Sec 1 Maths Qns
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Hi All, Pls help me and thks for your time. Q1 The HCF and LCM of three numbers are 12 and 2^3 x 3^2 x 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?
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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.
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Originally posted by Chocolate Cake:
Hi All,
Pls help me and thks for your time.Q1
The HCF and LCM of three numbers are 12 and 23×32×7 respectively. Given that two of the numbers are 24 and 36, find the third number.23 means 2 to the power of 3
32 means 3 to the power of 2
HCF : Highest common factor
LCM : Lowest common multiple24 = 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 = ...
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.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.
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?143143. calculator and press numbers. using 7,11,13 as factors
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Originally posted by Only-Way-4-Destiny!:
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?
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no adverts
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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 x 3 36 = 2^2 x 3^2 x = 2^a x 3^b x 7^c --------------------- HCF = 2^2 x 3 LCM = 2^3 x 3^2 x 7^1 => 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 x 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
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Originally posted by Worldlybusinessman:
smlj .
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Originally posted by SBS261P:
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 = 23×3
36 = 22×32
x = 2a x 3b x 7c
—-—-—-——
HCF = 22×3LCM = 23×32×71
=> 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.
Its sec 1 stuff. Go easy on him man. -
Qn1=84 only.
Note HCF is 12.
For qn2, it means that the 3 numbers are less than 300, not the sum. Therefore the answer is 244+245+246=735.