The Steenrod operations on mod 2 cohomology imply the vanishing of some characteristic numbers. Specifically, if $p(w_1,w_2,\ldots)\in H^k(M^n;Z/2)$ for $k\lt n$ then $0=\langle \sum_{i+j=n-k}w_i{\rm sum_{i+j=n-k}u_i{\rm Sq}^{j} p, [M^n]\rangle$. M^n]\rangle$where$u_i$is the Wu class of$M$(but take this with a grain of salt, since I'm quoting from memory here). Thom showed that all relations between characteristic numbers arise in this way. This allowed him to compute the bordism ring of unoriented manifolds exactly: it is$Z/2[m_k\vert k$not of the form$2^j-1]$. This gives you the tight bound you were asking for. But, as others have already pointed out, there are plenty of good books & papers on this, so you should probably just dive into the library. 2 corrected a mistake The Steenrod operations on mod 2 cohomology imply the vanishing of some characteristic numbers. Specifically, if$p(w_1,w_2,\ldots)\in H^k(M^n;Z/2)$for$k\lt n$then$0=\langle {\rm Sq}^{n-k\sum_{i+j=n-k}w_i{\rm Sq}^{j} p, [M^n]\rangle$. Thom showed that all relations between characteristic numbers arise in this way. This allowed him to compute the bordism ring of unoriented manifolds exactly: it is$Z/2[m_k\vert k$not of the form$2^j-1]$. This gives you the tight bound you were asking for. But, as others have already pointed out, there are plenty of good books & papers on this, so you should probably just dive into the library. 1 The Steenrod operations on mod 2 cohomology imply the vanishing of some characteristic numbers. Specifically, if$p(w_1,w_2,\ldots)\in H^k(M^n;Z/2)$for$k\lt n$then$0=\langle {\rm Sq}^{n-k} p, [M^n]\rangle$. Thom showed that all relations between characteristic numbers arise in this way. This allowed him to compute the bordism ring of unoriented manifolds exactly: it is$Z/2[m_k\vert k$not of the form$2^j-1]\$. This gives you the tight bound you were asking for.