The mean of a single number, 26, is 26.
So the question requires 3 numbers that equal 26. When added, multiplied, a combination?
In each case, the number of possibilities is infinite.
Addition:
(0,0,26), (0,1,25), 0,2,24), etc
and then add tenths to the second of the triplet:
(0,0.1,25.9), (0,1.1,24.9), (0,2.1,23.9), etc
and then add 0.2 to the second number, then 0.3, ad so on.
Next add 0.01, 0.02, 0.03 and so on.
And then decimals to 3 digit, 4 digit etc.
Finally start varying the first member - which has been 1 all along.
Multiplication:
Select ANY non-zero number B and let C = 26/B
then 1*B*C = 1*B*26/B = 26
Since the choice of B was arbitrary there are infinitely many such triplets.
Repeat with the first member of the triplet changed to any one of infinitely many non-zero numbers. In each case you will have infinitely many choices for B. All i all, the possibilities are endless.
How about: 1+2+3+20 = 26 out of many other combinations
11
whats 2 equal 3
(6 + 4 + 3)*2 = 26
there are 26 numbers and 26 letters in the alphbet, 3 repersents w
19 + 5 + 2
19 + 5 + 2
3 + 23 = 26 also 7 + 19 = 26 also 13 + 13 = 26
How about: 1+2+3+20 = 26 out of many other combinations
You have to prove it by showing what the mean of the five numbers is. First, you add up the five numbers, which is equal to 15. Then, you divide the sum by 5, which is 3, and not 8.
11
no 3 numbers add to equal 10 and multiply to equal 40.
3+26+24= 53
whats 2 equal 3
That means two numbers, which when added together equal 5, e.g. 2 + 3 = 5
It is simply: 3+23 = 26 because 3 and 23 are prime numbers
The numbers are: 3-sq rt of 3 and 3+sq rt of 3