document.write( "Question 1091348: From left to right, the first three digits of a 4-digit number add up to 6. Which digit could be in the ones place if the 4-digit number is divisible by 6? \n" ); document.write( "

Algebra.Com's Answer #705759 by greenestamps(13200) $\"\"$ $\"About$

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The prime factorization of 6 is
\n" ); document.write( " $\"6+=+2%2A3\"$

\n" ); document.write( "To be divisible by 6, a number must be divisible by both 2 and 3.

\n" ); document.write( "In your problem, the first three digits of a 4-digit number add to 6; to make the number divisible by 3, the last digit could be 0 (the sum of all four digits would still be 6, which is divisible by 3); or it could be 3 (sum of all four equal to 9, also divisible by 3), or it could be 6 (sum now 12, still divisible by 3); or it could be 9 (sum 15, again divisible by 3).

\n" ); document.write( "So we get a 4-digit number divisible by 3 if the last digit is 0, 3, 6, or 9.

\n" ); document.write( "But the number also has to be divisible by 2, which means the last digit must be even. So the last digit can't be 3 or 9; and we are left with only two possibilities for the last digit: 0 or 6. \n" ); document.write( "

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