1}$
• Smallest “Junction Number” with four generators is $\mathbf{1000000000000000000000102}$ or $\mathbf{1(0)_{21}102}$