Let S be the set of all positive integers that, when divided by 8, have a remainder of 5. What is the 76th number in this set?
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Got 613 but I did it the slow way (started at 5 and added 8 until I hit the 76th term). Is there a faster formula for these?
Yes, these are just an arithmetic sequence. First term 5, common difference 8, so the nth term is 5 + 8(n-1). Plug in n=76.
That's what I did after wasting a minute listing terms the first time lol
Careful: the first term is 5, not 13. I initially started at 13 and got the wrong answer. Once I used 5 + 8(75), it worked.
Is the 76th term just 5 + 8*76? I keep second-guessing whether the first term counts as term 1.
Yes, 5 is the 1st term. So it's 5 + 8*(76-1), not 5 + 8*76.
These remainder-sequence problems are so easy to overthink. Just remember a = first valid number, d = divisor. Done.
What it tests
Your ability to spot and extend arithmetic and geometric sequences and work with recursive definitions.
Common trap
Off-by-one errors in indexing (a₀ vs a₁) or confusing an arithmetic difference with a geometric ratio.