Wiki User
∙ 11y agoI'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
Wiki User
∙ 11y ago-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.
7 To make it a bit more intuitive, think of it like this: If you have a set of 7 elements, you can "turn it into" a set of 6 elements by removing one of the elements. So, in how many ways can you remove an element from the set of 7 elements, without making the same 6-element set more than once?
-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
It depends. If the numbers are only used once then it is 24. More precisely it is 4!. (4 factorial which is 4x3x2x1). The same would apply to any number of digits assuming they are used once each. If there is replacement involved (any number used any number of times like 7777 or 6688), then the answer would be 256. Or in other words, the number of possible numbers raised to the power of how many digit positions there are (44). So if the question was how many 5 digit numbers can you make from 1234567?75=16807