Vanishing of characteristic numbers vs vanishing of characteristic classes

Question

A famous result by Thom states that Oriented Bordism classes are determined by characteristic numbers; specifically, two closed manifolds are orientedly bordant if and only if they have the same Stiefel-Whitney and Pontryagin numbers (I'll just talk about Stiefel-Whitney for brevity). An immediate consequence is that if $M$ is a closed manifold which has a non-vanishing Stiefel-Whitney number involving $w_k$ for some $k$, then $w_k(N)\neq 0$ for any $N$ which is bordant to $M$; in other words this non-vanishing characteristic number provides a "bordism reason" for why a characteristic class should be non-zero.

My question concerns the converse. Given an $M$, suppose that for some $k$ every Steifel-Whitney number involving $w_k$ vanishes, so that there is "no bordism reason" for the class to be non-vanishing: is it then possible to find an $N$ which is bordant to $M$ and has $w_k(N)=0$? If so, is it possible to simultaneously eliminate all classes which have no bordism reason to be non-zero?

This seems like something which maybe shouldn't be expected since characteristic classes often provide obstructions to doing surgery, and two manifolds are orientedly bordant exactly when they differ by a finite sequence of surgeries. On the other hand the manifold $\mathbb{CP}^n\#\overline{\mathbb{CP}^n}$ is null-bordant even though it has many non-vanishing characteristic classes for $n>1$ (half the time it isn't even spin).

Answer 1

This is not a complete answer, but rather two specific instances where the answer is yes.

1. If every Stiefel-Whitney number of $M$ involving $w_1$ vanishes, then $M$ is cobordant to an orientable manifold $N$. See Proposition 4 of Wall, C. T. C., Determination of the cobordism ring, Ann. Math. (2) 72, 292-311 (1960). ZBL0097.38801.

2. If every Stiefel-Whitney number of $M$ involving $w_1$ and $w_2$ vanishes, then $M$ is cobordant to a spin manifold $N$. See Corollary 2.4 of Anderson, D. W.; Brown, Edgar H. jun.; Peterson, Franklin P., The structure of the Spin cobordism ring, Ann. Math. (2) 86, 271-298 (1967). ZBL0156.21605.

Answer 2

I suggest to consider $\mathbb CP^4 \# (\mathbb CP^2 \times \mathbb CP^2)$. Then all SW-numbers involving $w_4$ vanish, but on the other hand we have $\text{Sq}^2w_4 = w_2w_4 + w_6$ in $H^{\ast}(BSO;\mathbb F_2)$, so for any manifold with $w_4 = 0$ we also have $w_6 = 0$. But $\mathbb CP^4 \# (\mathbb CP^2 \times \mathbb CP^2)$ has nontrivial SW-number $w_2w_6$, so no manifold cobordant to it can have vanishing $w_6$.

EDIT: As pointed out in a comment below, this is wrong.