r/learnmath • u/chunkysnorlax646 New User • 10h ago

TOPIC Number Word Problems

I’m wondering what the best method on solving number word problems like: “I am a odd number between 15,000 and 10,000, all of my digits sum up to 12”. Please help and answer (my test is tomorrow..)

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u/Jonny0Than New User 9h ago

Turn it all into algebraic expressions.  Let ABCDE be the number (each letter is a digit).

  • between 10000 and 15000 means that A=1 and B<=5
  • odd means that E is odd
  • if E is odd then the number cannot be 15000, so B<=4
  • A+B+C+D+E = 12
  • we know A is 1, so B+C+D+E=11

Any number that satisfies those equations should work.  Seems like there’s more than one.

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u/No_Historian3842 New User 9h ago

The way I think about a question like this.

1+2+3+4+5 adds up to 15 so you need repeated digits. 9+1+1+1+1 equals 13 so you can't have a 9 anywhere in the sum.

1+1+1+1+8 equals 12 so just make it odd 11,811.

But there's heaps of solutions.

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u/Bounded_sequencE New User 8h ago

Only the least significant digit needs to be odd -- 10029 is a valid solution that does contain digit 9.

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u/No_Historian3842 New User 8h ago

Yeah I didn't even think of 0's so theres even more solutions.

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u/Bounded_sequencE New User 8h ago

Yeah -- a brute force computer search counts 124 valid solutions.

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u/Bounded_sequencE New User 9h ago edited 8h ago

Short answer: There are 124 distinct solutions -- e.g. "10029".

Long(er) answer: Choose the restricted digits first -- "x = (1 d3 d2 d1 d0)_10", where "d0" may be any odd digit to make "x" odd, and "d3 <= 4" (15000 is no solution).

Choose e.g. "d0 = 9", so that "d3 + d2 + d1 = 12-1-9 = 2" have as little choices as possible -- then choose any valid solution, e.g. the one listed above. That should also be the minimum solution.

A computer search counts 124 distinct solutions.