Determining $a+b$

Question

The sum of the digits of $x$ is $17$. When $(x^3+x^5)$ and $(x^2+7!)$ are divided by $3$ and $9$, the remainder outcomes as $a$ and $b$ respectively. Then determine $a+b$

I couldn't think of any way to solve this problem. Could I get help?

Regards!

Answer 1

The sum of the digits of $x$ has the same remainder modulo $9$ or $3$ than $x$ itself. So $$x\equiv 17\equiv -1\mod 3\text{ and }x\equiv 17\equiv -1\mod 9$$ Then $$x^3+x^5\equiv (-1)^3+(-1)^5\equiv 1\mod 3$$ and, because $7!\equiv 0\mod 9$, $$x^2+7!\equiv (-1)^2+0\equiv 1\mod 9$$ Therefore $a=b=1$ and $a+b=2$.

Answer 2

Since the sum of digits of x is 17. It should in the form of $3k+2$.

Now, $x^3 = 3k_1 + 8$, $x^5 = 3k'_2 +32 = 3k_2 +2$. So, when $x^3 + x^5$ is divided by $3$ we get a reminder of $1$ i.e, $a=1$.

Similarly, we can write x as $9k+8$. Which gives us $x^2 = 9k'_3+64 = 9k_3 +1$ and $7! $ is divisible by $9$. Reminder of $x^2 + 7!$ when divided by $9$ is $1$ i.e, $b=1$.

Hence $a+b = 2$.

Hope this helps!

Answer 3

THe is a well known (to some) theorem that if $s(x) = $ sum of the digits and $a \equiv x \mod n$ means that $a$ and $x$ have the same remainder when divided by $n$ that:

$x \equiv s(x) \mod 3$

and $x \equiv s(x) \mod 9$.

Always. For all $x$.

So if $s(x) = 17$ then $s(x) \equiv 2\equiv -1 \mod 3$ and $s(x) \equiv 8 \equiv -1 \mod 9$. (Note: $-1 = -1*3 + 2$ and $-1 = -1*9 + 8$ so $-1$ has remainder $2$ when divided by $3$ and $-1$ has remainder $8$ when divided by $9$

Now $7! = 1* 2*3*4*5*6*7 \equiv 0 \mod 3$ and $7! \equiv 0 \mod 9$.

And "remainder arithmetic" is consistent under addition. If $a$ has remainder $a'$ and $b$ has remainder $b'$ then $a + b$ will have the same remainder as $a' + b'$. So "remainder arithmetic" is consistent under subtraction (the inverse of addition) and multiplication (addition multiple times), NOT division, and exponential powers (multplication multiple times).

.....

So $s(x) =17\equiv -1 \mod 3,9$ then $x^3 + x^5 \equiv (-1)^3 + (-1)^5 = -1 -1 \equiv -2 \mod 3,9$ and $x^3 + x^5 \equiv -2\equiv 1 \mod 3$ and $x^3 + x^5 \equiv -2\equiv 7 \mod 9$.

And $x^2 + 7! \equiv (-1)^2 + 0 \equiv 1 \mod 3,9$.

Now I do not know what "When (x3+x5) and (x2+7!) are divided by 3 and 9, the remainder outcomes as a and b respectively" is supposed to mean.

$x^3 + x^5$ when divided by $3$ has the remainder $1$ and $x^3 + x^5$ when divide by $9$ has remainder $7$. And $x^2 + 7!$ when divided by $3$ or $9$ has remainder $1$. So I guess $a$ and $b$ are supposed to be $1$ and $7$ and $a+b=8$.

I guess. The question was not very clear.