Problem 12174
(American Mathematical Monthly, Vol.127, April 2020) Proposed by G. Galperin (USA) and Y. J. Ionin (USA).
(a) Let n be a positive integer, and suppose that the three leading digits of the decimal expansion of 4n are the same as the three leading digits of 5n. Find all possibilities for these three leading digits.
(b) Prove that for any positive integer k there exists a positive integer n such that the k leading digits of the decimal expansion of 4n are the same as the k leading digits of 5n.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. (a) Let N be the integer given by the k common leading digits of 4n and 5n. Then 10k−1≤ N < 10k and there are a, b ∈ N+ such that
N · 10a< 4n< (N + 1) · 10a and N · 10b< 5n< (N + 1) · 10b.
Both strict inequalities for the lower bounds are due to the uniqueness of the prime-factorization.
By multiplying the first inequality by the second one squared, we find N3· 10a+2b< 100n < (N + 1)3· 10a+2b which implies
103(k−1)≤ N3< 102n−a−2b< (N + 1)3≤ 103k. It follows that 2n − a − 2b ∈ {3k − 2, 3k − 1} and
N ∈ {⌊10k−2/3⌋, ⌊10k−1/3⌋}.
Therefore, for k = 3 the possibilities for the three leading digits are ⌊107/3⌋ = 215 and ⌊108/3⌋ = 464.
Both cases are attained
4712= 464 . . . , 5712= 464 . . . and 42·712= 215 . . . , 52·712= 215 . . .
(b) Since log10(2) is an irrational number, it follows that the sequence {n log10(2)+1/3 (mod 1)}n∈N
is dense in [0, 1). Hence, for any ε > 0 there exist n, m ∈ N such that m < n log10(2) + 1/3 < m + log10(1 + ε), that is
10m< 2n· 101/3< 10m(1 + ε).
Therefore
101/3< 4n
102m−1 < 101/3(1 + ε)2 and 101/3
1 + ε < 5n
10n−m < 101/3.
By letting ε = 1/10k, the above inequalities imply that that the first k leading digits of 4n and 5n
are the same, i. e. ⌊10k−2/3⌋.
