# Is there a 4-digit number with all its digits equal and which is the sum of the squares of 3 consecutive odd integers? If so find it (show?

Is there a 4-digit number with all its digits equal and which is the sum of the squares of 3 consecutive odd integers? If so find it (show your derivation and not just the answer).

### 7 Answers

- llafferLv 75 years agoFavourite answer
The sum of the squares of three consecutive odd integers, then the result has to be odd.

So that narrows down our possible answers to:

1111, 3333, 5555, 7777, and 9999.

So let's determine an expression for this and see what we get:

I'll call the middle of the three consecutive odd numbers "x", so the sum of the squares work out to be:

(x - 2)² + x² + (x + 2)² = k

simplify that a little:

x² - 4x + 4 + x² + x² + 4x + 4 = k

3x² + 8 = k

3x² = k - 8

So here we know that k - 8 has to be divisible by 3, and it's quotient is to be a perfect square. So since we only have 5 other candidates, let's brute force this the rest of the way to at least see if we can eliminate some options.

1111 - 8 = 1103 (not divisible by 3)

3333 - 8 = 3325 (not divisible by 3)

5555 - 8 = 5547 (is divisible by 3)

7777 - 8 = 7769 (not divisible by 3)

9999 - 8 = 9991 (not divisible by 3)

So we only have one number that is, let's see if the result is a square:

5547 / 3 = 1849

And the square root of that is:

√1849 = 43

So that is your middle number. The other two are:

41 and 45.

The sum of these numbers squared:

41² + 43² + 45²

1681 + 1849 + 2025

5555

Those are your numbers.

- Anonymous5 years ago
The number must be odd and equal 1111 times k, where k = k = 1, 3, 5, 7, or 9

The middle number would have to be close to √1111k

Substituting the values for k we can calculate that, then solve for the sum of the squares of the 3 possible odd numbers. Note that √1111k must give you the middle one.

k = 1 gives 19.2440467 17 19 21 1091=sum of 3 squares

k = 3 gives 33.33166662 31 33 35 3275

k = 5 gives 43.03099658 41 43 45 5555

k = 7 gives 50.91496178 49 51 53 7811

k = 9 gives 57.73214010 55 57 59 9755

- xyzzyLv 75 years ago
if such a number exists.

the sum of the squares of 3 consecutive odd integers

b = middle of the 3

(b-2)^2 + b^2 + (b+2)^2

3 b^2 + 8

if such a numbers exists... it is odd (as pointed out by others)

and aaaa - 2 is divisible by 3

aaa0 is always divisible by 3

so...

a - 2 is divisible by 3

5 is the only candidate

5555-8 = 5547

5547/3 = 1849

1849 is indeed a perfect square.

- ThomasLv 75 years ago
1111a=n^2+(n+2)^2+(n+4)^2

1111a=n^2+n^2+4n+4+n^2+8n+16

1111a=3n^2+12n+20 so we can say

3y^2+12y+20=1111x

3y^2+12y=1111x-20

y^2+4y=(1111x-20)/3 now by completing the square...

y^2+4y+4=(1111x-8)/3

(y+2)^2=(1111x-8)/3

y+2=[(1111x-8)/3]^(1/2)

y=[(1111x-8)/3]^(1/2)]-2, x can only have integer values of 1 through 9 so....

y is only an integer when x=5 so y=41

So the consecutive odd numbers are 41, 43, 45 with a sum of squares of 5555

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- DWReadLv 75 years ago
Let "aaaa" represent the digits of the number.

The value of "aaaa" is 1111a.

The consecutive odd integers can be represented by (2n+1), (2n+3), and (2n+5), where n is an integer.

Sum of their squares = (2n+1)² + (2n+3)² + (2n+5)²

= (4n²+4n+1) + (4n²+12n+9) + (4n²+20n+25)

= 12n² + 26n + 35

12n² + 26n + 35 = 1111a

There is no digit a for which n is an integer.

- Anonymous5 years ago
41, 43, 45

(41²) + (43²) + (45²) = 5555