Divisibility

Divide $b = 100$ by $a = 7$. The multiples of $7$ on either side of $100$ are $98 = 7\cdot14$ and $105 = 7\cdot15$. The quotient counts the whole sevens that fit without passing $100$, so $q = 14$; the remainder is what is left over, $r = 100 - 98 = 2$. So $100 = 7\cdot14 + 2$, and $0 \le 2 < 7$ as the theorem demands.

Divide $b = -50$ by $a = 6$. The remainder must be non-negative, so $aq$ has to be the multiple of $6$ at or BELOW $-50$. That is $-54 = 6\cdot(-9)$, not $-48$, which lies above. Then $r = -50 - (-54) = 4$, and $-50 = 6\cdot(-9) + 4$. The same answer comes from the formula $q = \lfloor -50/6 \rfloor = \lfloor -8.33\ldots \rfloor = -9$: rounding down moves left on the number line, even for negatives.

When $n$ is divided by $9$ the remainder is $5$. What remainder does $4n + 7$ leave? The information "$n$ leaves remainder $5$" means exactly $n = 9k + 5$ for some integer $k$, so substitute: $4n + 7 = 36k + 20 + 7 = 36k + 27 = 9(4k + 3)$. The remainder is $0$, whatever $k$ is. The $36k$ part is a multiple of $9$ and can never affect the remainder — the first glimpse of modular arithmetic, where we will drop such terms from the start.

Show that $\dfrac{3n + 5}{2n + 3}$ is in lowest terms for every positive integer $n$. Suppose $d$ divides both the numerator and the denominator. Choose multipliers that make $n$ cancel: $2(3n + 5) - 3(2n + 3) = 6n + 10 - 6n - 9 = 1$. By the combination rule $d \mid 1$, so $d = 1$: the two never share a factor. (Test $n = 4$: $\tfrac{17}{11}$.)

Find all positive integers $n$ with $n + 2 \mid n^2 + 10$. First divide the polynomial: $n^2 + 10 = (n + 2)(n - 2) + 14$. Since $n + 2$ divides $n^2 + 10$ and also divides $(n + 2)(n - 2)$, it divides their difference, $14$. The positive divisors of $14$ are $1, 2, 7, 14$, and $n + 2 \ge 3$, so $n + 2 = 7$ or $14$: $n = 5$ or $n = 12$. Check: $5^2 + 10 = 35 = 7\cdot5$ and $12^2 + 10 = 154 = 14\cdot11$.

Show that $7 \mid 2^{n+2} + 3^{2n+1}$ for all $n \ge 0$. First give both powers the same exponent: $2^{n+2} + 3^{2n+1} = 4\cdot2^n + 3\cdot9^n$. The factorization $x^n - y^n = (x - y)(x^{n-1} + x^{n-2}y + \cdots + y^{n-1})$ shows that $9 - 2 = 7$ divides $9^n - 2^n$. So split $3\cdot9^n = 3\cdot2^n + 3(9^n - 2^n)$, and the whole expression becomes $7\cdot2^n + 3(9^n - 2^n)$ — a combination of two multiples of $7$. (Check $n = 1$: $8 + 27 = 35$.)

Is $918{,}082$ divisible by $11$? Alternate signs along the digits: $9 - 1 + 8 - 0 + 8 - 2 = 22$, a multiple of $11$, so yes ($918082 = 11\cdot83462$). Starting the signs from the left or from the right only changes the sign of the result, which does not affect divisibility.

Is $7{,}352{,}416$ divisible by $8$? Since $1000 = 8\cdot125$, everything from the thousands digit up is a multiple of $8$: $7352416 = 7352\cdot1000 + 416$. So only $416$ matters, and $416 = 8\cdot52$. Yes.

Find all digits $a, b$ making the five-digit number $\overline{5a38b}$ divisible by $36$. Since $36 = 4\cdot9$ with $\gcd(4, 9) = 1$, test $4$ and $9$ separately. For $4$: the last two digits $\overline{8b}$ must be a multiple of $4$, so $b \in \{0, 4, 8\}$. For $9$: the digit sum $16 + a + b$ must be a multiple of $9$, so $a + b = 2$ or $11$. Now match: $b = 0$ needs $a = 2$; $b = 4$ needs $a = 7$; $b = 8$ needs $a = 3$. The numbers are $52380$, $57384$ and $53388$.