Prime numbers

The product of four prime numbers $a, b, c$ and $d$ is the sum of 55 consecutive numbers. Find the smallest possible value of $a+b+c+d$ .

#NumberTheory

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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.

Yatin Khanna - 4 years, 2 months ago

Pls help.. urgent...

Avanthi Vasudevan - 4 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$