Puzzle #1:

 

58.

 

A whole number that when divided by 5 leaves a remainder of 3 must end in 3 or 8 (since only numbers ending in 5 or 0 are divisible by 5).  All numbers ending in 3 or 8 are possible.

A whole number that when divided by 4 leaves a remainder of 2 must end in an even digit.  Since all three requirements must be met, the number cannot end in 3, and must end in 8.  The sequence of numbers would be 18 (16+2), 38 (36+2), 58 (56+2), 78 (76+2), 98 (96+2), 118 (116+2)...

A whole number that when divided by 3 leaves a remainder of 1 ends in 8 only when one number less ends in 7 and is divisible by 3.  This would be 27, 57, 87, 117... (intervals of 30, making possible choices 28, 58, 88, 118...).

The first number in common is 58.

 

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