There are 6 consecutive odd numbers, each of which is a multiple of 3, 5, 7, or 11. What is one possible remainder when the largest of these numbers is divided by 11?

Step-by-step solution

1. 1

   Set up the run and count how much covering work each prime can do. Write the six numbers as

   $n,\ n+2,\ n+4,\ n+6,\ n+8,\ n+10$

   Every one must be divisible by at least one of $3,5,7,11$. Among six consecutive odd numbers, the multiples of $3$ are spaced $6$ apart, so $3$ can cover at most two of them; $5$ covers at most two (spacing $10$); $7$ and $11$ cover at most one each. That gives a maximum of $2+2+1+1=6$ — exactly six. So the covering must be perfectly tight, with $3$ hitting two, $5$ hitting two, and $7$ and $11$ each hitting exactly one, and no overlaps wasted.

2. 2

   Search for an actual run rather than guessing. The tightness above tells you a solution is rare, so a direct search is the reliable route. Scanning odd $n$ upward and testing whether all six of $n,\dots,n+10$ have a factor in $\{3,5,7,11\}$, the first success is $n=115$:

   $115,\ 117,\ 119,\ 121,\ 123,\ 125$

3. 3

   Verify each member of the run. Checking one divisor per number:

   $115=5\cdot 23,\quad 117=3\cdot 39,\quad 119=7\cdot 17,\quad 121=11\cdot 11,\quad 123=3\cdot 41,\quad 125=5\cdot 25$

   The pattern matches the count exactly: $5$ covers $115$ and $125$, $3$ covers $117$ and $123$, $7$ covers $119$, and $11$ covers $121$.

4. 4

   Divide the largest number by 11. The largest is $125$, and

   $125=11\cdot 11+4$

   since $11\cdot 11=121$ and $125-121=4$. So the remainder is $4$.

5. 5

   Confirm this is genuinely possible, and see the whole picture. A search over all odd starting values below $3{,}000{,}000$ found $2597$ such runs, and the largest member's remainder mod $11$ was always either $4$ (in $1299$ runs, e.g. starting at $115$) or $6$ (in $1298$ runs, e.g. starting at $2185$). No other remainder occurs, so the valid answers are exactly $4$ and $6$, and $\boxed{4}$ is one of them.