This discussion board is a place to discuss our Daily Challenges and the math and science
related to those challenges. Explanations are more than just a solution — they should
explain the steps and thinking strategies that you used to obtain the solution. Comments
should further the discussion of math and science.
When posting on Brilliant:
Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.
Markdown
Appears as
*italics* or _italics_
italics
**bold** or __bold__
bold
- bulleted - list
bulleted
list
1. numbered 2. list
numbered
list
Note: you must add a full line of space before and after lists for them to show up correctly
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
Math
Appears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3
2×3
2^{34}
234
a_{i-1}
ai−1
\frac{2}{3}
32
\sqrt{2}
2
\sum_{i=1}^3
∑i=13
\sin \theta
sinθ
\boxed{123}
123
Comments
The condition is the same as:w+x≡y+z(mod20). Notice that if we can find w≡y,x≡z, then we are done. Hence it suffices to consider the case where there's at most one residue of 20 such that at most 3 numbers are congruent to it.
In this case, we have at least 7 distinct residues considering each integer in mod 20 . Note that there are at least (27)=21 ways to choose 2 of the numbers and then take their sum. Using the 20 mod 20 residues as pigeon holes, there exist a residue r with at least two sums i.e w+x≡y+z≡r(mod20)(we can't have w=y because that would imply x≡z which contradicts having distinct residues) and we are done.
The easy bash: The 10 possible sets of end digits of the 9 integers are [1,2,3,4,5,6,7,8,9],[0,1,2,3,4,5,6,7,8],[2,3,4,5,6,7,8,9,0],[3,4,5,6,7,8,9,0,1][4,5,6,7,8,9,0,1,2],[5,6,7,8,9,1,2,3,4,],[6,7,8,9,1,2,3,4,5][7,8,9,1,2,3,4,5,6],[8,9,1,2,3,4,5,6,7]and[9,0,1,2,3,4,5,6,7]It suffices to show that 1+9−2−8=0 and 8+0−7−1=0 as by Pigeonhole Principle, each of the other 8 sets is just a rearrangement of the first 2. Not very elegant but it works. Since this works for single digit numbers, it works for any other set because each number in that set is basically adding a multiple of 20 to the original 1-digit set. (Not clear, I know but basically, I'm saying that for 10+19−12−18, it works because it is 1+9−2−8+10+10−10−10, or 1+9+2+8+0.)
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
The condition is the same as:w+x≡y+z(mod20). Notice that if we can find w≡y,x≡z, then we are done. Hence it suffices to consider the case where there's at most one residue of 20 such that at most 3 numbers are congruent to it.
In this case, we have at least 7 distinct residues considering each integer in mod 20 . Note that there are at least (27)=21 ways to choose 2 of the numbers and then take their sum. Using the 20 mod 20 residues as pigeon holes, there exist a residue r with at least two sums i.e w+x≡y+z≡r(mod20)(we can't have w=y because that would imply x≡z which contradicts having distinct residues) and we are done.
The easy bash: The 10 possible sets of end digits of the 9 integers are [1,2,3,4,5,6,7,8,9],[0,1,2,3,4,5,6,7,8],[2,3,4,5,6,7,8,9,0],[3,4,5,6,7,8,9,0,1] [4,5,6,7,8,9,0,1,2],[5,6,7,8,9,1,2,3,4,],[6,7,8,9,1,2,3,4,5][7,8,9,1,2,3,4,5,6],[8,9,1,2,3,4,5,6,7]and[9,0,1,2,3,4,5,6,7]It suffices to show that 1+9−2−8=0 and 8+0−7−1=0 as by Pigeonhole Principle, each of the other 8 sets is just a rearrangement of the first 2. Not very elegant but it works. Since this works for single digit numbers, it works for any other set because each number in that set is basically adding a multiple of 20 to the original 1-digit set. (Not clear, I know but basically, I'm saying that for 10+19−12−18, it works because it is 1+9−2−8+10+10−10−10, or 1+9+2+8+0.)
one more question, given the series:1,3,5,7,9,11,13,15. now choose 5 digits from here whose sum is equal to 30.(u can take a number twice also)
Log in to reply
Odd×5=Odd