1) Let us assume required prime numbers are a, b and c.
Product of a, b and c= abc.
Sum of a, b and c = a+b+c.
"Their product is five times greater than their sum."
Therefore,
abc = 5(a+b+c) ----------------------equation (1)
Now, let us take first prime numbers 2 and second 5.
Plugging a=2 and b=5.
2×5×c = 5(2+5+c).
10c = 5(7+c)
10c = 35 +5c.
Subtracting 5c from both sides, we get
10c-5c = 35 +5c-5c.
5c = 35.
Dividing both sides by 5, we get
c=7.
Therefore, first "cool" triple is 2,5,7.
Let us check by taking a=2 and b=7.
Plugging a=2 and b=7 in equation (1), we get
2×7×c = 9(2+7+c).
c=9. But it's not a prime number.
Let us take a=2 and b=11, we get
2×11×c = 11(2+11+c).
c=13 (A prime)
If we take a=2 and b=17, we get
2×17×c = 17(2+17+c).
c=19 (A prime).
On the same way, if we continue the process, we can get many "cool" triples.
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2) We can get that number hit and trial method.
It is said that "two three-digit numbers and wrote one six-digit number, which is equal to seven times their product".
Let us check number 143.
If we write 143 two times without any sign in between , we get 143143 ( a six digit number).
But if we multiply 143 × 143 , we get 20449.
7 times of 20449 equals 143143.
Therefore, the required three digit number is 143.
