Help with math problem? Is it impossible?

Question

I try to figure out this problem and my whole family but we think its impossible. my whole class asked my teacher and she said ill give u the weekend to firgure out.

When open the website look at number 4. Also remember it says 4 DIFFERENT DIGITS

http://img152.imageshack.us/my.php?image=njmath22fu7.jpg

The word digit means any number from 0-9 so and negative numbers wouldnt work

Answer 1

Or one can use algebra:

The hundreds digit is 1/2 of the thousands digit. So, if we call the hundreds digit "x", the thousands digit would be "2x."

Then, the ones digit is 2 less than the sum of the hundreds and thousands digits, or using the above, it is 2 less than (x + 2x) or: 3x - 2.

If we call the tens digit "t", we can check the sum of all the digits:

2x + x + (3x - 2) + t = 15 or
6x - 2 + t = 15 or
t = 17 - 6x

Now, let's consider the possibilities. The tens digit cannot be negative, nor >9, by the usual conventions of our numbering system. So, 17 - 6x cannot be negative or greater than 9 so x cannot be 3 or larger (the tens digit would be negative) or less than 4/3 (the tens digit would be greater than 9). So that leaves only the digit 2 in the tens spot.

Does 2 work though, with that final restriction? No. For instance, we are sure the first two digits are different as x cannot equal 2x without x = 0 in which case, the ones digit is negative (3x - 2 = -2) and that cannot be. Perhaps a different pair are the same? 17 - 6x = x or 2x? They lead to x = 17/7 and 17/8, neither of which are integers and so cannot be the hundreds digit! Maybe 3x - 2 = 17 - 6x? That leads to x =19/9 with the same problem. This leaves one last combination pair to try: 3x - 2 = x or 2x which leads to x = 1/2 or x = 2. Oops... x = 2! So, 3x - 2 = 2x when x = 2 and we found that the only value x can have is 2. No luck then, we have shown algebraicly that the first and the last digit must equal each other with these relationships. Since such an outcome is forbidden, there is no solution.

(I wonder what question the workbook author thought he wrote? Perhaps you'll be able to tell when you get the workbook's answer on Monday.)

Answer 2

ok the hundreds are half the 1000's..that means the possible combinations for those 2 are ..just based on this rule

84
63
42
21


so those a the POSSIBLE first two numbers

next,

the ones digit is 2 less than the sum of the hundred and thousand.

2+1=3-2=1 so the combination can't be 21
8+4=12-2=10..so that cant be it...so 8+4 is out

6+3= 9-2=7 AHA.....it fits, but wait

we need 4 numbers to add up to 15 and so far we have
6+3=9+7=16 therefore the only possible combination of the

first two numbers is

42

4+2=6-2=4 OK we already have a 4. so no good.


Therefore, based on the parameters, the question is not solvable. The questioners made a mistake in the printing.

You can use my breakdown to backup your argument

I think the MISPRINT IS the numbers should add up to 16
in which case you would have
6307!!! At least have this for a backup answer!!! Its the only thing that makes sence because remember we ruled out 3 possible combinations 21, 84 and 42 .


the only OTHER possibility I can think of is in the 84 combination 8+4=12-2= 10 as a trick

8410. However,, that adds up to only 13


so the answer is TEACHER, there is a misprint in the question. THE ONLY POSSIBLE WAY this can meet ALL of the criteria stated, is if the question should say the numbers add up to 16. Either that, or there is another serious misprint.

Below me, even a friggin rocket scientist with a phd in math PROVED that there is no answer.

Answer 3

The problem is impossible with 4 different digits.
Here's the proof:
Let a b c d be the digits of the number
10³a+10²b+10c+d.
We know that a is positive and less than 10
and b,c,d are nonnegative and less than 10.
The given conditions yield the following
b= ½a, so a = 2b
d+2 = a+ b
a+b+c+d = 15.
Since a = 2b,
d + 2 = 3b
so d = 3b -2.
The last equation now yields
2b+b+c+3b-2 = 15.
6b+c = 17.
Note that b cannot equal 0 or 1, since that
would make c greater than 10.
Also if b > 2 then c would be negative.
So b = 2 and c = 5.
This finally gives us
a = 4 and d = 4.
So the only number satisfying all the
conditions is 4254, which does NOT have
distinct digits.

Answer 4

as the hundredth digit is half the thousands digit, there is only possiblity of 4 numbers in hundreth place
i.e 1,2,3,4
so thousands place possiblity is 2,4,6,8

so the possiblity for the one ones digit is
1+2-2=1
2+4-2=4
3+6-2=7
4+8-2=10

out of 4, the last is not possible,the last is also ruled out as the answer for the ones digit is 10, which is not possible

so we are left only 3 options
i.e thounds place 2,4,6
hundreds place 1,2,3
ones place 1,4,7

now the question says the sum of all 4 digits , so try this with both the options

first option
2+1+x+1=15
4+x=15
x=11

second option
4+2+x+4=15
10+x=15
x=5

third option
6+3+x+7=15
9+7+x=15
16+x=15
x=15-16
x=-1

the first & third option are not possible as the tens digit cant be double digit or negative.

so the tens digit no is 5
so the no is
4254

Answer 5

if the thousands is double the hundreds then we can determine that the thousands are even since all numbers are integers

if we take 6 as the thousands, then 3 will be the hundreds, so 6+3 -2 is 7, 6+3+7=16, so we know that the thousands has to be less than 6, which leaves 4 and 2, for 4 we get, 42x4, which has two 4's this is incorrect, for 2 we have 21x1 which has 2 1's, thus we can conclude that this has no answer.

Answer 6

i think it is impossible because the thousands and hundreds digit ha to b eeven so it can be half but then when u factor in the ones digit it is either a numbe that has already beend used of it equals more than 15.

Answer 7

4+2=6 and 2 less than 6 is not 0.

and i agree it's impossible, cuz i was sitting here with a piece of paper trying everything:D

Answer 8

Its 4254, it works. Hope I helped.

oh no wait. Its imposible. none of the digits can be the same

Answer 9

The answer is....

First Digit = 6
Second Digit = 3
Third Digit = -1
Fourth Digit = 7

Does that work

~PinoyPlaya

Answer 10

4254 that should be right, you can double check if you like.

[Edit] sorry didn't see that it had to be all different digits.


person is right there is no solution. Your teacher must have made a mistake.