Year 9 maths homework extension

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February 1st 2009, 10:16 PM
IndoorKid

Year 9 maths homework extension

Hey guys. I finished my regular homework and the teacher has given us 15 extension questions which I can't answer any of. Try them out, I think they are a bit beyond what a year 9 should know...

1. A 3 digit number is 629 times less than the sum of all other three digit numbers. What is this number?

2. 117% of David's mass equals 17% of Goliath's mass. What is the ratio of their masses?

3. Find the smallest possible integer divisible by 45, the sum of whose digits is 45.

4. The number 'a' can be obtained when the digits of the three digit number 'b' are written in the reverse order. If the product 'ab' is equal to 214,875 what are the values of 'a' and 'b'?

7. If a + b = 2 and a^2 + b^2 = 8 what is the value of 1/a + 1/b ?

8. Let a, b and c be distinct non-zero digits such that:
N = (abc)(bca)(cab) + 1 is divisible by 9
What is the smallest value a+b+c can have?

Thats just some, see if you can answer any...
February 1st 2009, 10:28 PM
Grandad

Problems

Hello IndoorKid

Quote:

Originally Posted by IndoorKid

1. A 3 digit number is 629 times less than the sum of all other three digit numbers. What is this number?

3 digit numbers start at 100 and finish at 999. So there are 900 of them altogether. They add up to 900 times (the average of the first and last) = 494500. So if the number you're looking for is $x$ :

$629x = 494500 - x$

$\Rightarrow x=785$

Quote:

Originally Posted by IndoorKid

2. 117% of David's mass equals 17% of Goliath's mass. What is the ratio of their masses?

Let David's mass be $d$ and Goliath's be $g$ . Then

$\frac{117d}{100} =\frac{17g}{100}$

$\Rightarrow 117d = 17g$

$\Rightarrow \frac{d}{g}=\frac{17}{117}$

$\Rightarrow d:g =17:117$

More later if no-one else can pick this up. Must go now!

Grandad
February 1st 2009, 10:35 PM
IndoorKid

thanks, there is no way i could have solved that with what i know but i will learn from it. Much appreciated
February 1st 2009, 11:14 PM
ADARSH

(Hi)a+b= 2
(a+b)^2= a^2+b^2+2ab =8+2ab = 4
so ab= (4-8)/2 = -2
hence
$<br /> \frac{1}{a} + \frac{1}{b} = \frac{a+b}{ab} = \frac{2}{-2} = -1 <br />$

I am sorry but your thread is an apt example of why you should not post many questions in the same thread* and why one should not double post
* this reduces the number of people who would watch your thread (Happy)
February 2nd 2009, 01:37 AM
Grandad

Problems

Hello IndoorKid

Quote:

Originally Posted by IndoorKid

3. Find the smallest possible integer divisible by 45, the sum of whose digits is 45.

Any number that's divisible by 45 will obviously also be divisible by 5 and 9, and any number that's divisible by 5 will end in 0 or 5. Also, any number whose digits add up to a mulitple of 9 is also a multiple of 9. (Did you know that?)

Now 99999 it's obviously the smallest number whose digits add up to 45 (because 9 x 5 = 45), but it doesn't end in 5 or 0. So we must be looking for at least a six-digit number. We could look at six-digit numbers ending in 0, but the only one whose digits then add up to 45 is 999990, which is a multiple of 45, but is it the smallest one?

Well, let's look at 6-digit numbers ending in 5. In this case, the first five digits need only add up to 40. For instance, 5 x 8, giving the number 888885, which again works. But is there a smaller one still?

Yes. Remember that we must keep the digits adding up to 45 with the last one being a 5, and as long as the digits add up to 45 (which is a multiple of 9), this will ensure that the number itself is a multiple of 9.

We need to make the number as small as possible. So let's make the first two digits 7's instead of 8's, leave the next one as an 8, and make the next two into 9's to keep the total = 45. This gives us 778995 (= 45 x 17311). And I think that's the answer!

Quote:

Originally Posted by IndoorKid

4. The number 'a' can be obtained when the digits of the three digit number 'b' are written in the reverse order. If the product 'ab' is equal to 214,875 what are the values of 'a' and 'b'?

