what is the probability that the sum of digit of a square of a number is 39
2006-07-12 19:33:23 · 11 answers · asked by champion 1 in Science & Mathematics ➔ Mathematics
1.Stupid!
Many have answered 39. . ..
Probability is always less than 1 guys! use ur brains
2.I give the answer its zero but try to justify it! it is very tricky not just 1/infinity!
Think
2006-07-12 19:47:48 · update #1
The sum of the digits of a square of an integer is never 39.
Use the casting out nines rule to show this. If the sum of the digits of N^2 were 39, the next step of the casting out nines rule would yield 3 + 9 = 12, then the next step after that would yield 1 + 2 = 3, implying N^2 is divisible by 3 but not by 9. Impossible for an integer N. So the probability is zero.
2006-07-12 19:46:16 · answer #1 · answered by ymail493 5 · 7⤊ 1⤋
Dan D is absolutely correct if by number you mean integer.
If not, then you need to prove some more.
Assume that some number is 11.2389, then we can say that the sum of it's digits is 1+1+2+3+8+9= 24. We can actually consider â99993. This number is obviously not an integer (by the argument that Dan made), but it still fulfills the requirements above.
To solve this problem in general, first notice that when not dealing with integer solutions, we don't care about the "square" part of the problem. We can find a 1-1 correspondence between the numbers that have sums of digits of their squares equal to 39 and numbers that have sums of their digits equal to 39. So let's find the probability that the sum of a numbers digits is equal to 39.
First of all, the set of all numbers such that the sum of their digits is equal to 39 is a subset of the set of all numbers with all but 39 digits of zero (if you have 40 or more non-zero digits, the sum of that numbers digits is ⥠40). This is a subset of numbers with all but finitely many digits being zero. This set has measure zero. Therefore any subset of it has measure zero, thus the set of numbers with the sum of their digits equal to 39 has measure zero (in the reals). Therefore the probability of this happening is zero.
2006-07-13 03:09:07 · answer #2 · answered by Eulercrosser 4 · 0⤊ 0⤋
39
2006-07-13 02:36:24 · answer #3 · answered by LindaLou 7 · 0⤊ 0⤋
If you are asking, "What is the probability that the sum of the digits of a number squared is equal to 39", then the answer is infinitely probable. Otherwise I have no clue what you are talking about.
2006-07-13 02:43:38 · answer #4 · answered by Chaosman 3 · 0⤊ 0⤋
0^2 = 0
1^2 = 1
2^2 = 4
3^2 = 0
4^2 = 7
5^2 = 7
6^2 = 0
7^2 = 4
8^2 = 1
9^2 = 0
all modulus(9) since we use base(10)
probly easier to list all outcomes and observe there
is no "3" on the right hand side
(I read the "cast-out-9's" answer and thought I'd
show it a bit easier than going the abstract route ...
that answerer was good)
2006-07-13 03:03:44 · answer #5 · answered by atheistforthebirthofjesus 6 · 0⤊ 0⤋
Well there is an infinite number of square numbers and only one can possibly have a sum of 39.
1 divided by infinity is 0.
Therefore 0
2006-07-13 02:38:36 · answer #6 · answered by Anonymous · 0⤊ 0⤋
your phrasing is bad! but what ever your question mean the answer is zero..because there are finite number of cases that fulfill your condition:(sum of digit of a square of a number is 39)
and there is infinite number of numbers(space)! right?
2006-07-13 04:51:54 · answer #7 · answered by mohamed.kapci 3 · 0⤊ 0⤋
hey!tricky. I think its 39!
2006-07-13 02:37:39 · answer #8 · answered by Rise__And__Shine__ 1 · 0⤊ 0⤋
my guess is 1out of infinite number is the probability You were asking for the probablity right?
2006-07-13 02:43:06 · answer #9 · answered by ladyviciousblade 2 · 0⤊ 0⤋
very low.
QED. I am a genius.
2006-07-13 02:42:02 · answer #10 · answered by Anonymous · 0⤊ 0⤋