12n > 36 n > 3
Oh, dude, you want to know the unit's digits of the product of the first 21 prime numbers? Well, let me casually tell you that the unit's digit of a product depends on the unit's digits of the numbers being multiplied. Since the unit's digit of all prime numbers greater than 5 is either 1, 3, 7, or 9, the product of the first 21 prime numbers will end in a unit's digit that is a result of multiplying these digits together. Cool, right?
Let one integer be n, then the other is 2n + 3 and n(2n + 3) = 90; solve this last equation for n: n(2n + 3) = 90 ⇒ 2n2 + 3n - 90 = 0 ⇒ (2n + 15)(n - 6) = 0 ⇒ n = 6 or n = -7.5 As n must be a (positive) integer, the solution n = -7.5 can be ignored, leaving n = 6, giving 2n + 3 = 15. Thus the two positive integers are 6 and 15.
38 or 46
12 * n = 12n That is the best you can do without knowing what n is!
24
n to the 3rd power is n x n x n
3n is the simplest, but you could also write it as 3*n or 3.n or 3xn. The problem with the last form is that it could represent the product of 3 and x and n.
3*(n+6)
The equation = 4n-3
3n < 50 if 3n/3 < 50/3 that is, if n < 50/3 = 162/3.
Let the three integers be, n, (n + 1), and (n + 2) Then at least one of these numbers is even and therefore has a factor of 2. And one of the numbers is divisible by 3 **. Therefore the product has factors of 2 and 3 and is thus divisible by 2 x 3 = 6. ** Either n is divisible by 3. Or, n leaves a remainder of 1 when divided by 3 in which case (n + 2) is divisible by 3. Or, n leaves a remainder of 2 when divided by 3 in which case (n + 1) is divisible by 3.
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6(n + 3) = 33
It is 3*(54n + 1.8)
12
4(n + 3)