Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**ZHero****Real Member**- Registered: 2008-06-08
- Posts: 1,889

Find a number (mathematically & logically) which is square of the sum of its digits!

How many digits can the number have?

"Zero is the real Hero, that's y i'm ZHero!"

*Last edited by ZHero (2008-06-08 12:26:56)*

If two or more thoughts intersect, there has to be a point!

Offline

**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

81 is one possible answer.

Offline

**ZHero****Real Member**- Registered: 2008-06-08
- Posts: 1,889

JaneFairfax wrote:

81 is one possible answer.

yes.. It OBVIOUSLY is the answer but how does one WORK IT OUT? Mathematically & Logically??

If two or more thoughts intersect, there has to be a point!

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

1 of course is the simplest non-zero answer. In order to solve for a solution, one must solve the equation:

Solving this equation for a, we get:

The trick now is to find b such that -36b + 100 is a perfect square. 0 works, but it leads to the trivial solution (i.e. 0) and a = 10. 1 works as well, which leads to both 81 and 1 itself.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Offline

**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

A faster method:

**0 = 0 (yes)****1 = 1 (yes)**

4 = 4 (no)

9 = 9 (no)

1 + 6 = 7 (no)

2 + 5 = 7 (no)

3 + 6 = 9 (no)

4 + 9 = 13 (no)

6 + 4 = 10 (no)**8 + 1 = 9 (yes)**

1 + 0 + 0 = 1 (no)

1 + 2 + 1 = 4 (no)

.

.

.

etc etc etc

*Last edited by JaneFairfax (2008-06-08 14:45:59)*

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Jane, that method isn't based in logic, it's just an algorithmic test by exhaustion.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Offline

**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Proof by exhaustion is still a logical proof.

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

I believe you're not interpreting his words properly. The way I interpreted them is that "logical" meant that you didn't go by exhaustion, that you used logic (math) to find them instead.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Offline

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Would it count as logical if you proved that no number bigger than, say, 50 could possibly fit the condition, and then tested all the others individually (treating them as special cases)?

Why did the vector cross the road?

It wanted to be normal.

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Typically, yes. However, this does not seem to be in the same sense that ZHero is using the word.

Offline

**ZHero****Real Member**- Registered: 2008-06-08
- Posts: 1,889

Hi guys!

Firstly, i'm new to the site & i'm really pleased to see so many people interested/engrossed in Mathematics!

Maths, no doubts, is most Beautiful, Puzzling & Fathomless of all things!

I think i've found a nice Mathematical way to crack the above problem but i'll find it a bit difficult to post it coz i don't know how to code for equations n i'm surfing through my Nokia6600 rather than computer..

Here it goes...

If two or more thoughts intersect, there has to be a point!

Offline

**ZHero****Real Member**- Registered: 2008-06-08
- Posts: 1,889

Let no. be 10x+y (x,y=0-9;x=!0).

Now..

10x + y = (x+y)^2

9x = (x+y)^2 - (x+y)

let's say x + y = z

so..

9x = z^2 - z = z(z-1)

consider rhs.. Its a product of two CONSECUTIVE numbers!

So, 9x should be multiple of two CONSECUTIVE numbers & x MUST be a ONE DIGIT number which leads to only consecutive number to 9 i.e. '8'!

So.. x = 8 is the only solution which in turn gives y = 1!

So the number is 81!

If two or more thoughts intersect, there has to be a point!

Offline

**ZHero****Real Member**- Registered: 2008-06-08
- Posts: 1,889

The required number can only be 'one/two digit' number can also be shown..

An 'n digit' number 'a' would satisfy the above question only if "square of the sum of the digits of the GREATEST 'n digit' number EQUALS/EXCEEDS the number'!

Greatest 'one digit' number is 9..

9^2 = 81 > 9... Satisfies!

Greatest 'two digit' number is 99..

9+9 = 18

18^2 = 324 > 99... Satisfies!

Greatest 'three digit' number is 999..

9+9+9 = 27

27^2 = 729 < 999.. Doesn't Satisfy!!

& so on...

Comments are most welcome!!

Regards...

