[content] evaluation/easy_problems.jsonl (jsonl)
{"input": [{"role": "user", "content": "Solve the following math problem. Make sure to put the answer (and only answer) inside \\boxed{}.

How many natural numbers greater than 6 but less than 60 are relatively prime to 15?"}], "output": "<think>
Okay, so I need to find how many natural numbers greater than 6 and less than 60 are relatively prime to 15. Hmm, let's break this down step by step. 

First, let me recall what \"relatively prime\" means. Two numbers are relatively prime if their greatest common divisor (GCD) is 1. So, I need to count all numbers between 7 and 59 (since it's greater than 6 and less than 60) that don't share any common factors with 15 except 1. 

15 factors into prime factors as 3 × 5. So any number that is not divisible by 3 or 5 will be relatively prime to 15. That means I need to exclude numbers in that range that are multiples of 3 or 5. 

First, let's figure out the total number of numbers we're considering. The numbers start from 7 and go up to 59. To find the count: 59 - 7 + 1 = 53 numbers. Wait, is that right? Let me check. If you subtract 6 from 59, you get 53, but since we start at 7, which is 6 + 1, yeah, the count is 53. So there are 53 numbers in total.

Now, I need to subtract the numbers that are multiples of 3 or 5 in this range. But I have to be careful not to double subtract those that are multiples of both 3 and 5 (i.e., multiples of 15). So this sounds like using the inclusion-exclusion principle.

First, find the number of multiples of 3 between 7 and 59. Then find the number of multiples of 5 in the same range. Then subtract those, but add back the multiples of 15 that were subtracted twice.

Let me start with multiples of 3. The smallest multiple of 3 greater than 6 is 3×3=9, and the largest multiple of 3 less than 60 is 59 ÷ 3 = 19.666..., so 19×3=57. Now the number of multiples is from 9 to 57, stepping by 3. The formula for the number of terms is ((last - first)/step) +1. So (57 -9)/3 +1 = (48/3)+1 =16 +1=17. So 17 numbers divisible by 3.

Next, multiples of 5. The smallest multiple of 5 greater than 6 is 10, and the largest less than 60 is 55. The numbers are 10,15,...,55. Number of terms: (55 -10)/5 +1 = (45/5)+1=9+1=10. So 10 numbers divisible by 5.

Now, multiples of 15 (since LCM(3,5)=15). The smallest multiple of 15 greater than 6 is 15, and the largest less than 60 is 60 -15=45? Wait, 15×3=45, 15×4=60, but 60 is not less than 60, so the largest is 45. So from 15 to 45, stepping by 15. Let's see: 15,30,45. How many? (45 -15)/15 +1 = (30/15)+1=2+1=3. So 3 numbers divisible by 15.

Applying inclusion-exclusion: the number of numbers divisible by 3 or 5 is (17 +10) -3=24. Therefore, the numbers not divisible by 3 or 5 (i.e., relatively prime to 15) would be total numbers (53) minus 24, which is 53 -24=29.

Wait, but wait a minute. Let me double-check the count of the multiples. Maybe I made a mistake there.

Starting with multiples of 3:

First multiple of 3 ≥7 is 9 (since 6 is excluded). Last multiple of 3 <60 is 57. The sequence is 9,12,...,57. The number of terms can be calculated by ((57 -9)/3) +1 = (48/3)+1=16 +1=17. So that's correct.

Multiples of 5:

First multiple of 5 >6 is 10. Last multiple of 5 <60 is 55. So numbers: 10,15,...,55. The count is ((55 -10)/5)+1=(45/5)+1=9+1=10. That's correct.

Multiples of 15:

First multiple of 15 >6 is 15. Last multiple of 15 <60 is 45. So numbers:15,30,45. So 3 terms. Correct.

So inclusion-exclusion gives 17+10-3=24. Total numbers 53. So 53-24=29.

Wait, but let's verify with another approach. Maybe list the numbers and count, but that's tedious. Alternatively, use Euler's totient function. Wait, Euler's totient function φ(n) gives the count of numbers less than n that are coprime to n. But here, we have numbers from 7 to 59. So the problem isn't exactly φ(15), but let's see.

Alternatively, if we think of the numbers from 1 to 59, then subtract the numbers from 1 to 6. But maybe this complicates.

Alternatively, φ(15) = 8, since numbers 1,2,4,7,8,11,13,14 are coprime to 15. So every 15 numbers, there are 8 coprime numbers. So perhaps the density is 8/15. But our range is 53 numbers. However, 53 divided by 15 is 3*15=45, so 3 full cycles of 15 and then 8 more numbers. Let me see if that's a useful approach.

