Katie had a
collection of red, green and blue beads. She noticed that the number of beads
of each colour was a prime number and that the numbers were all different.
She also observed that if she multiplied the number of red beads by the total number of
red and green beads she obtained a number exactly 120 greater than the number
of blue beads.
How many beads of each colour did she have?
SOURCE: Scottish Mathematical Challenge.
(In reply to
Puzzle Answer by K Sengupta)
Let the respective number of red, green, and blue beads be denoted by r, g, and b.
By the problem:
r(r+g) = 120 + b ........ (i)
If b is even, then it follows that b must be equal to 2, the only even prime.
Accordingly, r(r+g)=122, the only prime factorization of 122=2*61, implying inter-alia, that r=2
This is a contradiction, since by the given conditions, b and r must be different.
Therefore, it follows that b is odd.
If each of r and g are odd, then in that situation, the lhs of (i) is even, so that rhs being even, b must be odd.
r and g cannot both be even due to similar arguments.
Consequently, precisely one of r and g is even, while the other is odd.
If r is even, then rhs of (i) is even, so that b is even. Contradiction.
Accordingly, g is even - so that: g=2, giving:
r(r+2)=120+b
=> r^2+2r-120=b
=> (r+12)(r-10)=b ...... (ii)
If r>=12, then b is composite. Contradiction.
Also r<= 10 would force a non positive value for b. Contradiction.
Accordingly, r=11, so that b= 23 in terms of (ii)
Therefore, (r, g, b) = (11, 2, 23) is the only possible solution.
Consequently, Katie had 11 red beads, 2 green beads and 23 blue beads.
Edited on February 26, 2022, 1:54 am
Explanation to Puzzle Answer
