No.
Solving for the third integer:(a + b + c) / 3 = 37 (by the definition of "average")(18 + b + c) / 3 = 37 (assuming "a" is the least integer)18 + b + c = 111c = 111 - 18 - b (solving for the third integer)c = 93 - bNow, "b" must be at least 18, in which case:c = 93 - 18 = 75On the other hand, the largest "b" can be is when it is equal to "c" (since I am assuming that "c" is the largest integer):c = 93 - bc = 93 - c2c = 93c = 46.5Adjusting the numbers a bit (since we need integers), in this case we get the numbers 18, 46, 47Thus, the greatest integer can be anything in the range from 47 to 75.
Since 50 < n < 100, n is a 2-digit number. Since the sum of the digits is 11, n could be 56, 65, 74, 83, or 92, but since n is divided (evenly) by 2 and the quotient is a prime integer, then n must be 74 (74÷2 = 37).
Let the first odd integer be x, so the other consecutive odd integer is x + 2. So we have:x^2 + (x + 2)^2 = 74x^2 + x^2 + 4x + 4 = 742x^2 + 4x + 4 - 74 = 74 - 742x^2 + 4x - 70 = 0 divide by 2 to both sidesx^2 + 2x - 35 = 0(x + 7)(x - 5) = 0x + 7 = 0 or x - 5 = 0x + 7 - 7 = 0 -7 or x - 5 + 5 = 0 + 5x = -7 or x = 5Thus, the first odd integer would be -7 or 5, and the second consecutive odd integer would be -5 (-7 + 2) or 7 (5 + 2).Check:
0.030 is not an integer. So there would be no integer for it.
It is an integer.
74 is an integer and there is not a sensible way of representing it as a fraction. 74/1 is a silly, but mathematically correct answer.
Yes, 74 is divisible by 2. When you divide 74 by 2, you get 37, which is an integer. This means that 74 can be evenly divided by 2 without any remainder.
14 of them.
73.6
The square root of any negative number is imaginary. No integers nearby.
No, 74 is not a perfect square of an integer - its square root, rounded to two decimal places, is equal to 8.60. The closest square numbers to 74 are 81 (92) and 64 (82).
Exclusive is rather an unusual way to define an integer range! 125 - 74 - 1 = 50.
A perfect number is defined as a positive integer that is equal to the sum of its proper divisors, excluding itself. For 74, the proper divisors are 1, 2, 37, and their sum is 1 + 2 + 37 = 40, which is not equal to 74. Therefore, 74 does not meet the criteria to be classified as a perfect number.
Solving for the third integer:(a + b + c) / 3 = 37 (by the definition of "average")(18 + b + c) / 3 = 37 (assuming "a" is the least integer)18 + b + c = 111c = 111 - 18 - b (solving for the third integer)c = 93 - bNow, "b" must be at least 18, in which case:c = 93 - 18 = 75On the other hand, the largest "b" can be is when it is equal to "c" (since I am assuming that "c" is the largest integer):c = 93 - bc = 93 - c2c = 93c = 46.5Adjusting the numbers a bit (since we need integers), in this case we get the numbers 18, 46, 47Thus, the greatest integer can be anything in the range from 47 to 75.
Since 50 < n < 100, n is a 2-digit number. Since the sum of the digits is 11, n could be 56, 65, 74, 83, or 92, but since n is divided (evenly) by 2 and the quotient is a prime integer, then n must be 74 (74÷2 = 37).
The positive integer factors of 1184 are: 1, 2, 4, 8, 16, 32, 37, 74, 148, 296, 592, 1184
The positive integer factors of 1036 are: 1, 2, 4, 7, 14, 28, 37, 74, 148, 259, 518, 1036