Divisibility Quiz

To be divisible by 9 and 11, it must be divisible by 99. $98765 \div 99 = 997$ with remainder 62. Smallest number to add = $99 - 62 = 37$.

Last 3 digits (584) are divisible by 8 ($584/8 = 73$). Alternating sum of digits: $4-8+5-3+7 = 5$, not divisible by 11.

Least 6-digit number is 100000. $100000 \div 12$ leaves remainder 4. To be divisible, we add $12 - 4 = 8$. So, $100000 + 8 = 100008$.

First number after 200 divisible by 15 is 210 ($14 \times 15$). Last number before 500 is 495 ($33 \times 15$). Total count = $33 - 14 + 1 = 20$.

Greatest 5-digit number is 99999. $99999 \div 18$ leaves remainder 9. Greatest number = $99999 - 9 = 99990$.

Check options: $16 \div 6 = 2$ rem 4; $16 \div 9 = 1$ rem 7. So 16 is the least such number.

Sum of digits: $5+4+x+3+2 = 14+x$. For divisibility by 9, $14+x$ must be 18, so $x = 4$.

Grouping by 3 digits: $789 - 456 + 123 = 456$. $456 \div 13 = 35$ rem 1.

Since 2, 3, and 5 are prime to each other, the number must be divisible by their LCM, which is $2 \times 3 \times 5 = 30$.

The number must divide $(1200-4) = 1196$ and $(1800-6) = 1794$. $\text{HCF}(1196, 1794) = 598$.

$999,999 \rightarrow 999 - 999 = 0$. Since 0 is divisible by 7, 999999 is divisible by 7.

$\text{LCM}(8, 12, 15) = 120$. Smallest number = $120 + 5 = 125$.