This is a fun problem from my 4th grader homework that I can't solve. In fact, my son went back to school yesterday admitting to his teacher that neither he nor his mom could solve it. The teacher mused that it may be too easy for The Math Mom, dissolving my hopes for something being misprinted. I am intrigued and perplexed. Can you help?
Problem: A student saw that an addition problem had every digit from 1 to 9. Unfortunately she spilled ink on her paper and only 4 digits remained. Can you reconstruct the addition problem?
3 4 *
* * 9
+ ------
5 * *
Even worse, a note at the bottom of the paper says: How many solutions can you find?
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7 comments:
348
+ 219
-----
567
Wow!
One more point for Kim and a punch for TheMathMom. How did I not see this?
I should have gone for a jog or a swim. What strategies do you use when you are stuck on one approach and could not see the larger picture?
This is called the nine digit puzzle. The objective is to use the dine digits (1,2,3,4,5,6,7,8, and 9) to make a correct three digit addition problem.The sum of the digits in the solution is always 18. (except for these three sums: 684, 756, and 765. Knowing this helps you to find solutions. There are over 300 possible solutions.
This was fun! OK, let's separate this into cases:
Case 1: Let's say the hundreds digit is 2
Then we better not "carry" a 1 from the tens place, otherwise we won't end up with 5 in the answer, so:
Case 1a: If the ones digit is zero, then the tens digit could be 0-5, and we'd still end up with our "5" as we need for the hundreds digit in the answer. That's *6* solutions.
Case 1b:
If the ones digit is anything from 1-9, then we'll be carrying a 1 to the tens digit, which means our options for the missing tens digit has been reduced to 0-4, just five digits This gives us 5 X 9 = *45* solutions.
Total for this "case 1" is 6 + 45 = 51 solutions.
Case 2: Let's say the missing hundreds digit is 1
Now we need to carry from the tens, in order to end up with 5 in the answer!
Case 2a
Again, first let's consider the missing ones digit to be zero. Then we'd need the missing tens digit to be 6 or higher, that is; 6-9. This gives us *4* new solutions.
Case 2b
Now we'll consider if the missing ones digit is 1-9. This means we carry a 1 to the tens digit, and this "helps" our case, meaning that our options for the missing tens digit are now 5-9. this gives us 5 X 9 = *45* solutions.
So case 2 gives us 4 + 45 = 49.
I don't see any other possible cases, which means that the answer is:
51 + 49 = 100 solutions!
I hope you see how to generate as many as 100 solutions from this discussion. For example,
341
+169
----
510
Thanks for the puzzle! It was fun!
Mandy emailed to me the following comment:
The answer to the addition problem where only 4 known digits are visible is as follows.
348
+ 219
567.
There are no other possibilities.
I agree with Mandy. In the comment above D.M. forgot that every digit from 1 to 9 should be used and only once. Thank you for the clarification, Mandy, and welcome to the marathon.
The combinations are ... 738+216=954
478+215=693 183+276=459 327+519=846 there is more also
I have to do one in seminar. 28*+**4=***
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