
By: geniuskids (offline) Monday, February 24 2014 @ 06:06 AM CST (Read 7296 times)



geniuskids 
Hi there, would appreciate anyone who can help provide solutions to the following questions:
(1) List the
(a) smallest integer such that 5<21. (Ans =  4)
(b) greatest prime number such that 2p1< or = 14. ( Ans = 7 )
(2) by inserting the operation signs +, , x , divide and brackets, make the following a
' true' sentence. You may use any of the operation signs more than once)
0.1 0.2 0.3 0.4 0.5 = 1
* Please note that there are various answers to Q2, but is there a method to solve such
questions.
(3) What is the smallest positive integer value of n for which 56n is a multiple of 42.
( Ans 3)
(4) Write down the smallest value of n for which 294n is a square number.
(Ans 6)

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Registered: 11/12/11 Posts: 169





By: jo sarah (offline) Monday, February 24 2014 @ 06:47 AM CST



jo sarah 
Hi,
In the following, I give solution after each question, so have used to help highlight the solutions.
(1) List the
(a) smallest integer such that 5<21. (Ans =  4)
Solution:
Do you mean smallest integer x such that 5 < x < 21; else the statement 5<21 is merely a true statement. Or, the smallest integer between 5 and 21, both not inclusive?
If so, the answer is 4 because on the no. line, 5 is left of 4...[difficult to represent no. line here, but try...]
.................. 5....4....3...2...etc.........20....21
.................___________..............._____
the symbol < means the required integer is greater than 5,
and the smallest integer value greater than 5 is 4....
(b) greatest prime number such that 2p1< or = 14. ( Ans = 7 )
Solution:
first solve 2p  1 <= 14..........[<= represents < or =]
i.e., 2p <= 15
so that, p <= 7.5
and the prime no. <= 7.5 is 7...
(3) What is the smallest positive integer value of n for which 56n is a multiple of 42.
Solution:
prime factorise 42 & 56: so, 42 = 2 x 3 x 7, and 56 = 7 x 2^3......[2^3 means 2 to the power of 3]
and you want 56n = 42k, where k is small as possible.
that is:........7 x 2 x 2 x 2 x n = 7 x 2 x 3 k
dividing thro by 7 x 2, you get:
2 x 2 n = 3 x k
since you want n to be smallest, n = 3.... [which gives 56n = 42 x 4]
(4) Write down the smallest value of n for which 294n is a square number.
Solution:
Square nos. have powers which are even,
so first prime factorise 294: 294 = 3 x 2 x 7 x 7
factors 2 and 3 above have power of 1, while factor 7 is squared.
therefore the smallest n such that 294n is square no. is 2 x 3 = 6.....
you will find that if no restricted you may have n = 2^3 x 3^3; or 2^3 x 3^7; or 2 x 3 x 7^4, etc (so long as you have for n the powers of 2 and 3 odd). BUT question wants smallest n so answer is 6.

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Registered: 03/20/12 Posts: 111





By: geniuskids (offline) Monday, February 24 2014 @ 07:54 AM CST



geniuskids 
thanks jo sarah.
So sorry, for question 1a) the question should be:
List the smallest integer such that 5t < 21
I miss out the (t).
Fully understand Q(1b) Q3, Q4, thank you so much
But, I still does not understand Q2, pls provide solution as I don't get what you say.
Many thanks

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Registered: 11/12/11 Posts: 169





By: jo sarah (offline) Tuesday, February 25 2014 @ 05:55 AM CST



jo sarah 
Hi,
for this: List the smallest integer such that 5t < 21
You do it as:
5t < 21
therefore, t > (21/5),...............[I divide by (5) on both sides]
that is to say, t > 4.2
and the smallest integer is 4.
You may ask why the inequality sign is changed in direction. Whenever you Multiply or Divide by a Negative no., the inequality is reversed. The following should help you to appreciate it:
It is true that 2 < 5
now, multiply/divide both sides by (1) [or any negative no.], and you'll get
2 > 5, right? This latter statement is true.
trust this clears
As for no. 2, let me see how to put it, and will try to come back again.
thanks.

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Registered: 03/20/12 Posts: 111





By: geniuskids (offline) Saturday, March 01 2014 @ 09:11 AM CST



geniuskids 
More help needed pls.
Pls refer to attached file and appreciate some help on the questions.
Pls also help solve :
(1) What is the smallest whole number that can be multiplied by 28 such that the product is a perfect cube?
(Answer=98)
Many thanks!

