There is a list of 7 distinct positive integers less than 10. One can not deduce whether their sum is a multiple of 3 from their product. What is their product?

Sunday, August 21, 2011

Please email your solutions to puzzles@cotpi.com.

### 4 comments

### Credit

This is an original cotpi puzzle inspired by:

- Andreescu, Titu; Andrica, Dorin; Feng, Zuming. 104 Number Theory Problems: From the Training of the USA IMO Team. Boston: Birkhäuser, 2007. p. 4. Example 1.4.

### Further reading

The following is a list of resources on related topics:

- Pal, Susam. "Solving the Impossible Puzzle." Susam Pal. 21 Mar. 2010. 9 Sep. 2011 <http://susam.in/blog/solving-impossible-puzzle/>.
- Weisstein, Eric W. "Modular Arithmetic." Wolfram MathWorld. 29 Aug. 2011. Wolfram Research. 9 Sep. 2011 <http://mathworld.wolfram.com/ModularArithmetic.html>.

## Gemini6Ice solved this puzzle:

9! / 18 = 20160.

There are 9 possible numbers under 10. Two are missing from the set of 7.

(2, 9) and (3, 6) are the only two pairs of missing integers where one pair sums to a multiple of 3 and the other doesn't. And they have a product where we cannot certainly determine the seven numbers.

## Rajesh Balakrishnan solved this puzzle:

Product: 20160.

2 numbers need to be skipped from {1, 2, …, 9}. There are some pairs of factors that lead to the same product. e.g. 2 × 6 = 3 × 4, 1 × 6 = 2 × 3, 4 × 6 = 3 × 8, etc. But only 2 + 9 and 3 + 6 offer sums where one is a multiple of 3 and the other is not.

## Ken solved this puzzle:

The product can be got by more than one of the 7-member sets. So, the missing two numbers in each set must have the same product. Possible absentee pairs are (1, 6) and (2, 3); (1, 8) and (2, 4); (2, 6) and (3, 4); (2, 9) and (3, 6); (3, 8) and (4, 6).

Sums of digits in pairs are 7 and 5; 9 and 6; 8 and 7; 11 and 9; 11 and 10.

Sums of rest of sets are 38 and 40; 36 and 39; 37 and 38; 34 and 36; 34 and 35.

Only one pair of sets is mod-3-ambiguous, 1 × 3 × 4 × 5 × 6 × 7 × 8 = 1 × 2 × 4 × 5 × 7 × 8 × 9 = 20160

## Dan Hoey solved this puzzle:

It is easier to solve this problem by considering the two positive integers from {1, 2, 3, …, 9} that are not among the seven given. The product of the two numbers is in one-to-one correspondence with the product of the seven given, and the sum of the two numbers mod 3 is the negative of the sum of the sum of the seven given. Therefore, a necessary and sufficient condition is: one cannot deduce whether the sum of the two is a multiple of three from their product.

If the product were 1 (mod 3), then the two numbers would either both be 1 (mod 3) or both be 2 (mod 3), so one could deduce that the sum is not a multiple of 3.

If the product were 2 (mod 3), then the two numbers would be 1 (mod 3) and 2 (mod 3), so one could deduce that the sum is a multiple of 3.

Therefore, the product is divisible by 3. If the product were not divisible by 9, then one of the numbers would be a multiple of 3 and the other a non-multiple of 3, so one could deduce that the sum is not a multiple of 3.

Therefore the product is divisible by 9, and we can't tell from their product whether the two numbers are {x, 9y} or {3x, 3y}. In order that the numbers be less than 10 and distinct, x = 2 and y = 1.

Therefore, the seven given integers are either {1, 3, 4, 5, 6, 7, 8} or {1, 2, 4, 5, 7, 8, 9}. Their product is 20160.