Maths Olympiad Prep

Track / Stage 7 / 157 of 300 #1557 of 1964

Problem 1557

National olympiad second round; IMO P1/P4
Number theory Difficulty 7.3 Find the answer

Let f:NRf:\mathbb{N}\mapsto\mathbb{R} be the function f(n)=k=11lcm(k,n)2.f(n)=\sum_{k=1}^\infty\dfrac{1}{\operatorname{lcm}(k,n)^2}. It is well-known that f(1)=π26f(1)=\tfrac{\pi^2}6. What is the smallest positive integer mm such that mf(10)m\cdot f(10) is the square of a rational multiple of π\pi?

A number or a short expression. Spacing, $ signs and \frac vs / are all fine.

Official solution

1. We start with the given function:
f(n)=k=11lcm(k,n)2 f(n) = \sum_{k=1}^\infty \frac{1}{\operatorname{lcm}(k,n)^2}
and we need to find f(10) f(10) .

2. We decompose the sum based on the values of kmod10 k \mod 10 :
f(10)=k=11lcm(k,10)2 f(10) = \sum_{k=1}^\infty \frac{1}{\operatorname{lcm}(k,10)^2}
=i=0(1lcm(10i+1,10)2+1lcm(10i+2,10)2++1lcm(10i+10,10)2) = \sum_{i=0}^\infty \left( \frac{1}{\operatorname{lcm}(10i+1, 10)^2} + \frac{1}{\operatorname{lcm}(10i+2, 10)^2} + \cdots + \frac{1}{\operatorname{lcm}(10i+10, 10)^2} \right)

3. We calculate the least common multiples for each residue class modulo 10:
lcm(10i+1,10)=10,lcm(10i+2,10)=10,lcm(10i+3,10)=10,lcm(10i+4,10)=10 \operatorname{lcm}(10i+1, 10) = 10, \quad \operatorname{lcm}(10i+2, 10) = 10, \quad \operatorname{lcm}(10i+3, 10) = 10, \quad \operatorname{lcm}(10i+4, 10) = 10
lcm(10i+5,10)=10,lcm(10i+6,10)=10,lcm(10i+7,10)=10,lcm(10i+8,10)=10 \operatorname{lcm}(10i+5, 10) = 10, \quad \operatorname{lcm}(10i+6, 10) = 10, \quad \operatorname{lcm}(10i+7, 10) = 10, \quad \operatorname{lcm}(10i+8, 10) = 10
lcm(10i+9,10)=10,lcm(10i+10,10)=10 \operatorname{lcm}(10i+9, 10) = 10, \quad \operatorname{lcm}(10i+10, 10) = 10

4. We substitute these values back into the sum:
f(10)=i=0(1100(1(10i+1)2+1(10i+3)2+1(10i+7)2+1(10i+9)2) f(10) = \sum_{i=0}^\infty \left( \frac{1}{100} \left( \frac{1}{(10i+1)^2} + \frac{1}{(10i+3)^2} + \frac{1}{(10i+7)^2} + \frac{1}{(10i+9)^2} \right) \right.
+125(1(10i+2)2+1(10i+4)2+1(10i+6)2+1(10i+8)2)+14(1(10i+5)2)+1(10i+10)2) \left. + \frac{1}{25} \left( \frac{1}{(10i+2)^2} + \frac{1}{(10i+4)^2} + \frac{1}{(10i+6)^2} + \frac{1}{(10i+8)^2} \right) + \frac{1}{4} \left( \frac{1}{(10i+5)^2} \right) + \frac{1}{(10i+10)^2} \right)

5. We recognize that each of these sums is a partial sum of the Riemann zeta function:
i=01(10i+k)2=π26102 \sum_{i=0}^\infty \frac{1}{(10i+k)^2} = \frac{\pi^2}{6 \cdot 10^2}
for k=1,2,,10 k = 1, 2, \ldots, 10 .

6. We combine these results:
f(10)=1100(4π26102)+125(4π26102)+14(π26102)+1102(π26102) f(10) = \frac{1}{100} \left( 4 \cdot \frac{\pi^2}{6 \cdot 10^2} \right) + \frac{1}{25} \left( 4 \cdot \frac{\pi^2}{6 \cdot 10^2} \right) + \frac{1}{4} \left( \frac{\pi^2}{6 \cdot 10^2} \right) + \frac{1}{10^2} \left( \frac{\pi^2}{6 \cdot 10^2} \right)
=π26102(4100+425+14+11) = \frac{\pi^2}{6 \cdot 10^2} \left( \frac{4}{100} + \frac{4}{25} + \frac{1}{4} + \frac{1}{1} \right)
=π26102(4100+16100+25100+100100) = \frac{\pi^2}{6 \cdot 10^2} \left( \frac{4}{100} + \frac{16}{100} + \frac{25}{100} + \frac{100}{100} \right)
=π26102(145100) = \frac{\pi^2}{6 \cdot 10^2} \left( \frac{145}{100} \right)
=145π26104 = \frac{145 \pi^2}{6 \cdot 10^4}

7. We need mf(10) m \cdot f(10) to be the square of a rational multiple of π \pi :
m145π26104=(aπb)2 m \cdot \frac{145 \pi^2}{6 \cdot 10^4} = \left( \frac{a \pi}{b} \right)^2
m145π26104=a2π2b2 m \cdot \frac{145 \pi^2}{6 \cdot 10^4} = \frac{a^2 \pi^2}{b^2}
m1456104=a2b2 m \cdot \frac{145}{6 \cdot 10^4} = \frac{a^2}{b^2}

8. To make the left-hand side a perfect square, m m must be chosen such that:
m=6104b2145 m = \frac{6 \cdot 10^4 \cdot b^2}{145}

9. The smallest m m that makes this a perfect square is when b=1 b = 1 :
m=6104145=60000/145=414.48 m = \frac{6 \cdot 10^4}{145} = 60000 / 145 = 414.48

10. Since m m must be an integer, we need to find the smallest integer m m such that m1456104 m \cdot \frac{145}{6 \cdot 10^4} is a perfect square. The smallest such m m is 42.

The final answer is 42 \boxed{42} .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.