Generations apart
A woman and her grandson have the same birthday. For six consecutive birthdays, she is an integral multiple of his age. How old is the grandmother at the sixth of these birthdays?
Labels: mathschallenge
posted by Chris at
1:59 PM
12 Comments:
66
Grandma is 60 years older than grandson.
Their ages after his birth are:
1 & 61
2 & 62
3 & 63
4 & 64
5 & 65
6 & 66
so she is 66 at the 6th year.
There may be other solutions.
A dead heat, Ragknot!
and on his 10th, the multiple will be 7!
Oh, Wiz, what time is it there?
Just kidding. I was kidding Chris once about posting using local time.
It's 5:30 PM here.. just got home from work.
It's 10.50 am on Saturday here. I'm long retired and haven't worked for years.
Problem boils down to:
Find two consecutive prime numbers with a gap of >6, with 6 consecutive factors
Prime numbers<100 with gap >=6
23,31,47,53,61,73,83,89
All candidates except for 89 are exactly 6 gaps thus they must start with 1 as the age of the grandson since the first year will be a prime for grandma.
e.g.
31, 31/1=31,32/2=16,33/3=11,34/4=8.5 FAIL
only 61 passes for the 6 gaps <100.
for 89. The series must contain either 91 or 94. 91 has factors 7,13 and 94 has factors 2,47. 92/8=11.5 FAIL, 92/14=6.57.. FAIL. 95/3=31.66 FAIL.
Only series that pases <100 is that starting with 61 and with factors starting at 1.
i.e.
61,62,63,64,65,66
1,2,3,4,5,6
Answer:
Her final age is 66.
This is the only answer for <100
Cam
66 it is. I don't believe that (in general) that there is a nice analytical way to solve this sort of problem (for the minimum age); the gaps between the primes are, for practical purposes, random.
Gaps between primes random ?
Only to the extent that they randomly have gaps between the following bounds.
gn >(c log n log log n log log log log n)/(log log log n)^2
where c=2e^(Euler–Mascheroni constant )
and gn < 2*sqrt(n)+1
Cam
Cam. I had an idea that you might provide that info. I carefully phrased my comment because of you.
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