Find natural numbers $q, r$ with $0 \leq r < 3$ such that $n = 3q + r$
$n=22$
$n$ is a natural number
So I saw there are two unknown variables there and I believe the $0 \leq r < 3$ is the key to solving it but I'm not sure how. Unless I am supposed to test it from $0$ to $2$.
In which case only when $r=1$ would we get a $q$ as a natural number. Am I correct or is this supposed to be solved another way?
If the sum of the digits of a number is divisible by $3$, then the number is divisible by $3$.
Chose $r$ such that $n$ is divisible by $3$.
$n=22$ Since sum of it's digits is $1$ more than $3$, therefore choose $r=1$ .
$n-r=21$ is divisible by $3$.
Therefore $r=1$ and $q= \dfrac{21}{3}=7$
You are expected to divide $2459449$ by $3$. $q$ is the quotient and $r$ is the remainder. There is no reason to try different values of $r$ or do the divisibility test as $r$ will pop out.
try considering $n-r$ for each $r$ in 1,2,3. For every option, you can sum the digits, over and over. Then check if this sum is a multiple of 3, then the original number $n-r$ is also. Finally, $q=(n-r)/3$.
Yes r is remainder and in this question remainder is 0, 1 or 2.
Case 1 -
If r = 0
22 = 3q
q is in fraction. So rejected.
Case 2 -
If r = 1
22 = 3q + 1
3q = 21
q = 7
Case 3 -
If r = 2
22 = 3q + 2
3q = 20
q is in fraction. So rejected.
So we have q = 7 and r = 1.
