What inequality does the units digit of a two digit number less than $40$ satisfy if the sum of the units digit and tens digit equals $8$?

Question

For a two digit number , the sum of the "units digit" and "tens digit" is $8$ and the two-digit number is less than $40$.

Denoting that the "units digit" of the two-digit number to be $x$, form an inequality and solve for $x$.

I'm not sure how to find the "tens digit" as I have problems understanding the question . Thanks for helping ..

If a two digit number is less than $40$, then the tens digit is at least $1$ and at most $3$. — N. F. Taussig, Feb 25 '16 at 15:59

score 0 · Answer 1 · answered Feb 25 '16 at 16:02

0

You have two things here... let $x$ be the units digit and let $w$ be the tens digit: you know that

$$w+x=8$$ $$10w+x<40$$

Thus

$$9w+w+x<40\Rightarrow 9w+8<40\Rightarrow 9w<32\Rightarrow w<3.6$$

So $w$ can be $1,2,3$. That gives possibilities....

answered Feb 25 '16 at 16:02

Eleven-Eleven

8,806

Why did u multiply w by 10? Sorry a little bit of understanding problem here .. – user307640 Feb 25 '16 at 16:06
when you have a two digit number, like say 31, this is the same as $3\times 10+1$... – Eleven-Eleven Feb 25 '16 at 16:06
any number can be written as a sum of coefficients times powers of 10. For example again, $123=1\cdot 10^2+2\cdot 10+3\cdot 1$. – Eleven-Eleven Feb 25 '16 at 16:08

score 0 · Answer 2 · answered Feb 25 '16 at 16:06

Let $t$ denote the tens digit; let $x$ denote the units digit. Since the sum of the tens digit and the units digit is $8$, $$t + x = 8$$ Since the two-digit number is less than $40$, $1 \leq t \leq 3$. Since $$x = 8 - t$$ you can determine the possible values of $x$.

What inequality does the units digit of a two digit number less than $40$ satisfy if the sum of the units digit and tens digit equals $8$?

2 Answers2