# Please explain the logic behind the Capture Group Challenge

Tell us what’s happening:

I would like someone to walk me through the answer, please. This is how I thought my current answer would work:

1. `^(\d+)\s`- At the beginning of the string, find any digit followed by a space.
2. `\1` - Repeat that pattern, which is `(\d+)\s` . So currently I would have `['42 42 '].`
3. ` \d+\$` - At the end of the string, find any digit. So I’d assume I would have `['42 42 42']`. I’d have three digits because there were three \d+'s.
4. I didn’t call on the pattern again.

I’m wrong. The solution is `/^(\d+)\s\1\s\1\$/;`. I understand it up until the next `...\s\1\$/;` I understand that `/^(\d+)\s\1` would give me` ['42 42 ']` because the pattern is called two times, once explicitly and another using \1, which is what I put. So I’m thinking we’d be up the string at the third digit. But why is there another space followed by the same pattern? Wouldn’t the regex try to find `['(42 space) (42 space) space (42 space)']`?

``````
let repeatNum = "42 42 42";
let reRegex = /^(\d+)\s\1\d+\$/; // Change this line
let result = reRegex.test(repeatNum);

``````

User Agent is: `Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/84.0.4147.135 Safari/537.36`.
The number will only match the capture groups (expressions in the parenthesis), so just `\d+`. Hope this helps.