How many prime numbers less than 100 and greater than 3 are of the form: 4x + 1, 5y – 1?
The Correct Answer and Explanation is:
To find how many prime numbers less than 100 and greater than 3 are of the form 4x+14x + 14x+1 and 5y−15y – 15y−1, we need to first identify the prime numbers in this range and then check which ones satisfy both equations.
Step 1: List all prime numbers greater than 3 and less than 100
The prime numbers greater than 3 and less than 100 are:
5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97.5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97.
Step 2: Determine which of these prime numbers are of the form 4x+14x + 14x+1
We need to check each prime number to see if it can be written as 4x+14x + 14x+1 for some integer xxx. This means the prime number must leave a remainder of 1 when divided by 4.
Performing the division:
- 5÷4=15 \div 4 = 15÷4=1 remainder 1, so 5 is of the form 4x+14x + 14x+1.
- 7÷4=17 \div 4 = 17÷4=1 remainder 3, so 7 is not of the form 4x+14x + 14x+1.
- 11÷4=211 \div 4 = 211÷4=2 remainder 3, so 11 is not of the form 4x+14x + 14x+1.
- 13÷4=313 \div 4 = 313÷4=3 remainder 1, so 13 is of the form 4x+14x + 14x+1.
- 17÷4=417 \div 4 = 417÷4=4 remainder 1, so 17 is of the form 4x+14x + 14x+1.
- 19÷4=419 \div 4 = 419÷4=4 remainder 3, so 19 is not of the form 4x+14x + 14x+1.
- 23÷4=523 \div 4 = 523÷4=5 remainder 3, so 23 is not of the form 4x+14x + 14x+1.
- 29÷4=729 \div 4 = 729÷4=7 remainder 1, so 29 is of the form 4x+14x + 14x+1.
- 31÷4=731 \div 4 = 731÷4=7 remainder 3, so 31 is not of the form 4x+14x + 14x+1.
- 37÷4=937 \div 4 = 937÷4=9 remainder 1, so 37 is of the form 4x+14x + 14x+1.
- 41÷4=1041 \div 4 = 1041÷4=10 remainder 1, so 41 is of the form 4x+14x + 14x+1.
- 43÷4=1043 \div 4 = 1043÷4=10 remainder 3, so 43 is not of the form 4x+14x + 14x+1.
- 47÷4=1147 \div 4 = 1147÷4=11 remainder 3, so 47 is not of the form 4x+14x + 14x+1.
- 53÷4=1353 \div 4 = 1353÷4=13 remainder 1, so 53 is of the form 4x+14x + 14x+1.
- 59÷4=1459 \div 4 = 1459÷4=14 remainder 3, so 59 is not of the form 4x+14x + 14x+1.
- 61÷4=1561 \div 4 = 1561÷4=15 remainder 1, so 61 is of the form 4x+14x + 14x+1.
- 67÷4=1667 \div 4 = 1667÷4=16 remainder 3, so 67 is not of the form 4x+14x + 14x+1.
- 71÷4=1771 \div 4 = 1771÷4=17 remainder 3, so 71 is not of the form 4x+14x + 14x+1.
- 73÷4=1873 \div 4 = 1873÷4=18 remainder 1, so 73 is of the form 4x+14x + 14x+1.
- 79÷4=1979 \div 4 = 1979÷4=19 remainder 3, so 79 is not of the form 4x+14x + 14x+1.
- 83÷4=2083 \div 4 = 2083÷4=20 remainder 3, so 83 is not of the form 4x+14x + 14x+1.
- 89÷4=2289 \div 4 = 2289÷4=22 remainder 1, so 89 is of the form 4x+14x + 14x+1.
- 97÷4=2497 \div 4 = 2497÷4=24 remainder 1, so 97 is of the form 4x+14x + 14x+1.
So, the prime numbers less than 100 that are of the form 4x+14x + 14x+1 are:
5,13,17,29,37,41,53,61,73,89,97.5, 13, 17, 29, 37, 41, 53, 61, 73, 89, 97.5,13,17,29,37,41,53,61,73,89,97.
Step 3: Determine which of these prime numbers are of the form 5y−15y – 15y−1
Next, we need to check which of these numbers are of the form 5y−15y – 15y−1, meaning they must leave a remainder of 4 when divided by 5.
Performing the division:
- 5÷5=15 \div 5 = 15÷5=1 remainder 0, so 5 is not of the form 5y−15y – 15y−1.
- 13÷5=213 \div 5 = 213÷5=2 remainder 3, so 13 is not of the form 5y−15y – 15y−1.
- 17÷5=317 \div 5 = 317÷5=3 remainder 2, so 17 is not of the form 5y−15y – 15y−1.
- 29÷5=529 \div 5 = 529÷5=5 remainder 4, so 29 is of the form 5y−15y – 15y−1.
- 37÷5=737 \div 5 = 737÷5=7 remainder 2, so 37 is not of the form 5y−15y – 15y−1.
- 41÷5=841 \div 5 = 841÷5=8 remainder 1, so 41 is not of the form 5y−15y – 15y−1.
- 53÷5=1053 \div 5 = 1053÷5=10 remainder 3, so 53 is not of the form 5y−15y – 15y−1.
- 61÷5=1261 \div 5 = 1261÷5=12 remainder 1, so 61 is not of the form 5y−15y – 15y−1.
- 73÷5=1473 \div 5 = 1473÷5=14 remainder 3, so 73 is not of the form 5y−15y – 15y−1.
- 89÷5=1789 \div 5 = 1789÷5=17 remainder 4, so 89 is of the form 5y−15y – 15y−1.
- 97÷5=1997 \div 5 = 1997÷5=19 remainder 2, so 97 is not of the form 5y−15y – 15y−1.
Step 4: Final Answer
The prime numbers less than 100 and greater than 3 that satisfy both 4x+14x + 14x+1 and 5y−15y – 15y−1 are:
29,89.29, 89.29,89.
Thus, there are 2 prime numbers that satisfy both conditions.