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Registered: 11/12/11 Posts: 169





By: jo sarah (offline) Sunday, March 02 2014 @ 06:59 PM CST



jo sarah 
hi,
What is the smallest whole number that can be multiplied by 28 such that the product is a perfect cube?
Solution:
For perfect cube, the powers of the prime factors must all be multiples of 3
[compare to perfect squares, where powers must all be multiples of 2, so therefore even.
If you need explanation on this, let me know.]
So, since you are required to find smallest integer k such that 28k is perfect cube.
first, 28 = 2^2 x 7
[one factor has power 2, other has 1; both not multiple of 3:
so you are to 'up' both to their respective next (because "smallest" multiple of 3;
in this case to make both 3]
So, 28k = 2^3 x 7^3 = (2^2 x 7) x (2 x 7^2) = 28 x k
therefore, smallest k = 2 x 7^2 = 98
Math 1:
I thought this question would be a calculator question?
Else it will be solved step by step [not repeating question here]:
[27  {2  (2)^2}] / [29  9] = [27  {2  4}] / 10
= [27 + 2] / 10 = 25 / 10 = 2.5
Math 2:
You should have already prime factorised 225 and 216 in earlier parts of question, giving:
225 = 3^2 x 5^2
216 = 2^3 x 3^3
So, to find HCF using prime factored forms [which you probably know how to do now?]*
write thus:
225^2 = ......3^4 x 5^4.................["..." are for alignment]
216 = 2^3 x 3^3
HCF = ........3^3 = 27
LCM = 2^3 x 3^4 x 5^4..............[just added for your info. ]
* if you are still not sure, do let me know.

of the previous no. 2 (just for interest for now):
if you are allowed to shuffle the nos. (which I had assumed you are not), then:
(0.1 / 0.2) * 0.4 + 0.3 + 0.5 = 1

Regular Member
Registered: 03/20/12 Posts: 111





By: jo sarah (offline) Sunday, March 02 2014 @ 07:06 PM CST



jo sarah 
hi geniuskids,
What is the smallest whole number that can be multiplied by 28 such that the product is a perfect cube?
Solution:
For perfect cube, the powers of the prime factors must all be multiples of 3
[compare to perfect squares, where powers must all be multiples of 2, so therefore even.
If you need explanation on this, let me know.]
So, since you are required to find smallest integer k such that 28k is perfect cube.
first, 28 = 2^2 x 7
[one factor has power 2, other has 1; both not multiple of 3:
so you are to 'up' both to their respective next (because "smallest" multiple of 3;
in this case to make both 3]
So, 28k = 2^3 x 7^3 = (2^2 x 7) x (2 x 7^2) = 28 x k
therefore, smallest k = 2 x 7^2 = 98
Math 1:
I thought this question would be a calculator question?
Else it will be solved step by step [not repeating question here]:
[27  {2  (2)^2}] / [29  9] = [27  {2  4}] / 10
= [27 + 2] / 10 = 25 / 10 = 2.5
Math 2:
You should have already prime factorised 225 and 216 in earlier parts of question, giving:
225 = 3^2 x 5^2
216 = 2^3 x 3^3
So, to find HCF using prime factored forms [which you probably know how to do now?]*
write thus:
225^2 = ......3^4 x 5^4.................["..." are for alignment]
216 = 2^3 x 3^3
HCF = ........3^3 = 27
LCM = 2^3 x 3^4 x 5^4..............[just added for your info. ]
* if you are still not sure, do let me know.

of the previous no. 2 (just for interest for now):
if you are allowed to shuffle the nos. (which I had assumed you are not), then:
(0.1 / 0.2) * 0.4 + 0.3 + 0.5 = 1

Regular Member
Registered: 03/20/12 Posts: 111





By: jo sarah (offline) Sunday, March 02 2014 @ 07:07 PM CST



jo sarah 
Hmm, I am puzzled. I have typed a reply and submitted but twice they land in a blank page!
will have to find out from admin, and repost.
[updated:] Somehow managed to do something and got it in, please go to page 2 ...continuation of this.

Regular Member
Registered: 03/20/12 Posts: 111





By: jo sarah (offline) Sunday, March 02 2014 @ 07:33 PM CST



jo sarah 
hi geniuskids,
What is the smallest whole number that can be multiplied by 28 such that the product is a perfect cube?
Solution:
For perfect cube, the powers of the prime factors must all be multiples of 3
[compare to perfect squares, where powers must all be multiples of 2, so therefore even.
If you need explanation on this, let me know.]
So, since you are required to find smallest integer k such that 28k is perfect cube.
first, 28 = 2^2 x 7
[one factor has power 2, other has 1; both not multiple of 3:
so you are to 'up' both to their respective next (because it's to be smallest) multiple of 3;
in this case to make both 3]
So, 28k = 2^3 x 7^3 = (2^2 x 7) x (2 x 7^2) = 28 x k
therefore, smallest k = 2 x 7^2 = 98

Regular Member
Registered: 03/20/12 Posts: 111





By: jo sarah (offline) Sunday, March 02 2014 @ 07:39 PM CST



jo sarah 

Regular Member
Registered: 03/20/12 Posts: 111



