How many different counting numbers will each leave a remainder of 5 when divided into 47?
For a number to leave a remainder of 5 when divided into 47, it must satisfy two conditions:

1. it must divide exactly into $47 - 5$, or 42, and

2. it must be greater than 5, because the divisor is always greater than the remainder.

We list all divisors of 42 in pairs. They are 1 and 42, 2 and 21, 3 and 14, 6 and 7. Of these, only 42, 21, 14, 6, and 7 are greater than 5. There are $\boxed{5}$ different counting numbers that will leave a remainder of 5 when divided into 47.
5
Level 5
Number Theory
Problem: How many different counting numbers will each leave a remainder of 5 when divided into 47?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the counting numbers that leave a remainder of 5 when dividing 47, we first recognize that if a number \( n \) leaves a remainder of 5, then \( 47 \) can be expressed in the form \( 47 = kn + 5 \) for some integer \( k \). Rearranging this gives \( kn = 47 - 5 = 42 \). This means \( n \) must be a divisor of 42.

Next, we identify the positive divisors of 42. The prime factorization of 42 is \( 2 \times 3 \times 7 \). The divisors of 42 can be calculated as follows: 

1. The divisors are \( 1, 2, 3, 6, 7, 14, 21, 42 \).
2. We need to filter these divisors to find those that are greater than 5, since a remainder of 5 implies that \( n \) must be at least 6.

The divisors of 42 that are greater than 5 are \( 6, 7, 14, 21, \) and \( 42 \). 

Now, we count these valid divisors. There are exactly 5 numbers: \( 6, 7, 14, 21, \) and \( 42 \). 

Thus, the final answer is 5.<|END_OF_LATENT|> The final answer is $5$.