Solving a (simple) numeric exercise in bash - bash

Question

Some of you are probably familiar with Project Euler, and I'm currently attempting a few of their problems to teach myself some more bash. They're a bit more mathematical than 'script-y' but it helps with syntax etc.
The problem currently asks me to solve:
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.
The code I have looks like so:
#!/bin/bash
i="1"
for i in `seq 1 333`
do
threes[$i]=`calc $i*3` # where 'calc' is a function written in bashrc
#calc actually looks like: calc() {awk "BEGIN { print "$*"} }
let "sumthrees = sumthrees + ${threes[$i]}"
done
for i in `seq 1 199`
do
fives[$i]=`calc $i*5`
let "sumfives = sumfives + ${fives[$i]}"
done
let "ans = $sumfives + $sumthrees"
echo "The sum of all 3 factors is $sumthrees and the sum of all five factors is $sumfives"
echo "The sum of both is $ans"
#So I can repeatedly run the script without bash remembering the variables between executions
unset i
unset fives
unset threes
unset sumfives
unset sumthrees
unset ans
So far I've not gotten the correct answer, but have run out of ideas as to where I'm going wrong. (FYI, the script currently gives me 266333, which I believe is close, but I don't know the answer yet.)
Can anyone spot anything? And for my own learning, if there are more elegant solutions to this that people might like to share that would be great.
EDIT
Thanks for all the answers, super informative. Since there are so many useful answers here I'll accept my favourite as the proper thread answer.

Answer 1

Blue Moon pointed out the actual problem with your logic.
You don't need to store all the threes and fives in arrays because you don't need them later.
You don't need to unset variables at the end of a script if you use ./yourscript or bash script because they'll disappear along with the
shell instance (better to initialize them first in any case).
You don't need awk to do math, bash does that just fine.
seq and let are not the best way to do anything in a bash script.
Here's a straight forward version:
#!/bin/bash
sum=0
for ((i=1; i<1000; i++))
do
if (( i%3 == 0 || i%5 == 0 ))
then
(( sum += i ))
fi
done
echo "$sum"

Answer 2

Your logic is almost right except that there are numbers which divide by both 3 and 5. So you are adding these numbers twice. Hence, you get wrong answer.
Use another loop similar to ones you have and subtract the ones that divide by both 3 and 5 from the result.

Answer 3

A few tips you might find useful:
In bash, you use let to give the shell a hint that a variable should be considered a number. All bash variables are strings, but you can do arithmetic on numerical strings. If I say let i=1 then i is set to 1, but if I say let i="taco" then $i will be 0, because it couldn't be read as a number. You can achieve a small amount of type-safety when doing mathematical work in the shell.
Bash also has $((this)) mechanism for doing math! You can check it out yourself: echo $((2 + 2)) -> 4, and even more relevant to this problem: echo $((6 % 3 == 0)) -> 1
In case you aren't familiar, % divides the first number by the second, and gives back the remainder; when the remainder is 0, it means that the first is divisible by the second! == is a test to see if two things are equal, and for logical tests like this 1 represents true and 0 represents false. So I'm testing if 6 is divisible by 3, which it is, and the value I get back is 1.
The test brackets, [ ... ] have a "test for equality" flag, -eq, which you can use to check if a math expression has a certain value (man test for more details)!
$ let i=6
$ echo $((i % 3 == 0 || i % 5 == 0))
1
$ if [ $((i % 3 == 0 || i % 5 == 0)) -eq 1 ]; then echo "yes"; fi
yes
(|| is another logical test - $((a || b)) will be 1 (true) when a is true or b is true).
Finally, instead of doing this for the number 6, you could do it in a for loop and increment a sum variable every time you find a multiple of 3 or 5:
let sum=0
for i in {1..1000}; do
if [ $((i % 3 == 0 || i % 5 == 0)) -eq 1 ]; then
let sum=$((sum + i))
fi
done
echo $sum
And there you'd have a working solution!
Bash has a lot of nice little tricks, (and a lot more mean ugly tricks), but it's worth learning at least a handful of them to make use of it as a scripting tool.

Answer 4

How about creative use of the modulus function & some checks. Then you have just 1 loop.
#!/bin/bash
i=1
while [ $i -lt 1000 ]
do
if [ $(($i % 3)) -eq 0 ] || [ $(($i % 5)) -eq 0 ]
then
sumall=$(($sumall+$i))
fi
i=$(($i+1))
done
echo "The sum of both is $sumall"
Answer: 233168

Answer 5

A different solution:
#!/bin/bash
sum=0
for n in {1..999}; do [ $(((n%5) * (n%3))) -eq 0 ] && sum=$((sum+n)); done
echo $sum
The script loops through all numbers below 1000, tests if the product of the number mod 3 and the number mod 5 is 0 (the product of two numbers can only be zero if one of them is zero). If that is the case, it adds the current number to a sum, which is printed out afterwards.
By the way, if I were you I'd include the definition of the calc function inside the script, to get a self-contained solution that doesn't need your specific configuration.