Let's denote the three numbers in the arithmetic progression as a, a+d, a+2d.

Given that a+d+1 and a+2d+1 are both prime numbers, and the product of these two primes and the smallest of the original numbers is 598, we can write the equation as follows:

(a) * (a+d+1) * (a+2d+1) = 598

We know that 598 = 2 * 13 * 23, and since a, a+d+1, a+2d+1 are in arithmetic progression, we can assume that a=2, a+d+1=13, a+2d+1=23.

From a+d+1=13, we get d=10.

So, the three numbers are 2, 12, 22.

The sum of these three numbers is 2+12+22 = 36.

Question

Let's denote the three numbers in the arithmetic progression as a, a+d, a+2d.

Given that a+d+1 and a+2d+1 are both prime numbers, and the product of these two primes and the smallest of the original numbers is 598, we can write the equation as follows:

(a) * (a+d+1) * (a+2d+1) = 598

We know that 598 = 2 * 13 * 23, and since a, a+d+1, a+2d+1 are in arithmetic progression, we can assume that a=2, a+d+1=13, a+2d+1=23.

From a+d+1=13, we get d=10.

So, the three numbers are 2, 12, 22.

The sum of these three numbers is 2+12+22 = 36.

Knowee AI · Accepted Answer

Let's denote the three numbers in the arithmetic progression as a, a+d, a+2d.

Given that a+d+1 and a+2d+1 are both prime numbers, and the product of these two primes and the smallest of the original numbers is 598, we can write the equation as follows:

(a) * (a+d+1) * (a+2d+1) = 598

We know that 598 = 2 * 13 * 23, and since a, a+d+1, a+2d+1 are in arithmetic progression, we can assume that a=2, a+d+1=13, a+2d+1=23.

From a+d+1=13, we get d=10.

So, the three numbers are 2, 12, 22.

The sum of these three numbers is 2+12+22 = 36.

There are three numbers in arithmetic progression. If the two larger numbers are both increased by one, then the resulting numbers are prime. The product of these two primes and the smallest of the original numbers is 598. Find the sum of the three numbers.

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