let $ab$ be A two-digit number such that $ab=8(a+b)$ then what is $ab$

Question

let $ab$ be A two-digit number such that $ab=8(a+b)$ then what is $ab$

My Try :

$$x=ab \to x=10a+b$$

so we have :

$$10a+b=8(a+b)\\10a+b=8a+8b \\ 2a=7b \\ \frac{a}{b}=\frac{7}{2}$$

now what ?

Answer 1

$$2a=7b\Longrightarrow 7\mid 2a \Longrightarrow 7\mid a\Longrightarrow a=7\;{\rm since}\; a\in \{1,2,...,9\}$$ and $b=2$.

Answer 2

write your equation in the form $$a=\frac{7}{2}b$$ sorry the equation has changed

Answer 3

$\frac{a}{b}=\frac{7}{2}$ is the same thing as $a:b=7:2$, i.e. the digits $a$ and $b$ are in the ratio $7:2$. Which two digits can be in the ratio $7:2$? Well, $7$ and $2$...

Answer 4

Oh for goodness sake. I just realized a much better answer:

You have $\frac ab = \frac 72$ so there is some integer $k$ so that $\frac ab = \frac 72 = \frac {7k}{2k}$ where $a = 7k$ and $b = 2k$.

If $k \ge 2$ then $a = 7k \ge 14$ which is not acceptable. So $k \le 1$.

Now $a,b$ are both non-negative and $a$ can't be $0$ so $k > 0$. So $k = 1$ and $a = 7$ and $b = 2$

=== read on for a more extensive answer ====

You have one equation and two unknowns so you have an dependent relationship that

$2a = 7b$. There are an infinite number of real valued solutions. But we don't want all real solutions. (Example: If $b = 27$ and $a = \frac {7*27}2$ is obviously not acceptable.)

We have the further restrictions $a,b \in \mathbb Z$ and $0\le b \le 9$ and $1 \le a \le 9$. ($a \ne 0$ as a two-digit number doesn't [well, shouldn't] begin with $0$).

You have $\frac ab =\frac 72$ so clearly $a = 7$ and $b = 2$ is ONE solution. Are there any more?

Well go back to $2a = 7b$ As $1 \le a \le 9$ we know that $2*1 \le 2*a \le 2*9$ so $1 \le 2a = 7b \le 18$. So $\frac 27 \le b \le \frac {18}{7}$.

So $\frac 27\le b \le 2\frac 47$. As $b$ is an integer $1 \le b \le 2$. so $b = 1,2$.

We have $2a = 7b$ so $a = \frac {7b}2$. As $7$ is odd, you must have $b$ is even so $b = 2$ is the only possible answer and $2a = 7*2$ so $a = 7$.

So $a = 7$ and $b= 2$ and $72 = 8(7+2)$ is the only solution.

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Or does $ab$ mean $a\times b$? I really doubt this but:

$ab = 8(a + b)$

$ab - 8a = 8b$

$a(b-8) = 8b$.

If $b -8=0$ then $b = 8$ and then $8b = a(b-8) = 0$ and $b = 0$ which is a contradiction. So $b - 8\ne 0$.

So

$a = \frac {8b}{b-8}$ and we have $ab = b*\frac {8b}{b-8} = k$ where $10 \le k \le 99; k \in \mathbb Z$.

Or $8b^2 = kb - 8k$ or $8b^2 -kb + 8k = 0$ or $b = \frac{k \pm \sqrt{k^2 -64k}}{16}$.

$k^2\ge64k$ so $64 \le k \le 99$ so there are $72$ possible real valued values for $b$ and $a = \frac {8b}{b-8}$. (No-one ever said $a$ and $b$ where integers.)

Answer 5

Almost 1960.

Your answer is ok:

$2a=7b$.

Assume:

$1\le a,b \le 9 :$

$2a= 7b$ implies :

$2| b$ (Euclid's Lemma):

$b=2,4,6,8$; recall $1\le b \le 9$.

We have $a=7$ for $b=2$, all other choices of $b$ do not satisfy $1\le a \le 9.$

Hence $a=7 , b=2.$