The key to solving this problem is to find the prime factors of 214,875. They are:

3 x 3 x 5 x 5 x 5 x 191

Now you have to re-arrange these into two groups, in such a way that the product of each group is a 3-digit number, and one product is the same as the other with the digits reversed.

You don't have to look very far, because the only possibilities for the group containing 191 are 191 x 3 or 191 x 5. Any other combination makes the product more than 999.

So there's your answer: 191 x 3 = 573 and 3 x 5 x 5 x 5 = 375.

Grandad
February 2nd 2009, 01:45 AM
ADARSH

Quote:

Originally Posted by Grandad

The key to solving this problem is to find the prime factors of 214,875. They are:

3 x 3 x 5 x 5 x 5 x 191

Now you have to re-arrange these into two groups, in such a way that the product of each group is a 3-digit number, and one product is the same as the other with the digits reversed.

You don't have to look very far, because the only possibilities for the group containing 191 are 191 x 3 or 191 x 5. Any other combination makes the product more than 999.

So there's your answer: 191 x 3 = 573 and 3 x 5 x 5 x 5 = 375.

Grandad

So we need to use hit and try over here , Are there any other ways Grandad(Happy)?
February 2nd 2009, 02:04 AM
IndoorKid

Grandad I've been working on the 45 question and was following your line of thinking but my answer is different

I worked out that it should be a 6 digit number ending in 5 (99999 is the only 5 digit number that can add to 45 but doesn't divide to give an integer).

So from there I thought - If the last number is a 5 than the other 5 numbers must add to give 40. From here I worked out what was the smallest number that I could physically use that would allow the sum to be 40. It was a 4 (4+9+9+9+9 = 40). From here I was just lucky as 499995 divides by 45 and the digits summate to 45
February 2nd 2009, 02:41 AM
Grandad

Problems

Hello IndoorKid

Quote:

Originally Posted by IndoorKid

Grandad I've been working on the 45 question and was following your line of thinking but my answer is different

I worked out that it should be a 6 digit number ending in 5 (99999 is the only 5 digit number that can add to 45 but doesn't divide to give an integer).

So from there I thought - If the last number is a 5 than the other 5 numbers must add to give 40. From here I worked out what was the smallest number that I could physically use that would allow the sum to be 40. It was a 4 (4+9+9+9+9 = 40). From here I was just lucky as 499995 divides by 45 and the digits summate to 45

You are quite right! Why didn't I think of that?

But it's not just luck that it works, of course. Again, it's because it ends in 5 and the digits add up to a multiple of 9.

In answer to ADARSH's question: I don't really think that there's any other realistic way of tackling problems like this. If you try to do it algebraically - e.g. suppose that the first number is 'xyz' and the second 'zyx' you end up with the equation

(100x + 10y + z)(x + 10y + 100z) = 214875

and where do you go from there?

But there wasn't much trial and error, was there? There were really only two possibilities once we had found the prime factors.

Grandad
February 2nd 2009, 03:33 AM
Bruce

Quote:

Originally Posted by Grandad

Hello IndoorKid3 digit numbers start at 100 and finish at 999. So there are 900 of them altogether. They add up to 900 times (the average of the first and last) = 494500. So if the number you're looking for is $x$ :

$629x = 494500 - x$

$\Rightarrow x=785$

Hi Grandad.

I thought that the sum of a set was calculated like this:

Sum n = n/2(2a + (n - 1)d)

n = number of numbers, so 900
a = starting number, so 100
d = increase every time, so 1

900/2(2x100 + (900 - 1)1)
450(1099)
494550

Your answer was 494500, 50 less than mine.

Explain?
February 2nd 2009, 04:11 AM
Grandad

Typo!

Hello Bruce

Quote:

Originally Posted by Bruce

Hi Grandad.

I thought that the sum of a set was calculated like this:

Sum n = n/2(2a + (n - 1)d)

n = number of numbers, so 900
a = starting number, so 100
d = increase every time, so 1

900/2(2x100 + (900 - 1)1)
450(1099)
494550

Your answer was 494500, 50 less than mine.

Explain?

