*I am a 2-digit whole number between 30 and 80.
*My tens digit is one more than my ones digit.
*I am a prime number.
What number am I
a b
a = b+1
3 </= a </= 8
has to be odd if prime so b is 1,3,5,7, (not 9 if a = b+1)
list from Google search:
31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79
43 is the only one I see where a=b+1 or 4 = 3+1
the number ends in 1,3,7,9
So the candidates are 21, 43, 89
Only 43 fits the description
LOL so ends in 1, 3, 7 ! not 5 OR 9
To find the number that satisfies all the given conditions, we can break down the problem into logical steps.
Step 1: Identify the range of numbers between 30 and 80 (inclusive).
Step 2: Determine the eligible numbers from the range that have a tens digit one more than the ones digit. For example, 31, 42, 53, etc.
Step 3: Identify the prime numbers among the eligible numbers.
Step 4: Output the final number that meets all the given conditions.
Let's go through each step in more detail:
Step 1: We know the number is a 2-digit whole number between 30 and 80. Hence, the possible numbers are 31, 32, 33, ..., 79, 80.
Step 2: From the list of numbers generated in Step 1, we need to identify the numbers where the tens digit is one more than the ones digit. For example, 31, 42, and 53, but not 64 or 75.
Step 3: Among the numbers from Step 2, we need to find the prime numbers. A prime number is a number greater than 1 that has no divisors other than 1 and itself.
Checking each number one by one, we find that 31 is a prime number.
Step 4: Based on all the conditions given, the number that satisfies the problem is 31.
Therefore, the answer is 31.