If two or more thoughts intersect, there has to be a point!

Offline

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

ZHero wrote:

The required number can only be 'one/two digit' number can also be shown..

An 'n digit' number 'a' would satisfy the above question only if "square of the sum of the digits of the GREATEST 'n digit' number EQUALS/EXCEEDS the number'!

I don't agree with this.

For example, take 199. 1+9+9 = 19 and 19² = 361 > 199, so even though the highest 3-digit number is greater than the sum of its digits, there are still some 3-digit numbers that are less (which means there's no reason why there couldn't be some that are equal).

Testing exhaustively shows that there aren't any, but you couldn't say that without trying them all.

However, you could use a similar argument to prove there aren't any 5-digit (or higher) numbers with that property though.

The highest sum-of-digits-squared number would be made by 99999.

5x9 = 45, and 45² = 2025.

The highest sum-of-digits-squared number is lower than the lowest 5-digit number (10000), and so there can't be any 5-digit numbers equal to the squared sum of their digits.

Why did the vector cross the road?

It wanted to be normal.

Offline

**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Well, then. It looks like what ZHero means by logically is somewhere between my proof by exhaustion and Rickys algebraic proof.

Offline

**ZHero****Real Member**- Registered: 2008-06-08
- Posts: 1,889

Very well said mathsyperson!

I'm thinking again..

There's ONLY ONE such number & hence it should be uniquely determinable!!

Consider for example..

Find a number which is square of its last digit (digit in unit's place)?

Ans (quite obvious) is 25 & 36 and these can be determined using almost the same logic as above, for solving the equation!

If two or more thoughts intersect, there has to be a point!

Offline

**ZHero****Real Member**- Registered: 2008-06-08
- Posts: 1,889

Consider a three digit number (100a+10b+c) ; a, b, c ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} and a≠0

(100a+10b+c) = (a+b+c)[sup]2[/sup]

⇒ 99a+9b = (a+b+c)[sup]2[/sup] - (a+b+c)

⇒ 9(11a+b) = (a+b+c)(a+b+c-1)

Since LHS of the equation is a Multiple of 9 and that the Maximum Value of RHS (a+b+c) can be (9+9+9) = 27, we have the following three cases...

CASE-I: 9(11a+b) = 9×8

11a+b = 8 ⇒ a=1 and b=-3 Not Possilbe (b can't be -ve)

CASE-II: 9(11a+b) = 18×17

11a+b = 34 ⇒ a=3 and b=1

a+b+c = 18

c = 18 - (a+b) = 18 - (3+1) = 14 Not Possilbe (c=14 is not possible)

CASE-III: 9(11a+b) = 27×26

11a+b = 78 ⇒ a=7 and b=1

a+b+c = 27

c = 27 - (a+b) = 27 - (7+1) = 19 Not Possilbe (c=19 is not possible)

Hence, we can conclude that NO SUCH THREE DIGIT NUMBER IS POSSIBLE!

I think it'd be much easier to show this for higher digit numbers!

If two or more thoughts intersect, there has to be a point!

Offline

**TheDude****Member**- Registered: 2007-10-23
- Posts: 361

You need to check 2 more cases.

CASE-IV: 9(11a+b) = 10x9

11a+b = 10 ⇒ a=1 and b=-1 Not Possible

CASE-V: 9(11a+b) = 19x18

11a+b = 38 ⇒ a=3 and b=5

a+b+c = 19

c = 19 - (a+b) = 19 - (3+5) = 11 Not Possible

Wrap it in bacon

Offline

**ZHero****Real Member**- Registered: 2008-06-08
- Posts: 1,889

TheDude wrote:

You need to check 2 more cases.

CASE-IV: 9(11a+b) = 10x9

11a+b = 10 ⇒ a=1 and b=-1 Not PossibleCASE-V: 9(11a+b) = 19x18

11a+b = 38 ⇒ a=3 and b=5

a+b+c = 19

c = 19 - (a+b) = 19 - (3+5) = 11 Not Possible

I always miss one thing or the other..........

If two or more thoughts intersect, there has to be a point!

Offline

Pages: **1**