The smallest number which when increased by 17 is exactly divisible by both 468 and 520 is obtained by subtracting 17 from the LCM of 468 and 520 .
Prime factorization of 468 and 520 is:
\(468=2^{2} \times 3^{2} \times 13\)
\(520=2^{3} \times 5 \times 13\)
LCM = product of greatest power of each prime factor involved in the numbers \(=2^{2}\) \(\times 3^{2} \times 5 \times 13=4680\)
The required number is \(4680-17=4663\).
Hence, the smallest number which when increased by 17 is exactly divisible by both 468 and 520 is 4663 .
