How many integers exist such that not only are they multiples of A = 20202020 but also are factors of B = 20202022?
Explanation:
The required number should be of the form A × x (where x is any positive integer)
Also, B should be divisible by A × x
Now, 20202022 = 20202020 × 20202
Number of possible values of x will be same as the number of factors of 20202.
20202 = 24 × 52 × 1012
Number of factors of 20202 = (4 + 1) × (2 + 1) × (2 + 1) = 45.
Hence, option (c).
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