24 is the smallest example, 86 is the highest (unless both digits may be the same when you could have 22 & 88)
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.
73 is the largest 2 digit number that is both prime and has prime numbers for both of its digits.
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.
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.
81
64
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.
73 is the largest 2 digit number that is both prime and has prime numbers for both of its digits.
10
If the digits go from the thousands place to the ones place then we need to use 4 digits. Because the digits are all even, we are forced to use the 4 even digits (2, 4, 6 and 8). As they decrease by 2 each time, the only option for ordering them is greatest to lowest. Therefore, the number described in the question is 8,642.
1) Look at the two numbers you are multiplying 2) Find both their number of significant digits 3) Multiply both numbers together normally 4) Round your answer to the same number of significant digits of the least number in the first two factors 250 x 185 250 had 2 sig. digits------ 185 has 3 sig. digits 250 has the least number of sig. digits (2) Final answer has to have 2 sig. digits Normal answer: 46250 With rounding of sig. digits: 46000
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.
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.
81
It is 2 which also is the only even prime number
Your are 28 or 64
6 = 2 x 3So it must satisfy the divisibility tests for both 2 and 3, namely:Any even number for which the sum of its digits is divisible by 3.