How to summarize routes

To summarized these you need two bits which gives you posibilities:
00
01
10
11

.4 in binary = 0000 0100
.5 in binary = 0000 0101
.6 in binary = 0000 0110
.7 in binary = 0000 0111

You see that the first six bits never change from .4 to .7 so you can summarize at .4 with 22 bits
172.1.4.0/22

How would you summarize:

10.1.1.0 /30
10.1.1.4/30
10.1.1.8 /30
10.1.2.0 /23
10.1.4.0 /24
10.1.5.0 /25

You can't.
Not cleanly anyway.

That's why networks need good planning before designing so that proper route summarization can be done.  Even with proper planning it may never come out perfect.

Of course you can summarize this! There are no statement in the question that the network is discontiguous. If the networks do not have any 10.X.X.X addresses anywhere else in the topology the summarization gets an easy 10.0.0.0/8. If the question states summarize with the longest mask possible with one summary then you will get 10.1.0.0/21 with three “borrowed” bits. This gets a summary from 10.1.0.0 to 10.1.7.255.

If the question states that addresses from 10.1.6.0 and up may be in other networks then we can make two summaries to shorten the routing table:
10.1.0.0/22
10.1.4.0/23

But never say that you can’t summarize without nowing all the facts and also never summarize without nowing all the facts.

Oh, common'
Of course it is discontinuous:

10.1.1.2/30 is missing
10.1.1.6/30 is missing
10.1.1.10 through 255 is missing

The only way you can cleanly summarize 10.1.2.0 /23, 10.1.4.0 /24, 10.1.5.0 /25 would be to include 10.1.5.128/25 and 10.1.6.0/24.  In which case you could use 10.1.2.0 mask 255.255.252.0.   But 10.1.5.128/25 and 10.1.6.0/24 is in fact missing so you cannot summarize it that way cleanly.

If you summarize segments into your summarization that do not actually exist you can have issues if those segment come into play elsewhere over different paths.
Now if you do not care about doing that,..then yes you can use 10.0.0.0/8 or 10.1.0.0/21, of course. I do that myself with ours by using 192.168.0.0/16 even though I don't actually use all 256 segments,..but I do so with the full knowledge that I may have to change that later.

So I am answering the question exactly literally based on what was literally listed on the list.   I even ended it with the qualification of "not cleaning anyway" and that is exactly true.

I agree with you but not with the answer "you can't"
Like I said it depends on the design and gather all the  facts without guessing...

I agree with you but not with the answer "you can't"

Well,..that was not the answer I gave,...the answer I gave was "You can't. Not cleanly anyway"

You sure you're not just trying to "one up" me because I gave a different original answer to the original question than you did? The original question was "Can you provide an easy way to summarize routes?"  I took note of the term "easy way" and told them to forget the binary and do it the quick easy method where you only deal with the decimal numbers,...which is a lot easier and does fine in most situations.  I'm just trying to answer what the guy asked in the context he asked it,...that's all.

I sorry I didn't mean to "up you"... but we experts must check and validate each others answers and this time I did not agree with you and I had to point that out.

But I agree with your first answer, it is clean (and notice my finger here ;-) ) and you can count like that IF you understand the rules that governs summarization/subnetting. I'm a certified instructor in these mather and you can't guess to how subnetting often look when the students do not understand the binaries that lines beneath and tries your method - big FAIL. But of course, I also uses it!

I understand. No problem.  :-)
I'm not an instructor, but am on the IT Advisory Committee at a local college.

I still refer back to binary on some things but I built some spreadsheets that shows all the logic in Decimal to make it easier.  I don't use IP Calculators because I always want to force myself to think it through and understand what I am really doing.