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A 2 digit number that is the product of 2 consecutive integers and the sum of its digits is greater than the product of the digits both digits are even?
The 2-digit number must be 20, because it is the only 2-digit number whose sum of its two even digits, 2 + 0 = 2, is greater than the product of its two even digits, 2 x 0 = 0. Moreover, 20 is a product of the two consecutive integers 4 and 5.
What is the largest 2 digit number that is both prime and has prime numbers for both of its digits?
73 is the largest 2 digit number that is both prime and has prime numbers for both of its digits.
What is the smallest digits prime number?
It is 2 which also is the only even prime number
What is 3 digits plus 3 digits and equals 3 digits only using numbers 1 through 9 without carrying a number?
There is no solution to this problem. If each digit can be used once only then we have 5 odd numbered digits (1,3,5,7,9) and 4 even numbered digits (2,4,6,8). To create the two numbers that are added together requires the following combinations of digits. 5 Odd & 1 Even ....when added these will generate 2 Even digits & 1 Odd digit but the remaining digits are 3 Even. 4 Odd & 2 Even. These will generate 3 Even digits OR 1 Even digit & 2 Odd digits but the remaining digits are 1 Odd & 2 Even. 3 Odd & 3 Even. These generate 3 Odd digits OR 2 Even & 1 Odd digits but the remaining digits are 2 Odd & 1 Even. 2 Odd & 4 Even. These generate 3 Even digits OR 2 Odd & 1 Even digits but the remaining digits are 3 Odd & no Even.
If n is an odd number then the number of digits in its square root will be?
I cannot see how odd or even will determine the number of digits in the square root. For example:100 is an even number, and its square root is 10 (2 digits)121 is an odd number, and its square root is 11 (2 digits)Maybe the question is not phrased properly for what the questioner is asking.
