hazelnut
For positive integers n and m \(n>m\), is the average (arithmetic mean) of \(4*10^n\), \(4*10^{n-1}\), ......, and \(4*10^{n-m}\) an integer?
1) \(m<6\)
2) \(n=10\)
Asked: For positive integers n and m \(n>m\), is the average (arithmetic mean) of \(4*10^n\), \(4*10^{n-1}\), ......, and \(4*10^{n-m}\) an integer?
Number of terms = n - (n-m) +1 = m+1
the average (arithmetic mean) of \(4*10^n\), \(4*10^{n-1}\), ......, and \(4*10^{n-m}\) = \(\frac{4*10^n + 4*10^{n-1} + ......+4*10^{n-m}}{(m+1)}\)
1) \(m<6\)
Let us take m=0
\(4*10^n/1 = 4*10^n\) Integer
Let us take m=1
\(\frac{4*10^n + 4*10^{n-1}}{2} = \frac{4*10^{n-1} ( 1+10)}{2} =\frac{4*10^{n-1} *11}{2}\) Integer
Let us take m=2
\(\frac{4*10^n + 4*10^{n-1} + 4*10^{n-2}}{3} = \frac{4*10^{n-2} ( 1+10+100)}{3} =\frac{4*10^{n-2} *111}{3}\) Integer
Let us take m=3
\(\frac{4*10^n + 4*10^{n-1} + 4*10^{n-2} + 4*10^{n-3}}{4} = \frac{4*10^{n-3} ( 1+10+100+1000)}{4} = \frac{4*10^{n-3} *1111}{4}\) Integer
Let us take m=4
\(\frac{4*10^n + 4*10^{n-1} + 4*10^{n-2} + 4*10^{n-3} + 4*10^{n-4}}{5} = \frac{4*10^{n-4} ( 1+10+100+1000+10000)}{5} = \frac{4*10^{n-4} *11111}{5}\) Integer
Let us take m=5
\(\frac{4*10^n + 4*10^{n-1} + 4*10^{n-2} + 4*10^{n-3} + 4*10^{n-4} + 4*10^{n-5}}{6} = \frac{4*10^{n-5} ( 1+10+100+1000+10000 +10^5)}{6} =\frac{4*10^{n-5} *111111}{6}\) Integer
SUFFICIENT
2) \(n=10\)
Since if we take n=10 and m=6
The expression will not be divisible by 7
NOT SUFFICIENT
IMO A