Originally Posted by

**tonio** ]

You can do it by "gentle brute force": from the conditions the multiple, let's call it x, is odd and ends either in 1 or 6 ==> it ends in 1 (why?)

So you must concentrate on multiples of 11 ending in 1; there aren't that many: 11, 121, 231 341,...,1001, 1111, 1221, 1331, etc.

Since it must be 1 + 3k, for some integer k, this rules out 11, 121,1001, etc.

Of course, you could read carefully the proof of the Chinese Remainder Theorem and thus construct your number.

Tonio