I was only able to get 13 but I think there can be more, this is what i got: 0+0+4=4 4+0+0=4 0+4+0=4 0+1+3=4 0+3+1=4 0+2+2=4 1+2+1=4 1+1+2=4 1+3+0=4 2+2+0=4 2+0+2=4 3+1+0=4 3+0+1=4
X^2 + 7X + 18 = 0 try the discriminant b^2 - 4ac 7^2 - 4(1)(18) 49 - 72 < 0 so, no real roots and no factors of 18 add to 7
14: Quarters Dimes Nickels 0 0 13 0 1 11 0 2 9 0 3 7 0 4 5 0 5 3 0 6 1 1 0 8 1 1 6 1 2 4 1 3 2 1 4 0 2 0 3 2 1 1
A huge number. 0 + 1 + 2 = 3 0 + 2 + 1 = 3 1 + 0 + 2 = 3 1 + 2 + 0 = 3 2 + 0 + 1 = 3 2 + 1 + 0 = 3 -0 + 1 + 2 = 3 -0 + 2 + 1 = 3 1 - 0 + 2 = 31 + 2 - 0 = 32 - 0 + 1 = 32 + 1 - 0 = 3 0 - 1 + 3 = 2 0 + 3 - 1 = 2 -1 + 0 + 3 = 2 -1 + 3 + 0 = 2 3 + 0 - 1 = 2 3 - 1 + 0 = 2 -0 - 1 + 3 = 2-0 + 3 - 1 = 2-1 - 0 + 3 = 2-1 + 3 - 0 = 23 - 0 - 1 = 23 - 1 - 0 = 2 0 - 2 + 3 = 1 0 + 3 - 2 = 1 -2 + 0 + 3 = 1 -2 + 3 + 0 = 1 3 + 0 - 2 = 1 3 - 2 + 0 = 1 -0 - 2 + 3 = 1-0 + 3 - 2 = 1-2 - 0 + 3 = 1-2 + 3 - 0 = 13 - 0 - 2 = 13 - 2 - 0 = 1 1 + 2 - 3 = 0 1 - 3 + 2 = 0 2 + 1 - 3 = 0 2 - 3 + 1 = 0 -3 + 1 + 2 = 0 -3 + 2 + 1 = 0 For each of these equations there is a counterpart in which all signs have been switched. For example 0 + 1 + 2 = 3 gives -0 - 1 - 2 = -3and so on. Now, all of the above equations has three numbers on the left and one on the right. Each can be converted to others with two numbers on each side. For example:the equation 0 + 1 + 2 = 3 gives rise to0 + 1 = 3 - 20 + 1 = -2 + 30 + 2 = 3 - 10 + 2 = -1 + 31 + 2 = 3 - 01 + 2 = -0 + 3-0 + 1 = 3 - 2-0 + 1 = -2 + 3-0 + 2 = 3 - 1-0 + 2 = -1 + 31 + 2 = 3 + 01 + 2 = +0 + 3 As you can see, the number of equations is huge!
1 1/2
In the binary number system 10001 = 17 and 10010 = 18. 16 8 4 2 1 1 0 0 0 1 = 17 1 0 0 1 0 = 18
18 cannot be a binary number as binary numbers are base two and each digit can only be 0 or 1. If you are asking what is 18 in base 10 converted to base 2, then: 18 ÷ 2 = 9 r 0 9 ÷ 2 = 4 r 1 4 ÷ 2 = 2 r 0 2 ÷ 2 = 1 r 0 1 ÷ 2 = 0 r 1 reading the remainders in reverse order → 1810 = 100102
18
Not sure what a "didget" is. It is possible to make 18 5-digit numbers.
18 = 16 + 2 = 1*24 + 0*23 + 0*22 + 1*21 + 0*20 = 100102.
1001001010101 2346 ÷ 2 = 1173 Rem. 0 1173 ÷ 2 = 586 Rem. 1 586 ÷ 2 = 293 Rem. 0 293 ÷ 2 = 146 Rem. 1 146 ÷ 2 = 73 Rem. 0 73 ÷ 2 = 36 Rem. 1 36 ÷ 2 = 18 Rem. 0 18 ÷ 2 = 9 Rem. 0 9 ÷ 2 = 4 Rem. 1 4 ÷ 2 = 2 Rem. 0 2 ÷ 2 = 1 Rem. 0 1 ÷ 2 = 0 Rem. 1
24, of which 6 will begin with zero. Excluding these, gives 18.
I'm going to list them all in sets of (Q, D, N, P), where Q = quarters, D = dimes, N = nickels, and P = pennies. (1, 0, 1, 1) (1, 0, 0, 6) (0, 3, 0, 1) (0, 2, 2, 1) (0, 2, 1, 6) (0, 2, 0, 11) (0, 1, 4, 1) (0, 1, 3, 6) (0, 1, 2, 11) (0, 1, 1, 16) (0, 1, 0, 21) (0, 0, 6, 1) (0, 0, 5, 6) (0, 0, 4, 11) (0, 0, 3, 16) (0, 0, 2, 21) (0, 0, 1, 26) (0, 0, 0, 31) Thus, there are 18 total combinations.
You would need to use 18 digits: possibly, 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F,G and H to represent the decimal numbers 0 to 17.Then you would need to express any number x, in the formx = a(n)*18^n + a(n-1)*18^(n-1) + ... * a(1)*18 + a(0) + a(-1)*18^(-1) + a(-2)*18^(-2) ...where the a(i) belonged to the set of 18 "digits" as defined above.The the octadecimal representation of x isa(n)a(n-1) ...a(1)a(0).a(-1)a(-2) ...Note that there is a octadecimal point between a(0) and a(-1).
Caribe Road - 2011 0-2 1-2 was released on: USA: 18 July 2011
9 x 2 = 18 18 + 1 + 5 = 24
The highest number you can make out of the digits 2 4 0 and 1 is 4210.