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A more elegant method than brute force is :

you write 5t +3 /8z+6 / 12p+10 = 13k multiple of 13
They have the same negative remainder (positive remainder less quotient ) which is -2

Thus they are 5t-2 / 8z-2 /12p -2

When these forms have same positive or negative remainder they can be expressed as LCD (5,8,12)M - 2 with M costant

Thus a number both divisible by 5,8,12 that has remainder -2 is 30M-2 =13k

You notice that if you plug numbers they all finish by 8 thus only BCD can be.

You start dividing by 13 at the center and find C.
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We can see that the difference between divisors (5,8,12) and the respective remainders (3,6,10) is constant. So, the least such number will be 2 less than the LCM of 5,8,12 =>120-2=118

The second such number will be, 118+120 = 238.
The third such number will be, 238+120 = 358....and so on.

118 leaves a remainder of 1 with 13.
120 leaves a remainder of 3 with 13.

To arrive at required number the remainder with 13 will be 0 or 13 = 1+3+3+3+3.

So the least such number will be, 118+(120*4) = 118+480 = 598

Option C
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