PRIME DECOMPOSITION

Nigel Nettheim

This item first appeared in Mathematics Magazine, Vol 44 No. 3, May 1971, p. 168.

Problem 769. [September, 1970] Proposed by Charles W. Trigg, San Diego, California.

A triangular number is composed of nine distinct digits in the decimal system. When it is sectioned into three triads, each triad is prime. Find the number and show it to be unique.

Solution by Nigel F. Nettheim, Toronto, Ontario, Canada.

The triangular number T(n) = ½ n(n+1) has nine digits, so that 14142 <= n <= 44720. Let

n == 2000j + 200k + 20l + m ,

where 0 <= m <= 19, 0 <= l <= 9, 0 <= k <= 9, 7 <= j <= 22, and write T(n) == T(jklm). Then we have

T(jkl + 1, m) –  T(jklm) =  (21 + 2n) · 10

so that the last digits of the numbers T(n) repeat in cycles of 20. Since the last triad is prime, the last digit is 1, 3, 7 or 9 so that m is an element of the set {1, 2, 6, 13, 17, 18}. Also

T(j + 1 , klm) –  T(jklm) =  (2001 + 2n) · 1000    (*)

so that the last triads repeat in cycles of 2000. Since, moreover, for 1 <= n <= 1998 we have

T(1999 – n) – T(n) =  (1999 – 2n) · 1000,

the last triads, for given j, occur symmetrically about the central value n = 2000j + 999½. Hence, to determine all the values of (·, k, l, m) which yield a prime last triad with distinct digits, it is only necessary to examine 5 · 10 · 6 = 300 cases and use the above-noted symmetry. This procedure yields exactly 77 · 2 = 154 sets, namely (·, 0, 2, 13), · · · (·, 9, 7, 6) or, setting j = 0 for convenience, n = 53, · · · 1946.

We proceed to examine the middle triad in each case for each possible value of j; from (*) it follows that the middle triad is increased by (2n + 1) as j is incremented by 1. Bearing in mind the requirement of distinct digits and that the first triad also be prime, we find the unique solution: T(36161) = 653827041.

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