Oh...is that Foxy Fives Wrksht 1
32-22=5 9-4=5
To solve the equation with the numbers 2, 9, 5, 9, and 4 to equal 22, you can use basic arithmetic operations. One possible solution is: ( 9 + 9 + 5 - 2 + 4 = 22 ). This combines addition and subtraction to reach the target number.
Assuming you are using "combinations" in the mathematical sense where order doesn't matter (if order does matter it would be "permutations"), there are 22C5 = 22!/5!17! = 26,334 possible combinations of 5 numbers from 22. They start {1, 2, 3, 4, 5}, {1, 2, 3, 4, 6}, {1, 2, 3, 4, 7}, ... and end ... {16, 19, 20, 21, 22}, {17, 19, 20, 21, 22}, {18, 19, 20, 21, 22}; I'll leave the 26,328 combinations in the middle for you to list.
5 and 11 are prime. The others are not (3 x 4 =12, 3 x 5 =15, 2 x 11 =22)
11 and 16 Because: 1+0 = 1 1+1 = 2 2+2 = 4 4+3 = 7 7+4 = 11 11+5 = 16 16+6 = 22 22+7 = 29 and so on ....
you add all of the numbers together, and then divide the sum by however many numbers there were. example: 2, 4, 4, 5, 7 2+4+4+5+7=22 22/5=4.4 average=4.4
32-22=5 9-4=5
4*5 + 6/3 = 20 + 2 = 22
there are 22
The LCM of 4 and 5 is 20, which is the multiple of the highest power of prime factors from both numbers (22 x 5 = 2 x 2 x 5 = 20).
Oh, dude, like, 1 and 220 go into 220. And, um, 2 and 110, and 4 and 55, and 5 and 44, and 10 and 22. So, yeah, those are the numbers that can slide into 220.
The reciprocal of the mixed number 4 2/5 is 5/22 The number 4 2/5 becomes (4 x 5) + 2 /5 = 22/5
Assuming you are using "combinations" in the mathematical sense where order doesn't matter (if order does matter it would be "permutations"), there are 22C5 = 22!/5!17! = 26,334 possible combinations of 5 numbers from 22. They start {1, 2, 3, 4, 5}, {1, 2, 3, 4, 6}, {1, 2, 3, 4, 7}, ... and end ... {16, 19, 20, 21, 22}, {17, 19, 20, 21, 22}, {18, 19, 20, 21, 22}; I'll leave the 26,328 combinations in the middle for you to list.
50
5 and 11 are prime. The others are not (3 x 4 =12, 3 x 5 =15, 2 x 11 =22)
11 and 16 Because: 1+0 = 1 1+1 = 2 2+2 = 4 4+3 = 7 7+4 = 11 11+5 = 16 16+6 = 22 22+7 = 29 and so on ....
If you are asking to use each of those 4 numbers, than we would simply say there are 4 choices for the first digit, 3 for the second etc. and there would be 4! possible numbers, or 24 of them. Here they are Think of each one of the entries as a number. This was done for you with the program Mathematica. {{2, 3, 4, 5}, {3, 2, 4, 5}, {4, 2, 3, 5}, {2, 4, 3, 5}, {3, 4, 2, 5}, {4, 3, 2, 5}, {5, 3, 2, 4}, {3, 5, 2, 4}, {2, 5, 3, 4}, {5, 2, 3, 4}, {3, 2, 5, 4}, {2, 3, 5, 4}, {2, 4, 5, 3}, {4, 2, 5, 3}, {5, 2, 4, 3}, {2, 5, 4, 3}, {4, 5, 2, 3}, {5, 4, 2, 3}, {5, 4, 3, 2}, {4, 5, 3, 2}, {3, 5, 4, 2}, {5, 3, 4, 2}, {4, 3, 5, 2}, {3, 4, 5, 2}} If we can use a number more than once or not at all than that changes it. For example, you do not see 2222 above. Look at each number, let's start with 2, it is either in the first digit or not, in the second or not, in the third digit or not,and in the fourth digit or not.-. there are 2 choices for the number 2, same for 3,4,and 5. Then there are 2^4 choices for each (2^4)^4 or 16^4 possible numbers. That is much bigger than 24 as you can see. It is 65536. BUT that includes numbers such as 2 or 22 where we include 2 twice but not the other numbers, and we need to subtract those if we are talking about 4 digit numbers. That makes it a little harder! Do you seen why? Let me know if you need more help.