$\orange{\text{A }} \green{\text{pattern }}$

Algebra Level 2

Examine the pattern below and notice that the $\orange{\text{orange digits}}$ increment by one each time:

$\large \begin{aligned} 500 \times 500 & = \orange{250}\green{000} \\ 501 \times 501 & = \orange{251}\green{001} \\ 502 \times 502 & = \orange{252}\green{004} \\ 503 \times 503 & = \orange{253}\green{009} \\ 504 \times 504 & = \orange{254}\green{016} \\ & \vdots \end{aligned}$

What is the first value larger than $504$ for which the "increased by one" pattern in the $\orange{\text{orange digits}}$ is broken?

Inspiration: this daily challenge

The answer is 532.

5 solutions

Siddharth Chakravarty
Jun 25, 2020

Suppose the 2-digit number after 500 is x. Then the number can be written as (500+x). If we square the number i.e (500+x)² and expand it using the identity a² + 2ab + b², the expansion is

$\boxed{500²+2 \times 500 \times x + x²}$

$\boxed{250000 + 1000 \times x + x²}$

Here the value of the 2nd and 3rd digit keep on changing normally as x increases, however, the pattern breaks when x² becomes a 4 digit number because then 1 is carried to the left side increasing the value of the 3rd digit by 1.

So the first number which has x² as a four-digit no. is 32 as 32² = 1024.

A Former Brilliant Member
Jun 20, 2020

If we see the green numbers, then $(500+x)^2=500^2+2\cdot x\cdot 500+x^2=(250+x)\cdot 1000+x^2$ . The first $x$ , whose square is bigger then $1000$ is $\left \lceil \sqrt{1000} \right \rceil=32$ .

Chew-Seong Cheong
Jun 20, 2020

Let the number be $N(n) = (500+n)^2$ , where $n$ is a non-negative integer. Then $N(n) = (500+n)^2 = 500^2 + 1000n + n^2 = 250000 + 1000n + n^2$ . The order is okay until $n^2$ is a 4-digit integer. Then smallest 4-digit $n = \left \lceil \sqrt{1000} \right \rceil = 32$ , because $32^2 = 1024$ . Then $N(32) = \boxed{532}^2 = \red{533}024$ .

@Mahdi Raza , you need to add & to align. I have put it before the = so that it aligns the = signs. Note that \vdots (vertical dots) also align with = signs. You need only to use one \large without the braces in front of \begin{align} so that all the lines are affect. Don't use \Large use \large is large enough. If not it is like a children's book especially with colors. I would not have used \large for this case.

Chew-Seong Cheong - 11 months, 3 weeks ago

Ok, i think you have edited it. Thank you!

Mahdi Raza - 11 months, 3 weeks ago

Yes, I have edited it. Can you see the changes?

Chew-Seong Cheong - 11 months, 3 weeks ago

@Chew-Seong Cheong – Yes when i go in edit problem, i see changes in latex

Mahdi Raza - 11 months, 3 weeks ago

Sahil Goyat
Jun 23, 2020

i=501
while 1:
    if int(str(i**2)[0:3])!=int(str((i-1)**2)[0:3])+1:
        break
    i+=1
print(i)

you should use #(comments) for better understanding

Razing Thunder - 11 months, 1 week ago

they are not needed for such an easy question

Sahil Goyat - 11 months, 1 week ago

A Former Brilliant Member
Aug 11, 2020

The first perfect square number to have a digit on thousandth place is $32^2$ .

A pattern \orange{\text{A }} \green{\text{pattern }} A pattern

The answer is 532.

5 solutions

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$\orange{\text{A }} \green{\text{pattern }}$