Lisa_Love:

does 2^2 mean 2*2?

2^2 means 2 square
2^4 means 2 to the power of 4

BigDevil:

The HCF of 2 numbers is 4, and their LCM is 240.

What are the numbers?

Is the answer simply 4 and 240? How do I show it in working?

No, there are more possibilities

HCF=4=(2^2)
LCM = 240=(2^4)*3*5

Let the two integers be A,B
So it means one integer should have two number of 2s, the other integer should have 4 number of 2s. Then one integer should have one number of 3, the other should have zero number of 3. One integer should have one number of 5, the other should have zero number of 5.

Combing all these
A = 2^2 , B = 2^4*3*5
OR
A = 2^2 *3, B = 2^4*5
OR
A = 2^2 * 5, B = 2^4*3
OR
A = 2^2 * 3*5, B = 2^4

You can swap values for A,B, but then they are essentially still the same numbers.

Hope it helps.

Jerry Guo
Math Program Director -- MaxiMath
We specialize in teaching IP school Math, including Euler Program (RI)
http://www.maximath.sg

cftan:

jieheng:

[quote=\"insanePaPa\"]Can someone help my ds on this?


The LCM of 6, 12 & n is 660. Find all possible values of n.

6 = 2 * 3

12=2² * 3

660 = 2² * 3 * 5 * 11

n must include 5 * 11

n = 5 * 11 = 55

n = 2 * 5 * 11 = 110

n = 2² * 5 * 11 = 220

n = 3 * 5 * 11 = 165

n = 2 * 3 * 5 * 11 = 330

n = 2² * 3 * 5 * 11 = 660

Can someone help to explain. Thanks.[/quote]

NOTE that the way to pick LCM Of several numbers is to take the Highest Number of each prime factor among the three, and then multiply them.

LCM, which is 660, contains two number of 2s, so all three numbers must contain <= two number of 2s, with at least one number having two number of 2s. Since 12 already got two number of 2s,
so n can have zero or one or two number of 2s

Same for 3, LCM contains one 3, so n can have zero or one 3.

LCM contains one 5, but the other two numbers does not have 5, so 5 must come from n. n must contain exactly one 5.
LCM contains one 11, the other two numbers does not have 11, so n must contain one 11.

Combining all these, n must be 5*11 or 5*11*2, 5*11*2*2, 5*11*3, 5*11*2*3, 5*11*2*2*3

This type of questions are quite common in RI and other IP school worksheets. Some are even tougher than this, where they can combine LCM and HCF

Like
\t\"A,B,C are three different integers. A=18,B=60. The lowest common multiple of A,B,C is 540. The highest common factor of A,B,C is 3. Find the possible values of C.\"

Students need to have a good understanding of how to pick LCM and HCF from prime factorization.


Jerry Guo
Math Program Director -- MaxiMath
We specialize in teaching IP school Math, including Euler Program (RI)
http://www.maximath.sg

Volcano123:

can someone help me with this questions

q13(ii) and Q14.
Thank you

Where is the questions?

One way is to write out all numbers until it start to repeat. The period is 71+23=94 (though maybe a factor of it). However, this way is not recommended due to its tediousness.

2nd way, T2 mod (94) has the same remainder regardless of adding 71 or -23. Same for T3, T4,… Tn. Therefore T2013 mod (94) = (1+712012) mod 94 =(1+7138) mod 94 =67. Further note 100-23<= Tn < 100+71, So 77<= Tn <= 170. Therefore T2013 = 67+94 = 161

A qn designed by me, roughly SMO difficulty.

A sequence starts from 1 as 1st term. For subsequent terms, if Tn < 100, then
T(n+1)=Tn+71. If Tn≥100,then T(n+1)=Tn-23.Find T2013.

For Q2.

The following scores are possible
0,3,5,6,8,9…,32 Starting from 8, every number can appear.
Therefore the only scores that are not possible to occur are 1,2,4,7.
The idea is to find all combinations without any restrictions first, subtract the number of combinations having either 1,2,4,7.
Without restriction:
If Tom’s score = 0, Jerry’s score can be 0-32 (33 possibilities)
If Tom’s score =1, Jerry’s score can be 0-31 (32 possibilites)
…
Paul’s score is deducible if the scores of the other two are determined.
Total number of combinations = 33+32+…+1 = 561.

Now, we need to further calculate the number of combinations having 1,2,4 or 7.
If Tom’s = 1, Jerry’s score has 32 possibilities
If Tom’s =2, Jerry’s score has 31 possiblities
If Tom’s = 4, Jerry’s score has 29 possibilities
If Tom’s = 7, Jerry’s score has 26 possibilites.
Total = 118
Jerry’s score can take 1,2,4,7 too, so does Paul’s
1183 = 354.
Note that we double counted cases of T,J or T,P or J,P both having 1,2,4, or 7. There are 44*3 = 48 cases here.
Note that T,P,J all having 1,2,4 or 7 will not appear.
By Ven diagram.
Total number of cases of having at least an 1,2,4 or 7 =354-48 =306
Therefore total number of possible combi = 561-306 = 255

For Q1,

3n+5 is not a multiple of 3
5n+4 is not a multiple of 5
So highest common factor of 3n+5 and 5n+4 is same as
highest common factor of (3n+5)*5 and 5n+4 =
highest common factor of (3n+5)*5 and (5n+4)*3=
highest common factor of 15n+25 and 15n+12
Note that if A is a multiple of C and B is also a multiple of C, then A-B is a multiple of C. (Euclidean algo)
So (15n+25) - (15n+12) is a multiple of highest common factor.
13 is a multiple of highest common factor.
So highest common factor = 13 only
Therefore 3n+5 = 13k (both n and k are intergers)
13k-5 is a multiple of 3.
For primary school students, they can try k = 1,2,3 etc.
k = 2,5,8,11 (the rest of k are having values of n falling outside of required bound)
n = 7,20,33,46

For those familiar with Euclidean algo, they can also try
HCF(3n+5, 5n+4)= HCF(3n+5, (5n+4)-(3n+5))
=HCF(3n+5,2n-1)= HCF((3n+5)-(2n-1),2n-1)
=HCF(n+6,2n-1)=HCF(n+6, (2n-1)-(n+6))=HCF(n+6,n-7)
=HCF(n+6,(n+6)-(n-7))=HCF(n+6,13) = 13 or 1(reject)
Therefore n = 7, 20, 33, 46

Hope to promote sharing and discussion of Math Olym questions.


Two questions from a recent China Math Competition for P6. Quite challenging for Pri students and I gave to my RI GEP student from my Sec 1 Math Olympiad class instead.

Q1) n is an integer <50. Find all possible n such that 3n+5 and 5n+4 is not coprime, i.e. they have a common factor greater than 1.

Q2) In a math competition, all questions carry equal marks. Student get 5 marks if the question is answered correctly, 3 marks if the question is unanswered, and 0 marks if the question is answered wrongly. If the total marks for Tom, Jerry and Paul is 32. In how many different combinations can their scores be? Note that Tom = 10, Jerry = 12, Paul = 10 and Tom =12, Jerry =10, Paul = 10 are considered two different combinations.

Ans will be posted later

MaxiMath Math Specialist Jerry
SMO Gold Prize winner