There are 1,963 such integers. Every factor of a number has a pair. The only time there will be an odd number of factors is if one factor is repeated, ie the number is a perfect square. So the question is really asking: how many positive integers less than 2008 (in the range 1 to 2007) are not perfect squares. √2007 = 44 and a bit (it lies between 44 and 45) So there are 44 integers less than (or equal to) 2007 which are perfect squares → 2007 - 44 = 1963 integers are not perfect squares in the range 1-2007 and have an even number of factors (divisors).
It can be proven that it is impossible to find a pair of integers, p and q, such that p/q = sqrt(14).
337
The square root of 73 is between 8 and 9.
A zero pair is when one pairs a positive counter and a negative counter.
-3 and 6 is one possible pair.
THere are two pair of consecutive integers. 52 = 25 < 30 < 36 = 62 So, one pair is 5 and 6 and the other pair is -6 and -5.
-10 and -9 is one possible pair.
sqrt(89) = -9.4 or +9.4 So one possible pair of consecutive integers between the square roots of 89 are 2 and 3.
The lowest common factor of any pair of positive integers is 1.
There is no unique pair of numbers that satisfies these requirements. Suppose a and b is such a pair, and sqrt(105) = x then you want a < x < b But a < (a+x)/2 < x < (b+x)/2 < b So that (a+x)/2 and (b+x)/2 are a closer pair. and you can then find a closer pair still - ad infinitum. The question can be answered (sort of) if it asked about "integers" rather than "numbers". 100 < 105 < 121 Taking square roots, this equation implies that 10 < sqrt(105) < 11 so the answer could be 10 and 11. But (and this is the reason for the "sort of") the above equation also implies that -11 < sqrt(105) < -10 giving -11 and -10 as a pair of consecutive integers. So, an unambiguous answer is possible only if the question specifies positive integers.
For each pair of such integers, find the difference between the absolute values of the two integers and allocate the sign of the bigger number to it.
There are 1,963 such integers. Every factor of a number has a pair. The only time there will be an odd number of factors is if one factor is repeated, ie the number is a perfect square. So the question is really asking: how many positive integers less than 2008 (in the range 1 to 2007) are not perfect squares. √2007 = 44 and a bit (it lies between 44 and 45) So there are 44 integers less than (or equal to) 2007 which are perfect squares → 2007 - 44 = 1963 integers are not perfect squares in the range 1-2007 and have an even number of factors (divisors).
There are infinitely many numbers, but a simple pair to remember is 1 and 139. Since 139 is a prime, this is the only solution with a pair of positive integers.
Assuming you mean consecutive numbers are integers, then: The integers will be the two integers between which the square root of 4160 lies. Calculating the square root of 4160 using the "long" division method: ________6__4 _____---------- ___6|_41_60 _______36 _______--- __124|__560 _________496 _________---- __________64 The square root of 4160 is 64.<something>, thus it lies between 64 and 65, so: 4160 = 64 × 65
Take any negative integers, say -5 and -10, their sum is -15 which is smaller than both of them. We could have used 0 as well, so I should have said any non-positive integers. To see that is does not work with positive integers, take 5 and 10 whose sum is 15 which is BIGGER than either one.
Every positive rational number and its negative are the two square roots of the same positive rational number.