1986-revealed-1

Let the number be a

Now, $a1986 \equiv 0 \pmod{1987}$ [where $a1986 \neq a*1986$ ]

We can write a1986 as $a*{10}^{4} + 1986$

$a*{10}^{4} + 1986 \equiv 0 \pmod{1987}$

$a*{10}^{4} - 1 \equiv 0 \pmod{1987}$

$a*{10}^{4} \equiv 1 \pmod{1987}$

Now, ${10}^{4} \equiv 65 \pmod{1987}$

$65*a \equiv 1 \pmod{1987}$

Now, use Euclid's algorithm,

$1987 = 65*30 + 37$

$65 = 37 + 28$

$37 = 28 + 9$

$28 = 9*3 + 1$

Hence,

$1 = 28- 9*3$

$= 28 - 3(37 - 28)$

$= 4(65-37) -3(37)$

$= 4(65) - 7(1987-65*30)$

$1 = 214(65) - 7(1987)$

Hence, 214(65) = 1 mod 1987

As a result a = 214, therefore smallest positive integer is $\boxed{2141986}$

Here's a simple python script that gives you the answer:

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n=1
while(True):
    if((n*10000+9999)%1987==0):
        break
    n = n + 1
print(n*10000+9999+1987)

using $Mathematica$

Select[Range@1000,Mod[FromDigits@Join[IntegerDigits@#,{1,9,8,6}],1987]==0&]

returns 214