Great Granddad
Great Granddad is very proud of his telegram from the Queen congratulating him on his hundredth birthday and he has friends who are even older than he is... When was he born?
Problem
Great Granddad is very proud of his telegram from the Queen congratulating him on his hundredth birthday and he has friends who are even older than he is. Great Granddad was born in the year A (where A is the product of 3 prime numbers), he was 20 years old in the year B (where B is the product of a prime number and a square number), he was 80 years old in the year C (where C is the product of two prime numbers) and he celebrated his 100th birthday in the year D (where D is even and the product of 4 prime numbers). When was he born?
Getting Started
D = 2pqr where p, q and r are primes and there are only a few cases to consider as p, q and r can't be very big.
Student Solutions
We received correct solutions from Julia Collins, from Langley Park School for Girls in Bromley, Ross Haines, Ben Young and Philip Bennington from Mosgiel Intermediate School, and Andrei Lazanu from School 205 in Bucharest, Romania. Well done to you all.
Julia Collins said:
"I think I've found a solution to this problem, although I'm sure it's not the best solution, so I will be interested to see what other people have come up with!
We know that A + 100 is an even number, so we know that A is even. Thus we know that one of the primes making up each of the numbers A, B, C and D is 2 (because any other combination of primes is odd).
B = A + 20 = the product of a prime number and a square number
Since A is even, let the prime be 2. Then the only square number that gives a sensible date is
31 2 = 961.
So B = 1922.
So I checked the other numbers to see if the solution was valid, and it was!
A = 1902 = 2 x 3 x 317 (three primes)
B = 1922 = 2 x 961 = 2 x 31 2 (prime and a square)
C = 1982 = 2 x 991 (two primes)
D = 2002 = 2 x 7 x 11 x 13 (four primes and even)
I don't like the solution because it involves a guess, and so doesn't show whether the solution is unique or not. Can it be done purely by logical thinking?"
Andrei Lazanu was able to show that this was a unique solution:
"First I observed that the data known for the problem (A, B, C, D) corresponds to A, A + 20, A + 80 and A + 100.
Because A + 100 is even, all the other terms are also even (because 20, 80 and 100 are even).
Now, I write the data I have:
A = 2 * prime * prime
B = A + 20 = 2 * x 2
C = A + 80 = 2 * prime
D = A + 100 = 2 * prime * prime * prime
I start by considering the condition for B and observe that x must be smaller than 32 (since 2 * 32 * 32 = 2048):
| x | B | A = B - 20 | C = A + 80 | D = A + 100 | Satisfies condition |
| 31 | 2*31 2=1922 | 1902=2*3*317 | 1982=2*991 | 2002=2*7*11*13 | YES |
| 30 | 2*30 2=1800 | 1780=2 2*5*89 | 1860=2 2*3*5*31 | 1880=2 3*5*47 | NO |
| 29 | 2*29 2=1682 | 1662=2*3*277 | 1742=2*13*67 | 1762=2*881 | NO |
| 28 | 2*28 2=1568 | 1548=2 2*3 2*43 | 1628=2 2*11*37 | 1648=2 4*103 | NO |
| 27 | 2*27 2=1458 | 1438=2*719 | 1518=2*3*11*23 | 1538=2*769 | NO |
| 26 | 2*26 2=1352 | 1332=2 2*3 2*37 | 1412=2 2*353 | 1432=2 3*179 | NO |
| 25 | 2*25 2=1250 | 1230=2*3*5*41 | 1310=2*5*131 | 1330=2*5*7*19 | NO |
| 24 | 2*24 2=1152 | 1132=2 2*283 | 1212=2 2*3*101 | 1232=2 4*7*11 | NO |
| 23 | 2*23 2=1058 | 1038=2*3*173 | 1118=2*13*43 | 1138=2*569 | NO |
| 22 | 2*22 2=968 | 948=2 2*3*79 | 1028=2 2*257 | 1048=2 3*131 | NO |
| 21 | 2*21 2=882 | 862=2*431 | 942=2*3*157 | 962=2*13*37 | NO |
| 20 | 2*20 2=800 | 780=2 2*3*5*13 | 860=2 2*5*43 | 880=2 4*5*11 | NO |
| 19 | 2*19 2=722 | 702=3 4*13 | 782=2*17*23 | 802=2*401 | NO |
| 18 | 2*18 2=648 | 628=2*2*157 | 708=2 2*3*59 | 728=2 3*7*13 | NO |
| 17 | 2*17 2=578 | 558=2*3 2*31 | 638=2*11*29 | 658=2*7*47 | NO |
| 16 | 2*16 2=512 | 492=2 2*3*41 | 572=2 2*11*13 | 592=2 4*37 | NO |
| 15 | 2*15 2=450 | 430=2*5*43 | 510=2*3*5*17 | 530=2*5*53 | NO |
| 14 | 2*14 2=392 | 372=2 2*3*31 | 452=2 2*113 | 472=2 3*59 | NO |
| 13 | 2*13 2=338 | 318=2*3*53 | 398=2*199 | 418=2*11*19 | NO |
| 12 | 2*12 2=288 | 268=2 2*67 | 348=2 2*3*29 | 368=2 4*23 | NO |
| 11 | 2*11 2=242 | 222=2*3*37 | 302=2*151 | 322=2*7*23 | NO |
| 10 | 2*10 2=200 | 180=2 2*3 2*5 | 260=2 2*5*13 | 280=2 3*5*7 | NO |
| 9 | 2*9 2=162 | 142=2*71 | 222=2*3*37 | 242=2*11 2 | NO |
| 8 | 2*8 2=128 | 108=2 2*3 3 | 188=2 2*47 | 208=2 4*13 | NO |
| 7 | 2*7 2=98 | 78=2*3*13 | 158=2*79 | 178=2*89 | NO |
| 6 | 2*6 2=72 | 52=2 2*13 | 132=2 2*3*11 | 152=2 3*19 | NO |
| 5 | 2*5 2=50 | 30=2*3*5 | 110=2*5*11 | 130=2*5*13 | NO |
| 4 | 2*4 2=32 | 12=2 2*3 | 92=2 2*23 | 112=2 4*7 | NO |
I stop here because from now on A is below 0.
So, 1902 is the unique year that matches all conditions for Great Granddad's birth ."
PS. Julia Collins wrote to us pointing out a slight flaw in Andrei's argument. They discussed this on the web-board, and their discussion can now be read at
http://www.nrich.maths.org/askedNRICH/edited/4087.html