The product of four prime numbers \(a, b, c\) and \(d\) is the sum of 55 consecutive numbers. Find the smallest possible value of \(a+b+c+d \).
6 months, 2 weeks ago
Note that the sum of 55 consecutive numbers will always be divisible by 55; (sum equals 55*n ; where n is the middle term)
Hence; 2 of the primes must be 5 and 11; then to minimise the sum; take the other 2 primes as 2 and 2 or if you want different primes; 2 and 3.
So; the sum will be 20 or 21 according to whether duplication of primes is allowed or not.
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Pls help.. urgent...