Maths experts please!
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Sorry, I'm stuck again. This question is from HABs maths paper 2006. This time I will type it out in its entirety!
Great Aunt Maud gives Fred some money for his birthday each year, which he decides to save. On his first birthday she gives him £1. On his second birthday she gives him £2, on his third birthday she gives him £4 and so on, doubling the amount each year. On which birthday will the total amount saved reach £16 383?
What's the formula to work this out please?
Thanks very much
Jess
Great Aunt Maud gives Fred some money for his birthday each year, which he decides to save. On his first birthday she gives him £1. On his second birthday she gives him £2, on his third birthday she gives him £4 and so on, doubling the amount each year. On which birthday will the total amount saved reach £16 383?
What's the formula to work this out please?
Thanks very much
Jess
I think the answer is:-
2**0=1 (** means to the power of)
2**1=2
2**2=4
2**3=8
Hence the equation is 2**(n-1)=16,383 where n is the Age
taking Log of both sides you get (n-1)Log2=Log16383
Hence (n-1)=Log16383/Log2
Happy to be corrected by a mathematician!
The other way is by trial and error using a calculator with increasing powers of 2 until you pass 16383.
2**0=1 (** means to the power of)
2**1=2
2**2=4
2**3=8
Hence the equation is 2**(n-1)=16,383 where n is the Age
taking Log of both sides you get (n-1)Log2=Log16383
Hence (n-1)=Log16383/Log2
Happy to be corrected by a mathematician!
The other way is by trial and error using a calculator with increasing powers of 2 until you pass 16383.
1st year: gives £1 making a total of £1.Jess wrote:Great Aunt Maud gives Fred some money for his birthday each year, which he decides to save. On his first birthday she gives him £1. On his second birthday she gives him £2, on his third birthday she gives him £4 and so on, doubling the amount each year. On which birthday will the total amount saved reach £16 383?
What's the formula to work this out please?
2nd year: gives £2 making a total of £3.
3rd year: gives £4 making a total of £7.
4th year: gives £8 making a total of £15.
...
nth year: gives £2^(n-1) making a total of £2^n - 1.
And as we all know, 16383 is 2^14 - 1, so that total is reached on Fred's 14th birthday.
As this is an 11+ paper and I assume that no calculators are allowed, then simple adding will work.
Each time you add a number on you are nearly doubling the total.
e.g.
1 + 2 = 3 + 4 = 7 +8 = 15 +16 = 31
So keep up the doubling until you get to 8192 and then count the years.
1 + 2 + 4 + 8 + 16 + 32 +64 +128 + 256 +512 + 1024 +2048 +4096 +8192 = 16383.
14 numbers added together, so 14th birthday.
Each time you add a number on you are nearly doubling the total.
e.g.
1 + 2 = 3 + 4 = 7 +8 = 15 +16 = 31
So keep up the doubling until you get to 8192 and then count the years.
1 + 2 + 4 + 8 + 16 + 32 +64 +128 + 256 +512 + 1024 +2048 +4096 +8192 = 16383.
14 numbers added together, so 14th birthday.
Nope, no calculators allowed. I get the adding up method, but it seems a bit slow (like me!). Am afraid I'm still struggling with the other methods...what does ^ mean?!
Memories of my Dad shouting at me when he couldn't get me to understand a maths methodology are flooding back...and is it me, or is this just ridiculously hard for a 10 year old?
Memories of my Dad shouting at me when he couldn't get me to understand a maths methodology are flooding back...and is it me, or is this just ridiculously hard for a 10 year old?
Same as Ken's **, "raise to the power":Jess wrote:Nope, no calculators allowed. I get the adding up method, but it seems a bit slow (like me!). Am afraid I'm still struggling with the other methods...what does ^ mean?!
2^5 = 2*2*2*2*2 = 32
The trick is to notice that the total is always a power of 2 minus one, and then you can do it by doubling until you reach the target plus one:
2 ,4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384
If you don't notice that, adding will get you there slower but sure.
