Find the smallest 3-digit number which gives the remainder 1 when divided by 3 and by 4.

Multiples of 3 plus 1 remainder are:

3n+1

Multiples of 4 plus 1 remainder are:

4n+1

Since these multiples must be the same number must be a multiple of 12 plus a remainder of one, which is:

12n+1

And the smallest such three digit number is:

12n+1>99 subtract 1 from both sides

12n>98 divide both sides by 12

n>8.17

Since n must be an integer, n=9

So then number is 12n+1

12 (9) + 1=109