silly arithmetic question

nanase

Junior Member
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guys I'm given this question that I thought is simple but quite confusing in the end

What is the sum of all three digit numbers which are multiples of 14 but not multiples of 21?

what is the easiest way to do it without calculator, its given verbally to me.
thanks in advance
 
guys I'm given this question that I thought is simple but quite confusing in the end

What is the sum of all three digit numbers which are multiples of 14 but not multiples of 21?

what is the easiest way to do it without calculator, its given verbally to me.
thanks in advance
 
It isn't a silly question, or even a simple one.

Please give it some thought and show us your ideas, so we can interact with them. Don't worry about the "easiest" way; just think about some way to do it without actually listing and adding all the numbers. If it's more complicated than it has to be, we can help you with that part. We need to see what relevant concepts you have learned.
 
Having no clue as to what you know, it is very difficult to know even how to begin helping you.

[MATH]7 * 14 = 98 < 100.[/MATH]
[MATH]72 * 14 = 1008 > 1000.[/MATH]
Thus one sum we are looking for is

[MATH]\sum_{j=8}{71} 14 * j = 14 * \sum_{j=1}^{71} j =[/MATH]
[MATH]14 * \left \{ \left ( \sum_{j=1}^{71} j \right ) - sum_{j=1}^7 j \right \} = 14 * \left ( \dfrac{71 * 72}{2} - \dfrac{8 * 9}{2} \right ) = ???[/MATH]
If that is meaningful to you, what might you do next?

If that is meaningless to you, we need to drop back a bit.

Oops. LaTeX seems to have stopped working.
 
@DrPeterson @JeffM
alright this is my approach, I make a list of multiples of 14 from 100 to 999 (since it must be 3 digits) and I do the same for multiples of 21. I get the numbers using calculators, but i realised it is too many and choosing which multiples of 21 that I should subtract is also a problem. so I'm stuck here
 
Having no clue as to what you know, it is very difficult to know even how to begin helping you.

[MATH]7 * 14 = 98 < 100.[/MATH]
[MATH]72 * 14 = 1008 > 1000.[/MATH]
Thus one sum we are looking for is

[MATH]\sum_{j=8}{71} 14 * j = 14 * \sum_{j=1}^{71} j =[/MATH]
[MATH]14 * \left \{ \left ( \sum_{j=1}^{71} j \right ) - sum_{j=1}^7 j \right \} = 14 * \left ( \dfrac{71 * 72}{2} - \dfrac{8 * 9}{2} \right ) = ???[/MATH]
If that is meaningful to you, what might you do next?

If that is meaningless to you, we need to drop back a bit.

Oops. LaTeX seems to have stopped working.
sir I am very weak in reading sum notation, like reading kanji words to me, but I tried to make some sense of it. I dont know where is 7114 coming from? why are you equating it?
 
What is the sum of all three digit numbers which are multiples of 14 but not multiples of 21?
what is the easiest way to do it without calculator, its given verbally to me.
This is a good example of a question for a course in mathematics for programmers. I taught such some twenty years ago.
The number \(\displaystyle \left\lfloor {\frac{{999}}{{14}}} \right\rfloor - \left\lfloor {\frac{{99}}{{14}}} \right\rfloor = 64\) is the number of three digits numbers divisible fourteen. But some of those are also divisible by twenty-one. See this
Because \(\displaystyle 14=2\cdot 7\) and \(\displaystyle 21=3\cdot 7\) any number divisible by both fourteen & twenty-one must be divisible by \(\displaystyle 42=2\cdot 3\cdot 7\).
Thus look at this

 
@DrPeterson @JeffM
alright this is my approach, I make a list of multiples of 14 from 100 to 999 (since it must be 3 digits) and I do the same for multiples of 21. I get the numbers using calculators, but i realised it is too many and choosing which multiples of 21 that I should subtract is also a problem. so I'm stuck here
Let's see if we can turn this into a method that doesn't require making a list. I'll also avoid sigma notation. (JeffM's work had some typos in it.)

What is the first three-digit multiple of 14? Specifically, 14 times what? How about the last?

Then if we ignore the part about multiples of 21, what would the sum of those multiples look like conceptually (again, without actually listing them all)? Something like 14*_ + ... + 14*_, which could be factored as 14(_ + ... + _). Now, do you know a way to sum what's in the parentheses without listing? (You might know about arithmetic series, for example.)

Once you've done that, think about what the numbers are that have to be excluded.

If the help your getting seems above you, that would be because you haven't followed instructions and told us your background, so we have no idea what to expect you to be able to follow. Please do so!
 
sir I am very weak in reading sum notation, like reading kanji words to me, but I tried to make some sense of it. I dont know where is 7114 coming from? why are you equating it?
I am sorry that my answer was not helpful. This is why it is so important that you tell us about your mathematical knowledge.

I was having trouble getting the math symbols to come out right last night. The symbol you did not understand should have been

[MATH]\sum_{8}^{71} 14 * j.[/MATH]
Let's start with a simple idea. I want to add up the following

[MATH](1 * 14) + (2 * 14) + (3 * 14) + (4 * 14).[/MATH]
One way is to do 4 multiplications and then do 3 additions

[MATH](1 * 14) + (2 * 14) + (3 * 14) + (4 * 14) = 14 + 28 + 42 + 56 = 140.[/MATH]
A different way is to do 3 additions and then 1 multiplication.

[MATH](1 * 14) + (2 * 14) + (3 * 14) + (4 * 14) = 14 * (1 + 2 + 3 + 4) = 14 * 10 = 140.[/MATH]
Notice that the second way has fewer computations, 4 versus 7.

There is a third way, which is to use a formula for adding up succesive integers. That way involves 2 multiplications and one division.

[MATH](1 * 14) + (2 * 14) + (3 * 14) + (4 * 14) = 14 * (1 + 2 + 3 + 4) = 14 * \dfrac{4 * 5}{2} = 14 * \dfrac{20}{2} = 14 * 10 = 140.[/MATH]
Notice that the third way has even fewer computations, 3 versus 4 versus 7.

The more products to be added up, the greater the disparity in computations among the three methods.

And your problem has a lot of products. So we use the third method, which will always require only 2 multiplications and 1 division.

So here is the general formula for adding up n integers starting with 1:

[MATH]\dfrac{n * (n + 1)}{2}.[/MATH]
Do you understand so far?

So how would I write the sum of all the multiples of 14 less than 1000? Well, I am DEFINITELY going to use the third method, but first I need to figure out how many numbers are involved. What is the largest multiple of 14 less than 1000. Pka showed you how to determine that by using floor notation.

[MATH]\dfrac{1000}{14} = 71 + \dfrac{6}{14} \implies \left \lfloor \dfrac{1000}{14} \right \rfloor = 71.[/MATH]
Let's check.

[MATH]14 * 71 = 994 \text { and } 994 + 14 = 1008.[/MATH]
So there are 71 multiples of 14 less than 1000. So how do I compute the sum of all the multiples of 14 under 1000?

[MATH](14 + 28 + ... 994) = 14 * (1 + 2 + ... 71) = 14 * \dfrac{71 * 72}{2} = 14 * \dfrac{5112}{2} = 7 * 5112 = 35784.[/MATH]
But I do not want to include multiples of 14 less than 100 in my total. What is their sum?
 
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