How many numbers between 30 and 150 leave a remainder of 1 when divided by 5?
The correct answer and explanation is:
To solve this, we need to find how many numbers between 30 and 150 leave a remainder of 1 when divided by 5.
The key is to recognize that numbers which leave a remainder of 1 when divided by 5 follow a specific pattern. These numbers can be represented in the form: 5n+15n + 1
Where nn is a non-negative integer. For example, the first number in the sequence that leaves a remainder of 1 when divided by 5 is 31, since: 31÷5=6 remainder 131 \div 5 = 6 \text{ remainder } 1
The next number in the sequence is 36, which can be calculated by adding 5 to the previous number (31). The sequence continues like this: 31, 36, 41, 46, 51, …, up to the largest number less than or equal to 150.
Step 1: Find the first number greater than or equal to 30 that leaves a remainder of 1
Starting from 30, the smallest number that leaves a remainder of 1 is 31.
Step 2: Find the largest number less than or equal to 150 that leaves a remainder of 1
The largest number less than or equal to 150 that leaves a remainder of 1 is 146, because: 146÷5=29 remainder 1146 \div 5 = 29 \text{ remainder } 1
Step 3: Determine the total number of terms in the sequence
The sequence of numbers that leave a remainder of 1 when divided by 5 is: 31,36,41,46,51,…,14631, 36, 41, 46, 51, \dots, 146
This is an arithmetic sequence where the first term is 31, the common difference is 5, and the last term is 146.
To find the number of terms in this sequence, we use the formula for the nth term of an arithmetic sequence: an=a1+(n−1)⋅da_n = a_1 + (n-1) \cdot d
Where:
- an=146a_n = 146 (the last term),
- a1=31a_1 = 31 (the first term),
- d=5d = 5 (the common difference).
Substituting these values into the formula: 146=31+(n−1)⋅5146 = 31 + (n-1) \cdot 5
Solving for nn: 146−31=(n−1)⋅5146 – 31 = (n-1) \cdot 5 115=(n−1)⋅5115 = (n-1) \cdot 5 n−1=23n-1 = 23 n=24n = 24
Thus, there are 24 numbers between 30 and 150 that leave a remainder of 1 when divided by 5.