Posted by Marcus on Thursday, September 7, 2006 at 11:08pm.
Solve. I have no digits the same. My thousands digit is less than my tens digit. My hundreds digit is twice my ones digit. If you add my digits, the sum would be 25. Who could I be? Could I be 2 numbers?
We can not figure this out. Can someone help please.
Let's find out Marcus
If we have an arbitrary number we use letters. I'm not sure if you're using letters for 4th grade math or not, so correct me if I use methods you don't know yet.
Let the number be denoted by
ABCD Where A is the thousands digit, B is the hundreds digit, C is the tens digit and D is the ones digit.
We're told that
(1) A<>B,A<>C,A<>D,B<>C,B<>D,C<>D
no digits are the same; this <> means 'does not equal'
(2) A < C the thousands digit is less than the tens digit
(3) B = 2*D the hundreds digit is twice the ones digit
(4) A + B + C + D = 25 the digits sum to 25
Using (3) we can eliminate B in (4) to get
A + 2*D + C + D = A + C + 3*D = 25
Now we need to make a simple observation: A + C must have a remaider of 1 when divided by 3. Do see why? It's because 25 has a remainder of 1 if we divide it by 3.
We know A is a number from 1 to 9 and C is a number from 0 to 9.
After testing the numbers that satisfy the requirements I found only two solutions: 6874 and 7693.
I should mentions that you did't state this had to be a 4 digit number. Those are those only ones I checked for.
I thought I would elaborate just a little on how I found the numbers. I didn't have much motivation for this last night.
If you use the letters I gave in the previous post
then we have a number ABCD where A is the thousands digit, B is the hundreds digit, C is the tens digit and D is the ones digit.
Each of the digits goes from 0 to 9 except A which must go from 1 to 8, since it's less than C.
C must go from 2 to 9 because it's greater than A.
B is an even number since B=2*D thus
B= 2,4,6 or 8 it cannot be 0 since D would = 0 too.
Therefore D=1,2,3 or 4 those are the only possible choices.
If D=1, then B=2, but A+C=25-B-D=23
This is impossible since the max for A is 8 and the max for C is 9. Therefore the largest sum of A+C is 17. We can also see that
if D=2, then B=4 and A+C=25-B-D=19 is not possible either.
If D=3, then B=6 and A+C=25-B-D=16
Now run through the numbers where
A<C and A+C=16 A=7 and C=9 is the only possible answer here so 7693 is a solution.
If D=4, then B=8 and 25-B-D=13
Now run through the numbers where
A<C and A+C=13
There are more possibilities here
4+9, 5+8,6+7
We exclude the 4,9 and 5,8 combinations because all the numbers must be different and we've already got a 4 and 8, so 6874 is the only other solution.
Thus the only solutions are 7693 and 6874.
Hopefully you can follow all of the reasoning. I am a little surprised that a problem like this would be given to 4th graders. I suspect that even 8th graders might find this problem just a little challenging.
Hopefully this helps.
Roger, I believe this boy is a 4th grader. He does not understand how you solved this simple problem. It would be better to put it in a language that a 4th grader would understand.
I don't know if he's the 4th grader or not. Sometimes a parent poses these questions too. Also, I'm not entirely sure how or what math is taught to 4th graders. I know they cover fractions, multiplication tables, and other important material, but I'm really quite surprised that this question would be asked at that grade level. Hopefully you're aware that posing even simple questions doesn't mean there are simple answers too. Elementary number theory is filled with questions like this that require college algebra to answer them.
I was doubtful that letters (or variables if you prefer) would be used in problems here, but the explanation will be much longer if I don't use them.
I also hope that if the person posting the question doesn't understand the solution that they'll ask questions. I'll try again if they want me to.
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