[[1]] is the smallest number that is divisible by both 33 and 39, and leaves the remainder 5.

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Detailed Solution

434 is the smallest number that is divisible by both 33 and 39, and leaves the remainder 5.
Prime factorization of 33 and 39,

33 = 3 \times 11

39 = 3 \times 13

Thus, the LCM of 33 and 39 is,

3 \times 11 \times 13 = 429

Therefore, 429 is the smallest number that is exactly divisible by 33 and 39.
Thus, the smallest number that is divisible by both 33 and 39, and leaves remainder 5 is,

429 + 5 = 434

Therefore, the required number is 434.

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