How to enter this program?

2014 10 25
3 个回答
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更新时间：2014 10 26
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0 个投票

Hello, I have tried unsuccessfully for a couple hours to enter a program that loops until a number that is even, divisible by both 13 and 16, and whose square root is greater than 120 is found. I have reached a wall, so I have come to ask for help. Would using a while statement be the first step? Please don't solve this problem, I only ask for help getting it going. Yes, this is basic stuff, but I still have little understanding on it.

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 采纳的回答

Roger Stafford
Roger Stafford 2014-10-25

1 个投票

You need to translate each of those conditions into a matlab logical proposition. For example if n is an integer you are investigating, try dividing it by 13, then comparing the 'round' of the quotient with the original quotient to see if it is still an integer after the division. And yes, the 'while' function is the right thing to use here. Keep it looping until all the conditions turn out to be true for the same integer, for which you can make great use of the "&" operation.

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Rick Rosson
Rick Rosson 2014-10-25
编辑：Rick Rosson 2014-10-25
You may also find that the mod function is helpful in this particular case. This function returns the modulus of two integers, which is the remainder that is left after dividing the first integer by the second integer. So, for example:
mod(45,13)
returns 6, whereas
mod(52,13)
returns 0.
This provides a very easy way to test for divisibility. For example:
if mod(n,13)
disp('n is not divisible by 13');
else
disp('n is divisible by 13');
end
Roger Stafford
Roger Stafford 2014-10-25
Yes, that's a little easier than writing
round(n/13) == n/13
Colton
Colton 2014-10-25
编辑：Colton 2014-10-25
Okay, so I said
a = 2;
while sqrt(a) <= 120 && mod(a,13) && mod(a,16)
a = a + 2;
end
But this returns a = 16, satisfying only the final condition. Could you point me in the right direction? Thank you so much for your help so far though.
Rick Rosson
Rick Rosson 2014-10-25
Try using || instead of &&.
Colton
Colton 2014-10-26
@Rick, thank you. That fixed my issues and I was able to find the correct value! And thanks everyone else for the help!

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更多回答（2 个）

Rick Rosson
Rick Rosson 2014-10-25
编辑：Rick Rosson 2014-10-25

1 个投票

Here is a faster and simpler approach:
a = 0;
while a < 120^2
a = a + 13*16;
end
per isakson
per isakson 2014-10-26
编辑：per isakson 2014-10-26

0 个投票

Try
>> factor( cssm )
ans =
2 2 2 2 2 5 7 13
where cssm is
function ix = cssm()
% chose a start value, which is larger than 120^2 and divisible by 16
ix = 120*120+16;
while ne( mod(ix,13), 0 )
ix = ix + 16;
end
end

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