What is the least number which when divided by the numbers 3, 5, 6, 8, 10 and 12 leaves in each case $a$ remainder 2 but which when divided by 13 leaves no remainder?

Step 1 (Formulate Modular Congruences): Let the least positive integer be $N$. We are given:

1. $N\equiv 2(\mathrm{mod}3)$

2. $N\equiv 2(\mathrm{mod}5)$

3. $N\equiv 2(\mathrm{mod}6)$

4. $N\equiv 2(\mathrm{mod}8)$

5. $N\equiv 2(\mathrm{mod}10)$

6. $N\equiv 2(\mathrm{mod}12)$

7. $N\equiv 0(\mathrm{mod}13)$

Step 2 (Combine the first six congruences): The first six congruences imply that $N-2$ is a common multiple of $3,5,6,8,10,12$:

Therefore:

Step 3 (Solve the Modulo 13 Congruence): We substitute $N=120k+2$ into the divisibility by 13 condition:

Since $120=13\times 9+3\equiv 3(\mathrm{mod}13)$:

Testing integer values for $k$:

• $k=1⟹3\not\equiv 11$

• $k=2⟹6\not\equiv 11$

• $k=3⟹9\not\equiv 11$

• $k=4⟹12\not\equiv 11$

• $k=5⟹15\equiv 2\not\equiv 11$

• $k=6⟹18\equiv 5\not\equiv 11$

• $k=7⟹21\equiv 8\not\equiv 11$

• $k=8⟹24\equiv 11\equiv 11(\mathrm{mod}13)$ (This is true!)

So the smallest positive integer $k$ is $8$.

Step 4 (Compute N):

Step 5 (Conclusion): The least such number is exactly 962.