Best Answer

118

Start with the numbers that are 1 less than a multiple of 5.

This produces, 4, 9, 14, 19, 24,...,89, 94, 99

Clearly the sequence 4, 14, 24 etc has to be a multiple of 4 or 2 greater than a multiple of 4 so these do not apply.

This leaves 9,19,29,39,49,59,69,79,89,99.

The prime numbers within this group are, 19, 29, 59, 79 & 89.

The numbers that are 1 greater than a multiple of 4 are 29 & 89.

29 + 89 = 118.

Using Modular Arithmetic we have :

x ≡ 1 (mod4),

x = -1 ≡ 4 (mod5) so we can write that, x = 5n + 4... (where n is a positive integer)

Thus, 5n + 4 ≡ 1 (mod4) : 5n ≡ -3 ≡ 1 (mod4) : 4n + n ≡ n ≡ 1 (mod4)

Consequently n = 1,5,9,13,17...

As, x = 5n + 4 , then x = 9, 29, 49, 69, 89.

Of these, only 29 & 89 are prime numbers. Hence, 29 + 89 = 118

Q: What is the sum of all prime numbers between 1 and 100 that are simultaneously one greater than a multiple of 4 and one less than a multiple of 5?

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There are infinitely many of them. The square of any multiple of four that is greater or equal to 12 will meet the requirement.

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