Non Multiple Remainder Puzzle
Suppose we have four integers, no two of which are congruent $ \pmod 6 $. Let $ N $ be the product of the four integers.
If $ N $ is not a multiple of $ 6 $, then what is the remainder of $ N $ when $ N $ is divided by $ 6 $?
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
Since no two of the four integers are congruent modulo 6, they must represent four of the possible residues $ 0,1,2,3,4,5 $.
None of the integers can be $ 0\pmod 6 $, since this would make their product a multiple of 6.
The remaining possible residues are $ 1,2,3,4,5 $. Our integers must cover all but one of these, so either 2 or 4 must be among the residues of our four integers. Therefore, at least one of the integers is even, which rules out also having a multiple of 3 (since this would again make $ N $ a multiple of 6). Any integer that leaves a remainder of 3 (mod 6) is a multiple of 3, so such integers are not allowed.
Therefore, the four integers must be congruent to $ 1,2,4, $ and $ 5\pmod 6 $. Their product is congruent modulo 6 to $ 1\cdot 2\cdot 4\cdot 5=40 $, which leaves a remainder of $ \boxed{4} $.
Freeflows