e

n = 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  
e = 0b1101  
c = 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

Solution

An important observation is that e = 13 and c is divisible by 2¹³, hinting that because of the small value of e and large value of n, there isn’t any wrap-around when finding c = me mod n. Thus, we can directly find m = c1/e to get the flag. I’m not aware of any easy methods to do this in Python as it doesn’t support floating-point values of such high precision, so I ended up calculating m using binary search:

st = 0
en = c
while st < en: 
	m = (st + en)//2 
	if m**e == c: 
            print(long_to_bytes(m)) 
            break 
        elif m**e < c: 
            st = m + 1 
        else: 
            en = m - 1

Flag

rgbCTF{lucky_number_13}