There is no pair of consecutive even integers whose sum is 264 - as their average is divisible by 4, that must mean that the sum of the unit digits of each of the pair of numbers could never be a multiple of 4 - which 264 is.
Factor pairs for 18 include 1&18, 2&9, and 3&6. Only one of these pairs adds up to 11.
17 & 18 (306)
There are 3 whose sum is 45 whose sum is 57 whose sum is 69 whose sum is 711 whose sum is 813 whose sum is 915 whose sum is 1017 whose sum is 1119 whose sum is 1219 whose sum is 1317 whose sum is 1415 whose sum is 1513 whose sum is 1611 whose sum is 179 whose sum is 187 whose sum is 195 whose sum is 203 whose sum is 211 whose sum is 22.
2 and 2
23 * 16 = 368
No pair of real numbers can do that.The numbers are4.5 + j 18.43234.5 - j 18.4323
It is precisely 368.
The pair of factors you are looking for is 23 x 47 = 1081. In fact, aside from itself and 1, these are the only two positive integer factors of 1081.
2 and 1 .their sum is 3 .difference is 1
There is no pair of consecutive even integers whose sum is 264 - as their average is divisible by 4, that must mean that the sum of the unit digits of each of the pair of numbers could never be a multiple of 4 - which 264 is.
10 and -5 are factors of -20.
20 and 20.
complementary
7 and 6
Factor pairs for 18 include 1&18, 2&9, and 3&6. Only one of these pairs adds up to 11.
answer it!S***