All About The Digits 2

Algebra Level 3

Let $a$ and $b$ be single-digit numbers and that $a<b$ .

$ab$ is a two-digit number.
$ab^{2}$ is a three-digit number.
$a^{2}b^{2}$ is a four-digit number.

Find the product of the integer solutions. How many digits does the product have?

Try Part 1

The answer is 19.

2 solutions

Kaizen Cyrus
May 29, 2019

The integer solutions are:

$a=4, \ b=8$
$a=4, \ b=9$
$a=5, \ b=7$
$a=5, \ b=8$
$a=5, \ b=9$
$a=6, \ b=7$
$a=6, \ b=8$
$a=6, \ b=9$
$a=7, \ b=8$
$a=7, \ b=9$
$a=8, \ b=9$

Getting all the integer solutions, we get $\small 4^{2}×5^{3}×6^{3}×7^{4}×8^{5}×9^{5}$ to find the product. The product is $2006958485274624000$ which has $\boxed{19}$ digits.

Just an observation: if $a^2 b^2$ has four digits then $a^2 b^2\ge1000$ so $ab>31$ . If $ab>31$ then each of $a$ and $b$ is at least $4$ , so $ab^2>100$ . Also $ab<100$ , $ab^2<1000$ and $a^2 b^2<10000$ so we never have too many digits.

This means that all we have to solve is $ab>31$ and $a<b$ , leading straight to the list you've given.

Rather than work out the full product, it's also a bit easier to get the number of digits taking logarithms base $10$ ; these just need to be added to get the answer.

Chris Lewis - 2 years ago

A Former Brilliant Member
May 31, 2019

Brute force and direct:

$\text{Length}\left[\text{IntegerDigits}\left[\text{Times}\text{@@} \\ \text{Flatten}\left[\text{Table}\left[\text{If}\left[10\leq a b\leq 99\land 100\leq a b^2\leq 999\land 1000\leq a^2 b^2\leq 9999,a b,\text{Nothing}\right],\{b,9\},\{a,0,b-1\}\right]\right]\right]\right] \\ \Rightarrow 19$

All About The Digits 2

The answer is 19.

2 solutions

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