# 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).

Relevance

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

• Anonymous
5 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

• 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.

• 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

• 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.

• Anonymous
5 years ago

41, 43, 45

(41²) + (43²) + (45²) = 5555

• Hmmm....this appears to be my PIN number. I find this quite disconcerting.