Yes, I agree, it doesn't' make sense if you look at it from price perspective.
I will try to write it again, with more what I have in mind.
According to intel the i9 13900kf max RAM speed support is 5600 MT/s
I was told that MT/s is not the same as MHz and that I will need 6000 MHz ram to get 5600 MT/s.
So if my current ram is 5200 MHz and the above is correct, then my build is not ussing 100% of possible speed. I'm talking about stock not overclocked possibility.
My question is (ignoring price) what ram will I have to have to benefit all the stock speed my build has?
A 5600 MT/s one as i wrote above?
I should have been a bit more clear that i ment 5600mt and that the mobo supports a max of 6400mhz. 6400 would count as OC (if i did not read it wrong)
Think is most stores market MT/s as the same as MHZ, so a 5600M/T one is marketed as 5600MHZ.
And then we have the thing with CL wich is a storry in on itself. (MTs to MHZ used to be the same numbers so thats where it comes from)
Kingston seem to list all their RAM on their site as MT/s while Corsair atleast in some cases just list a number without realy telling what it means. like 5600. On a store i checket the kingston 5600 MTs modules coresponded exactly to the stores listed 5600mhz of the same ram module.
If you realy want to Min/Max i sugest you read up on it and ask people allot more techsavy than me. You are fighting not only the marketing of RAM but allso the marketing of Motherbord supported speeds. In your case the Motherboard is rated to 6400MHZ but do they mean MHZ or MT? It's a **** sandwitch allot of people been complaining about for a long time but MHZ is easyer to market since it has been done for a long time, just like 0-100 (or 0-62 for all you weird people) says allmost nothing about how a car performes, but that is how they have markete cars for years and years so they keep doing it. You could look at kingstons homepage and see what module they list at your requested speed and then just buy the same one from a reseller, if you want Kingston that is.