Help me answer this problem

I need someone to tell me how to solve this It’s from Australia math comp

Adding 1 to the product of 4 consecutive integers always result in a perfect square. The first 2017 such square numbers can be found: $1 \times 2 \times 3 \times 4 + 1 = 25 = 5^2$ $2 \times 3 \times 4 \times 5 + 1 = 121 = 11^2$ $3 \times 4 \times 5 \times 6 + 1 = 25 = 19^2$

In the list of 2017 numbers $5,11,19,...,4074341$ whose squares are found in this way, how many have last digit equal one?

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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Comments

Say we take your expression and write it in the form: $x(x+1)(x+2)(x+3)+1$

Expanding it, we get $x^4 + 6x^3 +11x^2 + 6x+1$

Factoring the expression, we arrive at $(x^{2}+3x+1)^{2} ∎$

To solve the second part, we need to find out the last digit of $x^{2}+3x+1$ .

To do this, let’s write $x$ as $5a+b$ where $0<=b<5$

This means the expression can be written as $\textcolor{#3D99F6}{(5a+b)}^{2}+3\textcolor{#3D99F6}{(5a+b)}+1$

Simplifying, we get $25a^{2}+40a+10ab+(b^{2}+3b+1)$

We already know that $10ab$ is a factor of $10$ , so that leaves us with $25a^{2}+15a$

Factoring that, we get $5(a)(5a+3)$ .

In order for this to be a multiple of $10$ , we need either $a$ or $5a+3$ to be even. If $a$ is odd, $5a+3$ is even, and if $a$ is even, then, well, $a$ is even

Since $25a^{2}+15a+10ab$ will always be a factor of $10$ , the ones digit will just be $b^{2}+3b+1$

For $b$ going from $0$ throught $4$ , the ones digit will be $1,5,1,9,9$

This means the pattern will be $5,1,9,9,1,5,1,9,9,1, \ldots$

For every $5$ , there will be two numbers ending with a $1$ . For the last $2$ numbers, there will be $1$ number ending with $1$ .

That gives us a total of $403*2+1=\fbox{807}$

Δrchish Ray - 2 years ago

You didn’t expand the blue expression properly $(5a+b)^2 = 25a^2+10ab+b^2$

Kano Boom - 2 years ago

Sorry, I'll fix that right now

Δrchish Ray - 1 year, 12 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}`	$\boxed{123}$