A 6-digit number is formed by repeating a 3-digit number; for example, 256256 or 678678 etc. Any number of this form is always exactly divisible by

## Concept & Logic Numbers formed by repeating blocks of digits are always divisible by a number comprised of 1s and 0s corresponding to the block's place value gaps. ## Step-by-Step Solution * Let the 3-digit repeating block be represented as $xyz$. * The actual value of this block is $(100x + 10y + z)$. * The 6-digit number is $xyzxyz$. * Expand this 6-digit number using its base-10 place values: $$ 100000x + 10000y + 1000z + 100x + 10y + z $$ * Combine the like variables: $$ 100100x + 10010y + 1001z $$ * Factor out the greatest common divisor, which is 1001: $$ 1001(100x + 10y + z) $$ * Because 1001 is a fixed multiplier for any number of this form, the 6-digit number is always exactly divisible by 1001. ## Exam Strategy & Shortcut Bypass the algebra by taking any example from the question and dividing it by its repeating block. Take 256256 and divide it by 256: $$ 256256 \div 256 = 1001 $$ This immediately reveals the universal factor. ## Common Pitfall It is a mathematical fact that $1001 = 7 \times 11 \times 13$. Therefore, the number is divisible by 7, 11, and 13. However, choosing "7 only", "11 only", or "13 only" is incorrect because of the word "only". 1001 is the complete and most accurate answer. ## Final Answer **Therefore, the correct answer is 1001.**