What is the smallest number which when increased by 17 becomes exactly divisible by both 520 and 936?

The number which when increased by 17 becomes exactly divisible by both 520 and 936 will be 17 less than the Least Common Multiple (LCM) of 520 and 936.
First, let's find the LCM(520,936) using prime factorization:

520=10×52=2×5×4×13=2
3
×5×13
936=2×468=2
2
×234=2
3
×117=2
3
×9×13=2
3
×3
2
×13
Now, take the highest power of all prime factors involved:

LCM(520,936)=2
3
×3
2
×5×13
LCM(520,936)=8×9×5×13
LCM(520,936)=360×13=4680
Thus, 4680 is the smallest number exactly divisible by both 520 and 936.

The required number is:

Required Number=LCM−17=4680−17=4643