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Praetorian
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[#22] 2 Min. Challenge : Number 17 Mar 2004, 12:17
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1. Time yourself
2. Solve as fast as you can
3. Write your solution here and your time , please


How many factors are there for the number equal to 40^3-17^3- 23^3?
1. 32
2. 42
3. 64
3. 72
4. 80



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[#22] 2 Min. Challenge : Number 17 Mar 2004, 13:13
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64 is the answer.
2 minutes to solve.

The equation can be expressed as (23+17)^3 - 17^3-23^3

(a+b)^3 = a^3 + b^3 + 3a^2b + 3ab^2

so what you have is 3 a b ( a + b ) as a result = 3 * 23 * 17 ( 23 + 17 )
= 3 * 23 * 17 * 40 = 3 * 23 * 17 * 5 * 2^3
Factors = 2 * 2 * 2 * 2 * 4 = 64
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[#22] 2 Min. Challenge : Number 17 Mar 2004, 13:29
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Brilliant. Thanks Anandnk!
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[#22] 2 Min. Challenge : Number 17 Mar 2004, 13:38
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HOW DO YOU SOLVE THese kind of problems...I easily got until the value 3*17*23*40....When you want to find out factors..dow we have to find out separately and multiply all of them to get the answer????
please let me know the solution for these kind of problems
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[#22] 2 Min. Challenge : Number 17 Mar 2004, 13:45
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If x = a^p. b^q. c^r

where a, b, c are the prime factors. Then total no. of factors of x is calculated by:

(p+1)(q+1)(r+1).

In the above ques.

p, q, r, s each equals 1 and t = 3.

Hence the answer

(1+1)(1+1)(1+1)(1+1)(3+1) = 64.
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[#22] 2 Min. Challenge : Number 17 Mar 2004, 13:48
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# of factors for 3*17*23*40:

All besides 40 is prime.

Prime factorization of 40 = 2^3 * 5

So 3*17*23*40 = 2^3*3*5*17*23

To find the total # of factors, you take each factor and find the total possible exponents that could result from it. (Don't think I'm explaining it quite well, but...)

For 2^3, you can have 2^0, 2^1, 2^2, or 2^3. So altogether, you have 4 possibilities. (Remember that 2^3 = 8 and the factors of eight are 1, 2, 4, and 8 which are respectively equal to 2^0, 2^1, 2^2, and 2^3.)

3 = 3^0 and 3^1 = 2 possibilities
5 = 5^0 and 5 ^1 = 2 possibilities
17 = 17^0 and 17^1 = 2 possibilities
23 = 23^0 and 23^1 = 2 possibilities

So, to find the total number of combinations, you multiply. 4*2*2*2*2 = 64
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[#22] 2 Min. Challenge : Number 17 Mar 2004, 14:04
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anandnk
64 is the answer.
2 minutes to solve.

The equation can be expressed as (23+17)^3 - 17^3-23^3

(a+b)^3 = a^3 + b^3 + 3a^2b + 3ab^2

so what you have is 3 a b ( a + b ) as a result = 3 * 23 * 17 ( 23 + 17 )
= 3 * 23 * 17 * 40 = 3 * 23 * 17 * 5 * 2^3
Factors = 2 * 2 * 2 * 2 * 4 = 64

Nice one Anandnk. Factorisation of 40 was indeed key. Also Virtual, nice explanation! :-D
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[#22] 2 Min. Challenge : Number 17 Mar 2004, 20:28
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praetorian123
1. Time yourself
2. Solve as fast as you can
3. Write your solution here and your time , please


How many factors are there for the number equal to 40^3-17^3- 23^3?
1. 32
2. 42
3. 64
3. 72
4. 80


64 is correct.

if a+b+c = 0, then a^3 + b^3 + c^3 = 3abc
now, since 40 -23-17 = 0,
40^3 - 17^3 - 23^3 can be written as,
40*17*23*3 = 2^3*3*5*17*23, so the total number of factors wud be,
(3+1)(1+1)(1+1)(1+1)(1+1) = 64



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