1422, 1431, 1440
Since you didn't specify a single number, and all numbers are multiples of themselves, the five smallest multiples are the counting numbers 1 to 5.
The total number of integers that are multiples of both 55 and 19 is infinite. Here are the first three: 1045, 2090, 3135.
The total number of integers that are multiples of 25 and 6 is infinite. The first few are these: 150, 300, 450, 600, 750, 900 . . .
Multiply 5 by the first 12 integers: 1, 2, 3, ... ,12.
Algorithm: multiples input: two positive integers, m and n output: print first n multiples of m i = m; for j = 1 to n print i i = i + m; next j
The first two consecutive prime numbers that have a difference of 20 are the numbers 887 and 907.
First, we are not interested for the sign of the numbers, because the product of two negative numbers is always positive, so we need to find the two consecutive factors of 240, which are 15 and 16. Thus, the numbers are -16 and -15.
It's any set of consecutive integers that are composite. For instance, 8, 9, and 10 are consecutive composites.
The first ten consecutive composite positive integers are: 114 115 116 117 118 119 120 121 122 123
Consecutive means one after another.If the first is a, then the second is a+1.The sum of these isa + a + 1 = 2a+1 = -3772a = -378a = -189So the two numbers are -189 and -188
The sum of the first four non-negative, consecutive, even integers is 20.
The first integer is 17.
121
Their sum is 99.
First of all, you need to be more specific. It's three consecutive integers. And they are 2,999, 3,000, and 3,001
86
Since the average of the three integers will be 114/3 = 38, and the three numbers are consecutive, the numbers will be 36, 38 and 40. Another way to do this problem using algebra is to let the first integer be n, then the next two are n+2 and n+4. Their sum is 3n+6 and it equals 114 So 3n+6=114 and 3n=108 so n=36 then next two numbers must be 38 and 40 since they are consecutive even integers.