Iamblichus's sum

Take three consecutive integers, with the largest a multiple of 3. Form their sum. Compute the sum of its digits, do the same for the result until a one-digit number is obtained. Iamblichus of Chalis claimed that the one-digit number obtained will always equal 6. For example, the sum of 9997, 9998 and 9999 is 29 994. The sum of the digits of 29 994 is 33 and the sum of the digits of 33 is 6. Prove Iamblichus's claim.

#NumberTheory

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:

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
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# 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`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Claim: Finding the digit sum of a number, $x$ , is equivilant to $x \pmod{9}$ .

Proof: Let $x$ have five digits (for simplicity), $a,b,c,d,$ and $e$ .

$x$ can be written as $10000a+1000b+100c+10d+e$

This is the same as $(9999a+999b+99c+9d)+(a+b+c+d+e)$

Therefore, $x \pmod{9} \equiv a+b+c+d+e \pmod{9}$ Since the rest of the expression above is divisible by 9.

This can be done continually, until a number less than 9 has been attained.

Now back to the original problem.

The three numbers can be written as $3n+1, 3n+2$ and $3n+3$

Adding them up results to $9n+6$

The digit sum of that is $9n+6 \pmod{9} \equiv 6$ because $9n$ is divisble by $9$

Therefore the digit sum of $3$ consecutive integers, when the largest number is a multiple of $3$ is always $\fbox{6}$

Δrchish Ray - 2 years, 4 months ago

Ok.Then what?

chakravarthy b - 2 years, 4 months ago

Sorry, i had to eat dinner.

Δrchish Ray - 2 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$