Euler Problem 1 Script

#!/bin/bash
# Solution code for Project Euler problem 1
# Problem: 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.

# Set up our variables
SUM=23
CURR=11

#Check to see if CURR is less than 1000. If it is, then quit.
while [ "$CURR" -lt 1000 ]
do
    if   !(( "$CURR" % 3 )) || !(( "$CURR" % 5 ))
    then
        #echo $CURR is divisible by 3 or 5 # a little test code
        let "SUM+=$CURR"
    else
        # Do nothing
        echo > /dev/null
    fi
    # add 1 to the value of CURR
    let "CURR += 1"
done
echo The sum of all multiples of 3 or 5 that are below 1000 is $SUM.