$a\log_{10}2+b\log_{10}3+c\log_{10}5+d\log_{10}7= 2018.$

$a,b,c$ and $d$ are integers that satisfies the equation above. Find $a+b+c+d$ .

1009
4036
2017
2018
4034

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If $\log_{c}{b}=a$ then $c^a=b$

$\log_{m}{n}+\log_{m}{l}=\log_{m}{(n.l)}$

$a\log_{c}{b}=\log_{c}{b^a}$

So, given equation can be written as

$\log_{10}{(2^a.3^b.5^c.7^d)}=2018$

Removing logarithm

$10^{2018}=2^a.3^b.5^c.7^d$

$2^{2018}.5^{2018}=2^a.3^b.5^c.7^d$

Equating the exponents we get

$a=2018,b=0,c=2018,d=0$

So, $a+b+c+d=\boxed{4036}$