proofs make me want to jump off a cliff (1 Viewer)

constexpr

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Suppose that , where p and q are primes and p is not equal to q
a) Use combinatorics (a counting argument) to explain why n has (a+1)(b + 1) factors.
b) hence determine the number of factors of 80000.
 

cossine

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a) is quite easy. Use the rule of product axiom. For there to be factor you got to decide how many you will choose p and q.
 

s97127

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Suppose that , where p and q are primes and p is not equal to q
a) Use combinatorics (a counting argument) to explain why n has (a+1)(b + 1) factors.
b) hence determine the number of factors of 80000.
a) Factors of N has the form of p^x * q^y where x and y are integer
X ranges from 0 to a so there are a+1 ways to choose X
Y ranges from 0 to b so there are b+1 ways to choose Y

Therefore, N has (a+1)(b+1) factors. JR kids do this type of problem in year 7 :)

b) 80000 = 2^7 * 5^4 so it has 8*5 = 40 factors
 

s97127

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let’s be real there’s no way ruse students do it in y7 LOL
considering most students from ruse go to du, du defs wouldn’t be covering that in y7
You are underestimating jr kids. They do it in math olympiad training and most of the problems are harder than 4u math. They don't go to dr du until year 8.
 

kkk579

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You are underestimating jr kids. They do it in math olympiad training and most of the problems are harder than 4u math. They don't go to dr du until year 8.
Lol there’s no way, maybe a select few that are extremely talented at maths could but the whole grade? Very unlikely
 

s97127

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Lol there’s no way, maybe a select few that are extremely talented at maths could but the whole grade? Very unlikely
Anyone can sign up for it. You don't need to be good at math to do it.
 

kkk579

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It is just a counting problem with prime and composite numbers. Year 7 at top selective schools also learn it.
Oh wtf I take back what i said then, the title said proofs so i was assuming it was a 4u concept hence why i was hesitant on believing that all y7 ruse students were able to solve the q
 

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