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?

Age

11 to 14

Challenge level

Exploring and noticing Working systematically Conjecturing and generalising Visualising and representing Reasoning, convincing and proving

Being curious Being resourceful Being resilient Being collaborative

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?

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=23317	1982=2*991	2002=2711*13	YES
30	2*30 ²=1800	1780=2 ²589	1860=2 ²35*31	1880=2 ³547	NO
29	2*29 ²=1682	1662=23277	1742=21367	1762=2*881	NO
28	2*28 ²=1568	1548=2 ²3 ²43	1628=2 ²1137	1648=2 ⁴*103	NO
27	2*27 ²=1458	1438=2*719	1518=2311*23	1538=2*769	NO
26	2*26 ²=1352	1332=2 ²3 ²37	1412=2 ²*353	1432=2 ³*179	NO
25	2*25 ²=1250	1230=235*41	1310=25131	1330=257*19	NO
24	2*24 ²=1152	1132=2 ²*283	1212=2 ²3101	1232=2 ⁴711	NO
23	2*23 ²=1058	1038=23173	1118=21343	1138=2*569	NO
22	2*22 ²=968	948=2 ²379	1028=2 ²*257	1048=2 ³*131	NO
21	2*21 ²=882	862=2*431	942=23157	962=21337	NO
20	2*20 ²=800	780=2 ²35*13	860=2 ²543	880=2 ⁴511	NO
19	2*19 ²=722	702=3 ⁴*13	782=21723	802=2*401	NO
18	2*18 ²=648	628=22157	708=2 ²359	728=2 ³713	NO
17	2*17 ²=578	558=23 ²31	638=21129	658=2747	NO
16	2*16 ²=512	492=2 ²341	572=2 ²1113	592=2 ⁴*37	NO
15	2*15 ²=450	430=2543	510=235*17	530=2553	NO
14	2*14 ²=392	372=2 ²331	452=2 ²*113	472=2 ³*59	NO
13	2*13 ²=338	318=2353	398=2*199	418=21119	NO
12	2*12 ²=288	268=2 ²*67	348=2 ²329	368=2 ⁴*23	NO
11	2*11 ²=242	222=2337	302=2*151	322=2723	NO
10	2*10 ²=200	180=2 ²3 ²5	260=2 ²513	280=2 ³57	NO
9	2*9 ²=162	142=2*71	222=2337	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=2313	158=2*79	178=2*89	NO
6	2*6 ²=72	52=2 ²*13	132=2 ²311	152=2 ³*19	NO
5	2*5 ²=50	30=235	110=2511	130=2513	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

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