How Does The Four Numbers 1 1 9 And 9 Equal 10

How do you get numbers 1-20 using the numbers 4,5, and 9?

Whoa..... Good question I have used all 4,5 and 9 only once in the solution [math]1= (4+5)/9[/math][math]2= 4-5+ \sqrt{9}[/math][math]3= (5+4)/\sqrt{9}[/math][math]4= 9-(5!/4!)[/math][math]5= 5/(4- \sqrt{9})[/math][math]6= 5-\sqrt{4}+\sqrt{9}[/math][math]7= (9+5)/\sqrt{4}[/math][math]8= 9+4-5[/math][math]9= (5-4)*9[/math][math]10= 9+5-4[/math][math]11=  \sqrt{5!+4-\sqrt{9}}[/math][math]12= 9+\sqrt{4+5}[/math][math]13= 5* \sqrt{9}-\sqrt{4}[/math][math]14= 5*4-\sqrt{9}![/math][math]15= 5!*\sqrt{9}/4![/math][math]16= (9-5)*4[/math][math]17= 5*4-\sqrt{9}[/math][math]18= 9+4+5[/math][math]19= 5*\sqrt{4}+9[/math][math]20= 4!*5/\sqrt{9}![/math]I hope this helps........PS: Others might give simpler answers, but these are the ones that struck my head ; P

How many 10 digit numbers can be formed by using 0 to 9, including repeating of digits but 0 should not be the first digit of the 10 digit number?

Since zero cannot be at first place therefore are 9 Ways to put a number on first place(since we can put 1,2,3,4,5,6,7,8,9)… on second to last place there can be any no from 0 to 10Therefore the number that can be made will be the multiplication of the numbers at each placeTherefor it will be9 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10Which is equal to 9 x 10^9Hence number of numbers that can be formed is equal to 9 x 10^9Hope the answer helps you…

How many 4 digit numbers can be formed using 0 to 9?

+case 1 (with repitition):consider four blanks (digits) to be filled with the numbers from 0 to 9. There is only one constrained that 0 cannot be the first number.— — — — be the 4 blanksThe first blank can be filled with 9 digits (1,2,3,4,5,6,7,8.9) and the rest in 10 digits(including 0).9 ways*10 ways*10 ways*10 ways = 9000 numbers (By principle of multiplication)case 2(without repitition):Again consider 4 blanks to be filled but now without repitition,i.e., a digit can be used only once. In other words, if a digit is already used in forming the number it cannot be used again.therefore, again the first blank can be filled by 9 digits(excluding zero)The second blank has to be filled also by 9 digits because the digit used in the first blank is excluded and zero also can be used (9+1–1=9).The third blank can be filled by digits only . Because the 2 digits helped in forming the first 2 blanks now cannot be used in the third blank.similarly, 4th blank by 7 digits.Therefore, the total number of numbers are:9 ways* 9 ways * 8 ways * 7 ways = 4536 numbers (Again, by principle of multiplication).Hope it helped..!!

How can the numbers 1 through 10 be arranged on a circle such that the sum of any two adjacent numbers equals the sum of their two opposite numbers?

Lets make this question a little bit clear.To find solutions to the system of equations:[math]a+b = f +g[/math]...1[math]b+c = g+h[/math]....2[math]c+d = h+i[/math]....3[math]d+e = i+j[/math]....4[math]e+f = j+a[/math].....5From 1 and 2[math]a+h = c+f[/math]...6From 6 and 3[math]a+d = f+i[/math]...7 Now this system of equations doesn't have a unique solution. Consider a = 1 and f = 2. Then the other equations as follow[math]f  = a + 1[/math][math]b = g + 1[/math][math]h = c+ 1[/math][math]d = i+1[/math][math]j = e +1[/math]The numbers can fit into the letters(pizza slices) just following the equations above.Another solution can be following a = 1 and f = 6 then,[math]f=a+5[/math][math]b=g+5[/math][math]h = c+5[/math][math]d = i+5[/math][math]j = e+5[/math]Sample solution:(a,b,c,d,e,f,g,h,i,j) = (1,10,4,5,8,2,9,3,6,7)(a,b,c,d,e,f,g,h,i,j) = (1,7,3,9,5,6,2,8,4,10)These configurations are permutations of pairs (1,2),(3,4),(5,6),(7,8)(,9,10)and (1,6),(2,7),(3,8),(4,9),(5,10).Finding the number of different solutions to the problem.Consider a circular permutation. We can choose to place the first pair in two different ways ( [math]a> f[/math] or [math]a < f[/math]). The rest are then follow a pattern i,e. which is greater than which. No. of possibilities for first pair is 5. and decreases with further allocation as 4,3,2,1So total no of arrangements possible are [math]2*5! = 240[/math]For two different pairings the total no of solutions are [math]240*2 = 480[/math]We considered a = 1, f = 2 and f = 6.For other pairing method such as a gap of 2 we can not make pairs. (1,3),(2,4),(5,7),(6,8) (9,10) not possible. Similarly, for gap 3 (1,4),(2,5),(3,6),(7,10),(8,9) not possible.For gap 4 (1,5),(2,6),(3,7),(4,8),(9,10) not possible

How many 4 digit numbers can be formed from 0-9 without repetition?

The Question can be re-written as :How many 4-digit numbers are possible with the digits 0 to 9?(I)Digits cannot be repeatedSolution: There are 10-digits :0,1,2,3,4,5,6,7,8,9The digits to be formed =No.of places=4(I)Case I: Digits cannot be repeated:If 0 is placed in first place then it becomes a 3-digit number out of 4-places.Thus ,we can fill 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9 in the first place.Therefore,No.of possibilities in the first place =9Again,consider the second place.Here we can fill 0 and any of the eight digitsThus, No.of possibilities=9 (the digit 0 and 8 digits)Consider the third place.We can fill any of the 8 digits.Thus, No.of possibilities=8Consider the fourth place.Here we can fill any 7-digits.Thus ,the number of possibilities =7Hence the total number of possibilities to arrange the even numbers from 0 to 9 without repetition of any digits =9X9X8X7=4536 ways.

You have a combination lock with 3 digits, which has 10 numbers (1 to 9), but you forgot the code. How many maximum combinations you can make with it?

That’s actually only 9 numbers per position (1 to 9). With 0 to 9 or 1 to 10, you’d have 10.This is trivial to calculate. You have one of 10 (or 9) possibilities for the first number. Then there 10 (or 9) for each of those on the 2nd, so just multiply - 10 x 10 = 100 (or 9x9 = 81) possibilities for the first two numbers.Multiply again for the third, and you’ll see that you have 1000 possible combinations with 10 choices for each, or 729 possibilities with 9 choices for each place.