It's a system of equations. 2 equations, 2 unknowns. The number of boxes of 10. And the number of boxes of 100 are your unknowns. Call these x and y respectively. (Or your x1 x2)
The two equations come from the following conditions:
As an aside. This problem is dumb. Fuck brian. Fuck his paperclips.
Thanks.
The problem says “Some boxes hold 10 clips and some boxes hold 100. He has some clips left over.” I think it’s a poorly worded problem but let’s just suppose that “some” means 2 minimum. I read a comment that went like “If you’ll say that there are some donuts and I’ll find out there’s only 1 or 0 donut, I’ll be disappointed.” Sensible.
It’s assumed that Brian put clips in boxes as much as possible, so the number of leftovers is less than 10.
x2 = x1 + 3
n = x2 − 2
x2 should be 5 minimum
The total number of clips is 523, 634, 745, 856, 967, 1078 or 1189.
I asked Llama 3.1 (405b), Claude 3.5 sonnet and Chatgpt 4o after I solved it. I edited the problem to improve it. I was curious if any of those could solve it. Claude 3.5 sonnet and Chatgpt 4o did. Llama 3.1 (405b) got 523 but didn’t talk re the other answers. A follow-up of “Are there other answers?” yielded all the 7 answers.
You are getting close. You are getting so many answers because you introduced the variable n. This makes it 2 equations and 3 unknowns. You need to cast the 2nd equation in terms of the number of boxes and the number of paper clips in each box rather than an additional variable. Here's a hint. If:
x1 * 100 = 300 then x1=3
I.e. the total number of paperclips from the 100 paperclip boxes is 300 if there are 3 boxes of them