143 lines
1.2 KiB
Markdown
143 lines
1.2 KiB
Markdown
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## Forensics 101
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The main idea finding the flag using basic forensics training.
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#### Step-1:
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After we download the given image `95f6edfb66ef42d774a5a34581f19052.jpg` from the cloud, we just try simple techniques.
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<img src="95f6edfb66ef42d774a5a34581f19052.jpg">
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#### Step-2:
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I first tried a very basic `strings 95f6edfb66ef42d774a5a34581f19052.jpg`
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#### Step-3:
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It just gave me this output:
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```
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JFIF
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, #&')*)
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-0-(0%()(
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((((((((((((((((((((((((((((((((((((((((((((((((((
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L?~f
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:UwR
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y>2|
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*'?-
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yhH_&
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Lmz'
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+f[
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!"1$246B`35A
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au>
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~b*D
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F_X:D
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[ElC
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him8
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lr|.
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L{2^
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]]te
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tBfE
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j_s7Os/[i
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W="'
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"fkO
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G&,ke:
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eM_F
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8O:J
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9)/m>&
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[P{!
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}EI5#
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a~Wt,1
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]<e<
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g:rc
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"1Pa
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ujM^P
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P#3a
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vFGO
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ZniFi
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%a ?}
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2AQaq 0B#@r
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\xr>
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\37g
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4=i#N
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F:Jx
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.`Ot
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v[xU
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|='u
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{~T{@
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LX.9
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?mBx
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L1QG
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lIYB+
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|] %
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QTj?:
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$*,-
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pk4D
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4R gX
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$mmk
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_QSK
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b)^L
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a[T=
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mpj<N;
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!1AQaq
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x'<L
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we|C{
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iiR8
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"'A*
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aZ(4
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cj}+
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NL7'4
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u-kF3n
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x cd
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s}F3_Y
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$+Ym
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zrsq
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f}]@
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L&/F
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&F'$,
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| *
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ktWxn
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yLX*s]
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pT)J
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eR-3
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SG=3\Y
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JK%0
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h"(p
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!01Q
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%^cJ
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H|cO
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!10AQ
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'([P
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!1AQaq
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s`%"
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X`3d'
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'P]!
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;#_l
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ABx=~
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"r&"
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0YAP
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ch-h
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XoXrl
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l0Je
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V^_W
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xp7p
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v{*{8
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=k"$TW3G
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1)j!
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7y}U
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<~0GD
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n%CeoQ=m8
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`"n<P
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i}\D
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X`(
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8kF=
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~9%]Tn
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flag{wow!_data_is_cool}
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$lqU
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AG{u
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Xm*CnC
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@'hnQ
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ax+p
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bdQG
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D_ O
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```
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#### Step-4:
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Finally the flag becomes:
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`flag{wow!_data_is_cool}`
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