Sec 1 Maths Qns

Chocolate Cake SG

Hi All,
Pls help me and thks for your time.

Q1
The HCF and LCM of three numbers are 12 and 2^3×32×7 respectively. Given that two of the numbers are 24 and 36, find the third number.

2^{3 means 2 to the power of 3

3}2 means 3 to the power of 2
HCF : Highest common factor
LCM : Lowest common multiple

Q2
Using tests of divisibility of otherwise, find the sum of the 3 greatest consecutive integers less than 300 for which the least number has 4 as a factor, the next number has 5 as a factor and the greatest number has 6 as a factor.

Q3
Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 123123). Whatever digits you choose the number will always be divisible by 7, 11 and 13 (Try it out). Explain why. What other numbers can divide the 6 digit number?

Only-Way-4-Destiny! SG

I give up. a²=b²+c²-2bc cosA cannot be used sia. 1/2 ab sin C oso cannot be used too.

For the first one let third number be x and make equations and solve it. I think.

All lazy to think.

Garrick_3658 SG

Hi All,
Pls help me and thks for your time.

Q1
The HCF and LCM of three numbers are 12 and 2^3×32×7 respectively. Given that two of the numbers are 24 and 36, find the third number.

2^{3 means 2 to the power of 3

3}2 means 3 to the power of 2
HCF : Highest common factor
LCM : Lowest common multiple

24 = 2^3 x 3

36 = 2^2 x 3^2

So... since you need to have the 7 for LCM... and HCF is 12...

Number = ...

guess and check. usually this will suffice.

middle number has to be _0 or _5.but _0 is impossible since no numbers that have 9 as last digit can have 4 as a factor. so it leaves _5. Then try everything.

calculator on hand!

195. find the numbers yourself.

143143. calculator and press numbers. using 7,11,13 as factors

Garrick_3658 SG

I give up. a²=b²+c²-2bc cosA cannot be used sia. 1/2 ab sin C oso cannot be used too.

For the first one let third number be x and make equations and solve it. I think.

All lazy to think.

Then why did you post?

Worldlybusinessman SG

no adverts

SBS261P SG

if i’m nt wrong q1 can have many answers

Let x be the third number. It can only have 2, 3, and 7 as its prime factors.

24 = 2^3×3

36 = 22×3²

x = 2a x 3^{b x 7}c
—-—-—-——
HCF = 2^2×3

LCM = 23×3^2×71

=> a can be 2 or 3
=> b can be 1 or 2
=> c = 1
___________

q2. as discussed, middle number must end with 5.

Let the middle number be x.

3x < 300

=> x < 100

x must be divisible by 5 => x must be a multiple of 5.
x – 1 must be divisible by 4 => x – 1 can be 84, 64, 44…
x + 1 must be divisible by 6 => x + 1 can be 96, 66, 36…

From here, deduce x = 65, therefore the 3 numbers are 64, 65, 66.
___________

q3. 7×11 x 13 = 1001.

any 3 digit number repeated twice (e.g. abcabc) will be divisible by 1001 because abcabc = abc x 1000 + abc x 1 = abc x 1001.

for starters, an easy guide to divisibility tests can be found, as always, on wikipedia.

http://en.wikipedia.org/wiki/Divisibility_rule

93'guy SG

smlj .

seotiblizzard SG