2006-08-25 03:41:24 · 21 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
every multiple of nine consists of digits which sum to nine. so this can be changed to 3 times, 4 times etc...
2006-08-25 03:56:52 · answer #1 · answered by Gary L 2 · 0⤊ 0⤋
18
2006-08-25 12:48:07 · answer #2 · answered by CSUFGrad2006 5 · 0⤊ 0⤋
18
2006-08-25 10:47:05 · answer #3 · answered by Anonymous · 0⤊ 0⤋
18
2006-08-25 10:46:00 · answer #4 · answered by sebsilversmith 1 · 0⤊ 0⤋
18
2006-08-25 10:45:26 · answer #5 · answered by a_strzynski 1 · 0⤊ 0⤋
18
2006-08-25 10:45:25 · answer #6 · answered by Anonymous · 0⤊ 0⤋
4
2006-08-25 11:28:44 · answer #7 · answered by mark!!001 2 · 0⤊ 0⤋
4
2006-08-25 10:50:41 · answer #8 · answered by vineet s 1 · 0⤊ 0⤋
The number xy really means x*10 + y
To say that the number is the same as 2 times the sum of its digits would be written as:
10*x + y = 2(x+y)
or
8x = y
We know that x is a single digit so there is only an answer when y = 1.
2006-08-25 11:02:45 · answer #9 · answered by tbolling2 4 · 0⤊ 1⤋
10x + y = 2(x + y)
10x + y = 2x + 2y
-y = -8x
y = 8x
y = 8(1) = 8
10(1) + 8 = 10 + 8 = 18
since y = 8(2) = 16 would give you a 2 value in the ones place, and y = 8(0) = 0 would give you 0 as your value.
so the number is 18.
2006-08-26 02:13:28 · answer #10 · answered by 5abiKudi_USA 3 · 0⤊ 0⤋
Suppose the number ...fedcba (a, b, c, d, e, f... are digits 0-9).
...fedcba = a + 10b + 100c +1000d + ...
The sum of its digits is a + b + c + d + ...
Therefore,
a + 10b + 100c +1000d + ... = 2a + 2b + 2c + 2d + ...
=> -a + 8b + 98c + 9998d + ... = 0
or 8b + 98c + 9998d +... = a
If you remember a is a digit 0-9.
So, 8b + 98c + 9998d +... is a digit 0-9.
The case are 0 for b, c, d, ... = 0 (a=0)
and 8 for b = 1 and c, d, e, ... = 8 (a=8)
So, the cases are two (ONLY)!
0 and 18.
After, I have seen some of the answer, this is the only one that proves that there is a limited number of solutions. (By saying 18 you don't prove that it is the only solution).
2006-08-25 11:06:45 · answer #11 · answered by Christos :) 2 · 0⤊ 0⤋