121 - 16 = 105.
There are two other pairs.
Call the integer square roots of the specified pair of numbers l and g for lesser and greater respectively. Then, from the problem statement, g2 - l2 = 105. Possible values for l2 are successively 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, .... , and corresponding values for g2 are 106, 109, 114, 121, ... etc. The first of the latter series that is a perfect square number is 121. Therefore, the square numbers are 121 and 16.
4 is the only square number that is a factor of 84. No square number is a factor of 105.
56 and 105
One possible pair is 1 and 105.
52.5 x 2 = 105
Suppose x and y are any two co-prime integers. Then 105*x and 105*y will have a GCF of 105.
The second one.
the simple way is: 100+5=105 but there is ALOT of other ways to get 105
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.
1 and 105 are a factor pair of 105 since 1 x 105= 105 3 and 35 are a factor pair of 105 since 3 x 35= 105 5 and 21 are a factor pair of 105 since 5 x 21= 105 7 and 15 are a factor pair of 105 since 7 x 15= 105
21 and 5. 21 x 5 = 105. Simple math.
The same as the square root of positive 105, multiplied by i (the imaginary unit). So, the answer will be approximately 10.25i. If only real numbers are acceptable, then there is no solution.