SOLUTION: Steve was born in 1950. HIs grandmother was born in a year which is the product of two prime numbers- one of which is one less than twice the other. in what year was Steve's grandm

Let x be the other prime.

>>...one of which is one less than twice the other...<<

So the one prime = 2·(other prime) - 1 = 2x - 1

>>...His grandmother was born in a year which is the 
           product of two prime numbers...<<

So the year Steve's grandmother was born was x(2x - 1)

The grandmother had to be born before Steve, and since
Steve was born in 1950,

             x(2x - 1) < 1950

It is safe to say she was not more than 100 years old
when Steve was born, which means she must have been
born after 1850. 

So the grandmother had to be born in some year between
1850 and 1950

1850 < x(2x - 1) < 1950

Subtract 1850 from all three sides:

0 < x(2x - 1) - 1850 < 100

0 < 2x² - x - 1850 < 100

This amounts to

2x² - x - 1850 > 0 AND 2x² - x - 1850 < 100

The solution set will have to be the 
intersection of the two solution sets of
those. So we find the solution set of
both inequalities:

We solve the first inequality:

2x² - x - 1850 > 0

Find the critical values of 2x² - x - 1850
by setting it equal to 0:

2x² - x - 1850 = 0

Solving that by the quadratic formula we get

x = 30.66 and -30.16

Since x > 0 cannot be negative we discard the 
negative critical value.  So the 
critical value is 3.66.

This divides the number line into two
parts:

{x|030.66}

or in interval notation

(0,30.66), (30.66, oo)

We choose a test point in the first
interval, say 1, and substitute it 
into the inequality

2x² - x - 1850 > 0

2(1)² -(1) - 1850 > 0
      
            -1850 > 0

This is false, so we do not include (0, 30.66)

We choose a test point in the second
interval, say 31, and substitute it 
into the inequality

2x² - x - 1850 > 0

2(31)² -(31) - 1850 > 0
      
                103 > 0

This is true, so the solution set of

2x² - x - 1850 > 0

is (30.66, oo)

----------------------

Now we solve the second inequality:

2x² - x - 1850 < 100

Get 0 on the right:

2x² - x - 1950 < 0

Find the critical values of 2x² - x - 1950
by setting it equal to 0:

2x² - x - 1950 = 0

Solving that by the quadratic formula we get

x = 31.48 and -30.98

Since x > 0, x cannot be negative we discard the 
negative critical value.  So the 
critical value is 31.48.

This divides the number line into two
parts:

{x|031.48}

or in interval notation

(0,31.48), (31.48, oo)

We choose a test point in the first
interval, say 1, and substitute it 
into the inequality

2x² - x - 1950 < 0

2(1)² -(1) - 1950 < 0
      
            -1850 < 0

This is TRUE, so we do include (0, 31.48)

We choose a test point in the second
interval, say 32, and substitute it 
into the inequality

2x² - x - 1950 < 0

2(32)² -(32) - 1950 < 0
      
                66 < 0

This is false, so we do not include

(31.48, oo) in the solution set of

2x² - x - 1950 < 0

So the solution set of

2x² - x - 1950 < 0

is (0, 31.48)

--------------------

Now the solution set of

2x² - x - 1850 > 0 AND 2x² - x - 1850 < 100

must be the intersection of the two solution
sets or

(30.66, oo) n (0, 31.48)

which is the interval

   (30.66, 31.48)

There is only one integer in that interval, 31

and since x must be an integer, x can only be 31,

and yes, that is indeed a prime number.  

The other prime is one less than two times the other,

or 2x - 1 which = 2(31) - 1 = 62 - 1 = 61.

And yes, 61 is also a prime number.

So to see what year the grandmother was born, we
multiply 31×61 and get 1891.

So she was born in 1891.

It was not necessary to tell us they were prime
numbers, but only that they were integers,
although it is interesting that they are both
prime numbers.  

Edwin