Math Help for Algebra I (word problem)
Please help me, thank you!
Find the smallest positive integer containing only the digits $2$ and $3$, and at least one of each, that is divisible by both $2$ and $3$.
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Let $n$ such number, by hypothesis $2$ and $3$ divides $n$. This means that $n$ is even and since $n$ only can have digits $2$ and $3$, $n$ must end in $2$. On the other the sum of the digits of $n$ must be multiple of $3$. Because sums of numbers $3$ gives us a multiple of $3$, then the sum of numbers $2$ also has to be multiple of $3$. We have $2+2+2=6$, so $3$ is the minimum number of $2$'s we need to have a multiple $3$. Therefore the smallest number is $n=2232$.
If it's divisible by $2$ the last digit must be and even number. If the digits are only $2,3$ the last digit must be $2$.
If it is divisible but $3$ the digits must add up to a multiple of $3$. It has $2$ and $3$. Those add to $3+2=5$. That is not divisible by $3$ so we must add a third digit. If we add a $2$ we get $3+2+2 = 7$ and if we add a $3$ we get $3+2+3=8$. Neither of those are divisible by $3$ so we must add a third and a fourth digit.
If we add two $2$s, we have $3+2+2+2=9$. If we add two $3$s we get $3+2+3+3=11$. If we add a $2$ and a $3$ we get $3+2+3+2=10$. Only $3+2+2+2 = 9$ is divisible by $3$.
To get a number as small as possible we want as few digits as possible so we want the digits $3,2,2,2$. The smallest number with those digits is $2223$ but that doesn't end with $2$.
So the smallest number with those digits that ends in $2$ is $2232$.
$2232 = 2*1116$ and $2232 = 3*744$. This is the smallest such number.