How do I solve this algebra word divistion problem?
- Thread starter RedGuy12
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D
Deleted member 4993
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How many letters have you used for "unique" numbers? You have:
s, e, t, o, n, d, i, a, f, u
Now start writing equations.
s - d = u..................or...................10 + s - d = u
Continue.....
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Please share your work/thoughts about this assignment.
There is nothing in the puzzle description that says S or D cannot be zero. Can you prove those statements?
D
Deleted member 4993
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Those are the "Begining" (left-most) digits of the dividend, quotient, etc.
s * stone \(\displaystyle \to\) sssss if s=0
A number can have zero at the beginning (left-most-digit) - but that would be unusual!!
Steven G
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What does unusual mean in math? I've seen unusual things in math but finding a zero at the beginning of a number isn't even on the list.
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My point was that you can prove that S is not zero; there is no need to assert it based on an assumption about unstated rules. I personally enjoy puzzles and believe that they are good tools for teaching how to solve problems, but making unsupported assumptions defeats the purpose.
Although I did not find it the obvious way to start, you certainly may start by letting x = the number represented by STONE, y = the number represented by DISEASED, z = the number represented by SEE, and p = the number represented by SONUAO.
[MATH]\dfrac{y}{x} = z + \dfrac{p}{x} \implies y = xz + p. [/MATH]
[MATH]S \ne 0 \implies 0 < S. [/MATH] I shall let the OP prove this.
[MATH]\therefore x > 10,000S,\ z \ge 100S,\text { and } p > 100,000 \implies xz + p > 1,000,000S^2 + 100,000.[/MATH]
By the rules of the puzzle, D < S or D > S.
[MATH]D < S \implies y < 1,000,000(S - 1) + 999,999 = 1,000,000S - 1.[/MATH]
[MATH]1 \le S \implies S \le S^2 \implies 1,000,000S \le 1,000,000S^2 \implies[/MATH]
[MATH]y < 1,000,000S - 1 < 1,000,000S \le 1,000,000S^2 < 1,000,000S^2 + 100,000 < xz + p.[/MATH]
But that is a contradiction. So D > S.
Summarizing
[MATH]1 \le S < D \le 9 \implies 1 \le S \le 8, \ 2 \le D \le 9, \text { and } 1 \le D - S \le 8.[/MATH]
So the OP is correct, but for the wrong reason.
Although I did not find it the obvious way to start, you certainly may start by letting x = the number represented by STONE, y = the number represented by DISEASED, z = the number represented by SEE, and p = the number represented by SONUAO.
[MATH]\dfrac{y}{x} = z + \dfrac{p}{x} \implies y = xz + p. [/MATH]
[MATH]S \ne 0 \implies 0 < S. [/MATH] I shall let the OP prove this.
[MATH]\therefore x > 10,000S,\ z \ge 100S,\text { and } p > 100,000 \implies xz + p > 1,000,000S^2 + 100,000.[/MATH]
By the rules of the puzzle, D < S or D > S.
[MATH]D < S \implies y < 1,000,000(S - 1) + 999,999 = 1,000,000S - 1.[/MATH]
[MATH]1 \le S \implies S \le S^2 \implies 1,000,000S \le 1,000,000S^2 \implies[/MATH]
[MATH]y < 1,000,000S - 1 < 1,000,000S \le 1,000,000S^2 < 1,000,000S^2 + 100,000 < xz + p.[/MATH]
But that is a contradiction. So D > S.
Summarizing
[MATH]1 \le S < D \le 9 \implies 1 \le S \le 8, \ 2 \le D \le 9, \text { and } 1 \le D - S \le 8.[/MATH]
So the OP is correct, but for the wrong reason.
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D
Deleted member 4993
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I disagree. Assuming S & D are NOT 0 is not same as assuming S & D are (say) 1 & 2. If we come to a solution set with that assumption, for these types of problems, then that solution is valid. Then of course we may try to spend time in "proving" those assumptions.
Here of course those two jumped at me at the very beginning - with the "suspicion" (not assertion) that a number will not start with zero. Then of course s\(\displaystyle \ne\)0 can be proven easily (I did that looking at the given operation) and also d\(\displaystyle \ne\)0 can be proven by observing the first line of the division operation. And Jeff proved those in a different way.
My point is, I believe there is nothing wrong in acting on a hunch and getting a solution (Sherlock Homes did that - doctors of medicine do that) as long as it is not "blatantly" false - and if it is "bad" - the solution would not be reached (and if we get a "solution" with a bad hunch - that speaks non-uniqueness of the problem statement).
However, I do agree that after having and using the "hunch" - it needs to be proven beyond a "reasonable doubt" (must exist on Jomo's list of unusual things).
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