1. We start with the given function:
f(n)=k=1∑∞lcm(k,n)21
and we need to find f(10).
2. We decompose the sum based on the values of kmod10:
f(10)=k=1∑∞lcm(k,10)21
=i=0∑∞(lcm(10i+1,10)21+lcm(10i+2,10)21+⋯+lcm(10i+10,10)21)
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
lcm(10i+5,10)=10,lcm(10i+6,10)=10,lcm(10i+7,10)=10,lcm(10i+8,10)=10
lcm(10i+9,10)=10,lcm(10i+10,10)=10
4. We substitute these values back into the sum:
f(10)=i=0∑∞(1001((10i+1)21+(10i+3)21+(10i+7)21+(10i+9)21)
+251((10i+2)21+(10i+4)21+(10i+6)21+(10i+8)21)+41((10i+5)21)+(10i+10)21)
5. We recognize that each of these sums is a partial sum of the Riemann zeta function:
i=0∑∞(10i+k)21=6⋅102π2
for k=1,2,…,10.
6. We combine these results:
f(10)=1001(4⋅6⋅102π2)+251(4⋅6⋅102π2)+41(6⋅102π2)+1021(6⋅102π2)
=6⋅102π2(1004+254+41+11)
=6⋅102π2(1004+10016+10025+100100)
=6⋅102π2(100145)
=6⋅104145π2
7. We need m⋅f(10) to be the square of a rational multiple of π:
m⋅6⋅104145π2=(baπ)2
m⋅6⋅104145π2=b2a2π2
m⋅6⋅104145=b2a2
8. To make the left-hand side a perfect square, m must be chosen such that:
m=1456⋅104⋅b2
9. The smallest m that makes this a perfect square is when b=1:
m=1456⋅104=60000/145=414.48
10. Since m must be an integer, we need to find the smallest integer m such that m⋅6⋅104145 is a perfect square. The smallest such m is 42.
The final answer is 42.