Problem:
Let hnβ and knβ be the unique relatively prime positive integers such that
11β+21β+31β+β―+n1β=knβhnββ
Let Lnβ denote the least common multiple of the numbers 1,2,3,β¦,n. For how many integers with 1β€nβ€22 is knβ<Lnβ?
Answer Choices:
A. 0
B. 3
C. 7
D. 8
E. 10
Solution:
In assessing the denominator of the sum of rational numbers, the following principle is useful. Suppose that
q1βa1ββ+q2βa2ββ=q3βa3ββ,
where gcd(aiβ,qiβ)=1 for i=1,2,3. Let p be a prime number, and suppose that pr is the greatest power of p that divides q1β, that ps is the greatest power of p that divides q2β, and that r>s. Then pr is the greatest power of p that divides q3β. To see why this is true, let pt denote the greatest power of p that divides q3β. Multiply both sides of the displayed equation by pr. Then the resulting left-hand side is a rational number with no p in the denominator. The power of p in the factorization of the right-hand side is prβt. Hence tβ€r. Next, instead of multiplying the above equation by pr, now multiply by prβ1. Then p occurs to the first power in the denominator of the first term, p does not occur at all in the second denominator, and on the right-hand side the contribution of p to the denominator is prβ1βt. Thus tβ₯r. From these inequalities it follows that t=r.
Let
Hnβ=11β+21β+31β+β―+n1β=knβhnββ.
It is useful to note that
H1β=11β,H2β=23β,H3β=611β, and H4β=1225β.
The greatest power of 2 that divides knβ is 2r, where r is the greatest integer such that 2rβ€n. This is the same as the greatest power of 2 that divides Lnβ. For 3β€n<6 there is only one summand in Hnβ with a 3 in the denominator. For 6β€n<9, there are two such terms; they contribute
31β+61β=31βH2β=31ββ
23β=21β.
Hence for these n, the power of 3 in the denominator is less than the power of 3 in Lnβ. For 9β€n<18 there is only one term in Hnβ whose denominator has a factor 32. For 18β€n<27 there are two, and they contribute 91βH2β=61β. Thus the power of 3 in the denominator is less than the power of 3 in Lnβ for these n.\
For 5β€nβ€22, the summands that have a 5 in the denominator contribute one of
51βH1β,51βH2β,51βH3β, or 51βH4β.
None of h1β,h2β, or h3β is divisible by 5 , but 5β£h4β, so 5 does not appear in the denominator of Hnβ for 20β€n<25. For 7β€nβ€22, the contribution of 7 to the denominator of Hnβ is one of
71βH1β,71βH2β, or 71βH3β.
Because 7 does not divide any of h1β,h2β, or h3β, it follows that 7β£knβ for 7β€n<28, as in Lnβ. The contributions of 11 to the denominator of Hnβ is 111β or 223β, so 11β£knβ for 11β€nβ€22. The primes 13,17 , and 19 occur in at most one denominator in Hnβ for 1β€nβ€25, and therefore occur with the same multiplicity in knβ and Lnβ in this range.
Thus for 1β€nβ€22, the inequality knβ<Lnβ holds if and only if n=6,7,8,18,19,20,21, or 22 , a total of (D)8β values.
Note: The quantity Hnβ is known as a harmonic number.
The problems on this page are the property of the MAA's American Mathematics Competitions