document.write( "Question 248626: Palindromes are numbers that read the same backwards and forwards, like 5665. What is the least possible positive four-digit palindrome that is divisible by 3? \n" ); document.write( "

Algebra.Com's Answer #181165 by jsmallt9(3758) $\"\"$ $\"About$

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Let's call the first and last digit \"x\" and the middle two digits \"y\". So our 4-digit number is in the form: xyyx. In order for this to be divisible by three, the digits must add up to a number divisible by three. So:
\n" ); document.write( "x+y+y+x = 2x + 2y = 2(x+y)
\n" ); document.write( "must be divisible by three. I wrote it in factored form because it gives us a clue to the solution. Since 2 is not divisible by 3, it must be the (x+y) which is divisible by three. And what are the two small numbers that add up to something divisible by 3? Answer: (1 and 2) or (3 and 0)

\n" ); document.write( "So our possible 4-digit numbers are:
\n" ); document.write( "1221
\n" ); document.write( "2112
\n" ); document.write( "3003
\n" ); document.write( "(Note: x cannot be zero or else the number would not actually be a 4-digit number). As we can see, the smallest of these is the first one: 1221. \n" ); document.write( "

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