neuronxcc-2.15.128.0+56dc5a86/0_REGISTRY/0.0.25.dev0/training/llama/meta-llama/Llama-3.2-1B/00e01e2d36568340e936.json

{"architectures": ["LlamaForCausalLM"], "attention_bias": false, "attention_dropout": 0.0, "bos_token_id": 128256, "eos_token_id": 128257, "head_dim": 64, "hidden_act": "silu", "hidden_size": 2048, "initializer_range": 0.02, "intermediate_size": 8192, "max_position_embeddings": 131072, "mlp_bias": false, "model_type": "llama", "neuron": {"compiler_version": "2.15.128.0+56dc5a86", "input_specs": {"chosen": ["<|im_start|>user\nLet $n$ be an odd integer with exactly 11 positive divisors.  Find the number of positive divisors of $8n^3$.<|im_end|>\n<|im_start|>assistant\nTo find the number of positive divisors of $8n^3$, I need to factor it into prime factors and use the formula for the number of divisors of a product of primes.\nFirst, I factor 8 as $2^3$.\nThen, I factor $n^3$ using the fact that $n$ is odd and has exactly 11 positive divisors.\nSince $n$ is odd, it has no factors of 2, so its prime factorization must be of the form $n = p_1^{a_1}p_2^{a_2}\\cdots p_k^{a_k}$, where $p_1, p_2, \\ldots, p_k$ are odd primes and $a_1, a_2, \\ldots, a_k$ are positive integers.\nThe number of positive divisors of $n$ is given by the formula $(a_1 + 1)(a_2 + 1)\\cdots (a_k + 1)$.\nSince this is equal to 11, which is prime, it means that $n$ has only one prime factor, say $p$, and its exponent is 10.\nSo, $n = p^{10}$, where $p$ is an odd prime.\nThis implies that $n^3 = p^{30}$.\nNow, I can factor $8n^3$ as $8n^3 = 2^3p^{30}$.\nThe number of positive divisors of $8n^3$ is given by the formula $(3 + 1)(30 + 1) = 4 \\times 31 = 124$.\n# Answer\n\n124<|im_end|>\n"], "chosen_attention_mask": [1, 512], "chosen_input_ids": [1, 512], "chosen_labels": [1, 512], "prompt": ["Let $n$ be an odd integer with exactly 11 positive divisors.  Find the number of positive divisors of $8n^3$."], "prompt_attention_mask": [1, 512], "prompt_input_ids": [1, 512], "question": ["Let $n$ be an odd integer with exactly 11 positive divisors.  Find the number of positive divisors of $8n^3$."], "rejected": ["<|im_start|>user\nLet $n$ be an odd integer with exactly 11 positive divisors.  Find the number of positive divisors of $8n^3$.<|im_end|>\n<|im_start|>assistant\nSince $t(n) = 11$ is prime and is the product of 1 more than each of the exponents in the prime factorization of $n$, there can be only one exponent, and therefore one prime in the prime factorization of $n$.  This means $n = p^{10}$ for some odd prime number $p$, so $$ 8n^3 = 2^3 \\cdot p^{30} \\qquad \\Rightarrow \\qquad t(8n^3) = (3 + 1)(30 + 1) = \\boxed{124}. $$<|im_end|>\n"], "rejected_attention_mask": [1, 512], "rejected_input_ids": [1, 512], "rejected_labels": [1, 512], "source": ["prm_dpo_pairs"]}, "model_class": "PeftModelForCausalLM", "num_neuron_cores_per_node": 2, "pipeline_parallel_size": 1, "precision": "bfloat16", "tensor_parallel_size": 1, "training": true}, "num_attention_heads": 32, "num_hidden_layers": 16, "num_key_value_heads": 8, "pad_token_id": 128257, "pretraining_tp": 1, "rms_norm_eps": 1e-05, "rope_scaling": {"factor": 32.0, "high_freq_factor": 4.0, "low_freq_factor": 1.0, "original_max_position_embeddings": 8192, "rope_type": "llama3"}, "rope_theta": 500000.0, "tie_word_embeddings": true, "torch_dtype": "bfloat16", "use_cache": true, "vocab_size": 128258}