You are quite right. I got 494550 too. I just didn't type it very well, did I? You'll find if you solve my equation for x, the answer x = 785 uses 494550, not 494500.

Incidentally, my formula is the same as yours, but in a different format. You can say that the sum of an AP is $\frac{n}{2}(a + l)$ , where $l$ is the last term, which is equivalent to saying that it's the number of terms, $n$ , multiplied by the average of the first and last terms: $\frac{1}{2}(a + l)$ . I used this way of explaining it because I wasn't sure if the OP would have met AP's.

Thanks for checking my working!

Grandad
February 2nd 2009, 04:23 AM
Grandad

Problems

Hello again IndoorKid

Quote:

Originally Posted by IndoorKid

8. Let a, b and c be distinct non-zero digits such that:
N = (abc)(bca)(cab) + 1 is divisible by 9
What is the smallest value a+b+c can have?

I finally got round to looking at this one. I'm not sure that I'm understanding the question here, because I don't think it's possible that N can be a multiple of 9. Why do I say that? Well, it's all to do with the sum of the digits again. If by (abc)(bca)(cab) we mean a number like 125,251,512, where a = 1, b = 2 and c = 5, then it's definitely not possible, because these digits always add up to 3(a + b + c) which is a multiple of 3. Therefore the number itself is a multiple of 3. So if we add one to it, it can't possibly be a multiple of 3 (or 9), can it?

Grandad
February 2nd 2009, 05:08 AM
Bruce

Thanks for the clarification.
February 2nd 2009, 06:39 AM
Gusbob

Quote:

Originally Posted by Grandad

Hello again IndoorKidI finally got round to looking at this one. I'm not sure that I'm understanding the question here, because I don't think it's possible that N can be a multiple of 9. Why do I say that? Well, it's all to do with the sum of the digits again. If by (abc)(bca)(cab) we mean a number like 125,251,512, where a = 1, b = 2 and c = 5, then it's definitely not possible, because these digits always add up to 3(a + b + c) which is a multiple of 3. Therefore the number itself is a multiple of 3. So if we add one to it, it can't possibly be a multiple of 3 (or 9), can it?

Grandad

I'm pretty sure the numbers 137, 371, and 713 works for the formula.

N = (137)(371)(713) + 1
=36239651 + 1
=36239652
= 9(4026628)
=9n where n is a positive integer

I'm sorry if I had any misconceptions, its just that I'm not sure I made the connection between "...these digits always add up to 3(a+b+c) which is a multiple of 3. Therefore the number itself is a multiple of 3."

Isn't it really in the form 100a + 10b + c, and not a + b + c?
February 2nd 2009, 06:55 AM
ADARSH

n (n+2)(n+6)+1=n(n^2+8n+12)+1=n^3+8n^2+12n+1
Putting any number(integers) from 1-3 gives a multiple of 9 in above equation
and what you have found should most probably be the answer:D
February 2nd 2009, 09:37 AM
Grandad

Problems

Hello Gusbob

Quote:

Originally Posted by Gusbob

I'm pretty sure the numbers 137, 371, and 713 works for the formula.

N = (137)(371)(713) + 1
=36239651 + 1
=36239652
= 9(4026628)
=9n where n is a positive integer

I'm sorry if I had any misconceptions, its just that I'm not sure I made the connection between "...these digits always add up to 3(a+b+c) which is a multiple of 3. Therefore the number itself is a multiple of 3."

Isn't it really in the form 100a + 10b + c, and not a + b + c?

As I said, I didn't interpret the question in the same way as you. I assumed that (abc)(bca)(cab) was a nine-digit number formed from the digits a, b and c by re-arranging them as in the example I gave. Your interpretation may well be the correct one; namely the product of the three 3-digit numbers: (abc) x (bca) x (cab). It certainly provides possible solutions, which mine doesn't!

My explanation (which is valid for my interpretation!) is based on the fact that a number is divisible by 3 if and only if the sum of its digits is also divisible by 3. Thus, my 9-digit number, the sum of whose digits is 3(a + b + c), is always a multiple of 3.

In the example you gave - which certainly works! - a + b + c = 11. Have you thought about how you could find a smaller sum (or show that there isn't one) as the question required?

Grandad