\(10A=10^{2012}+10^{2013}+10^{2014}+...+10^{2019}+160\)
\(9A=10A-A=10^{2019}-10^{2011}+160-16\)
\(9A=10^{2011}\left(10^8-1\right)+9\cdot16\)
\(9A=10^{2011}.99999999+9.16\)
\(9A=10^{2011}.11111111.9+9.16\)
\(A=10^{2011}.11111111+16\)
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\(A⋮48\Rightarrow A⋮16;A⋮3\) (1)
\(10:3\) dư 1
\(10^2:3\) dư 1
...
\(\Rightarrow10^{2011}:3\) dư 1
\(11111111=11100000+11100+11\)
\(11100000⋮3;11100⋮3;11:3\) dư 2
\(\Rightarrow11111111:3\) dư 2
\(16:3\) dư 1
\(\Rightarrow A:3\) dư \(1.2+1=3\)
\(\Rightarrow A⋮3\) (2)
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\(10^{2011}=2^{2011}.5^{2011}=2^4.2^{2007}.5^{2011}⋮2^4=16\)
Vì \(10^{2011}⋮16\) \(\Rightarrow10^{2011}.11111111⋮16\)
\(16⋮16\)
\(\Rightarrow A⋮16\) (3)
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Từ (1), (2) và (3) suy ra: \(A⋮48\) (đpcm)