Jan24

Junction Combination Theorem

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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}