Wait, but 7 to 59 is 53 numbers. Let's see, 15*3=45, so 45 numbers in 3 cycles. Then 53-45=8. So perhaps there's 3 full cycles (each contributing 8 coprime numbers) and then 8 numbers. Let's check that.

But wait, the first number in the range is 7, which is 7 mod 15. So the cycle might not align. For example, the first 15 numbers from 1-15, next 16-30, etc. But our range is 7-59, which spans from 7-15 (9 numbers), 16-30 (15 numbers), 31-45 (15 numbers), 46-59 (14 numbers). Hmm, perhaps splitting the range into these intervals.

But maybe this complicates. Alternatively, using inclusion-exclusion as before, the answer is 29.

But let's verify by considering the counts.

Total numbers:53.

Numbers divisible by 3:17. Numbers divisible by 5:10. Numbers divisible by 15:3.

So total numbers not divisible by 3 or 5:53 -17 -10 +3=29. That's the inclusion-exclusion result. So seems correct.

But let's check a small example to see if inclusion-exclusion works. Suppose the problem was numbers greater than 6 and less than 16. So numbers from 7 to 15. Total numbers:9. Numbers coprime to 15: numbers not divisible by 3 or 5.

Numbers in 7-15:7,8,9,10,11,12,13,14,15.

Which of these are divisible by 3:9,12,15.

Divisible by 5:10,15.

Numbers to exclude:9,12,15,10. Wait, 15 is in both. So total excluded numbers:3 (multiples of 3) +2 (multiples of 5) -1 (multiple of 15)=4. So 9 total numbers -4=5 coprime numbers. Let's list them:7,8,11,13,14. Yes, those are 5 numbers, which is 9 -4=5. So 5 numbers. φ(15)=8, but since we started at 7, which is after 1-6, which includes 1,2,4,7 (but wait, φ(15) is 8 numbers from 1-14. Wait, 1-14: φ(15)=8. But in our small example, 7-15 is 9 numbers, but 15 is excluded. Wait, no, original small example was up to 15, which is included. Wait, in my hypothetical example: numbers greater than 6 and less than 16: so 7 to 15 inclusive. 15 is included? Wait, if the upper limit is less than 16, then 15 is included. Wait, the original problem is less than 60, so 59. So in the small example, numbers greater than 6 and less than 16:7-15. So 9 numbers. And among them, 15 is included and is divisible by both 3 and 5. So the count for that is 5 numbers coprime to 15. That works with inclusion-exclusion. So that method works. So our original method of inclusion-exclusion should be correct.

Therefore, for the original problem, 29 numbers. So the answer is 29. Let me just check once more.

Total numbers: from 7 to 59 inclusive. 59-7+1=53.

Multiples of 3: starting at 9 (3×3) up to 57 (3×19). Number of terms: (57-9)/3 +1=48/3 +1=16 +1=17. Okay.

Multiples of 5: starting at 10 (5×2) up to 55 (5×11). (55-10)/5 +1=45/5 +1=9+1=10. Okay.

Multiples of 15:15,30,45. That's three numbers. So inclusion-exclusion:17+10-3=24.

So 53-24=29. Seems correct.

I think that's the answer. So the number of natural numbers greater than 6 and less than 60 that are relatively prime to 15 is 29.
</think>

The natural numbers greater than 6 and less than 60 are from 7 to 59, inclusive. There are \\(59 - 7 + 1 = 53\\) numbers in total. We need to count numbers that are relatively prime to 15, which means they are not divisible by 3 or 5.

1. **Multiples of 3**: The smallest multiple of 3 greater than 6 is 9, and the largest less than 60 is 57. The sequence is \\(9, 12, \\ldots, 57\\), which has \\(\\frac{57 - 9}{3} + 1 = 17\\) terms.
2. **Multiples of 5**: The smallest multiple of 5 greater than 6 is 10, and the largest less than 60 is 55. The sequence is \\(10, 15, \\ldots, 55\\), which has \\(\\frac{55 - 10}{5} + 1 = 10\\) terms.
3. **Multiples of 15 (LCM of 3 and 5)**: The smallest multiple of 15 greater than 6 is 15, and the largest less than 60 is 45. The sequence is \\(15, 30, 45\\), which has \\(\\frac{45 - 15}{15} + 1 = 3\\) terms.

Using the inclusion-exclusion principle, the number of numbers divisible by 3 or 5 is \\(17 + 10 - 3 = 24\\). Subtracting these from the total gives \\(53 - 24 = 29\\).

\\boxed{29}", "category": "math", "license": "cc-by-4.0", "reasoning": "on", "generator": "Qwen-2.5-32B-Instruct, DeepSeek-R1", "used_in_training": "Ultra, Super, Nano", "version": "v1.1", "system_prompt": "detailed thinking on"}