I'm sure there will be simpler solutions to this one.
[(48 - 39) x 14] + 143 - [48 - 14 - 14 - 14] = 263
9 x 14 = 126
126 + 143 = 269
269 - [48 - 14 - 14 - 14] = 269 - 6 = 263
-1230
106 or a million.
There are 5*4*3 = 60 such numbers.
No. Not if the numbers are to be used only once. There is only one 5 (or a multiple of 5) in the numbers available. Using times and divide cannot produce any more of them. On the other hand 100 is divisible by 5*5 so at least two fives are required. If the numbers can be used more than once then 2*2*5*5 is one possible solution.
3! = 6.
-1230
106 or a million.
There are 5*4*3 = 60 such numbers.
i dont understand the question . sorry!and please make it more understangable. its just numbers. sorry once again!
Three hexagons comprise more than 7 lines so you cannot use only seven numbers, once each, to make all the lines add up to the same number!
No. Not if the numbers are to be used only once. There is only one 5 (or a multiple of 5) in the numbers available. Using times and divide cannot produce any more of them. On the other hand 100 is divisible by 5*5 so at least two fives are required. If the numbers can be used more than once then 2*2*5*5 is one possible solution.
If no number is repeated then every number that appears once is a mode. They appear once, which is more often than any numbers that do not appear at all.
A graph that used the same one of the numbers more than once.
he mode is the number showed more hen once . and it can be more the one number .
-9
if we do not want to use the same number more than once then the answer is: 49*48*47*46*45*44 however if we can use a number more than once the solution is: 49^6
3! = 6.