Hey guys - would someone mind helping explain the solution for the below problem? Thanks in advance!
For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100)+1, then p is...
a. between 2 and 10
b. between 10 and 20
c. between 20 and 30
d. between 30 and 40
e. greater than 40
Help on quant problem
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Important Concept: If k is a positive integer that's greater than 1, and if k is a factor (divisor) of N, then k is not a divisor of N+1lukemacdougall wrote:Hey guys - would someone mind helping explain the solution for the below problem? Thanks in advance!
For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100)+1, then p is...
a. between 2 and 10
b. between 10 and 20
c. between 20 and 30
d. between 30 and 40
e. greater than 40
For example, since 7 is a factor of 350, we know that 7 is not a factor of (350+1)
Similarly, since 8 is a factor of 312, we know that 8 is not a factor of 313
Now let's examine h(100)
h(100) = (2)(4)(6)(8)....(96)(98)(100)
Factor to get: h(100) = 2[(1)(2)(3)(4)....(48)(49)(50)]
Since 2 is in the product of h(100), we know that 2 is a factor of h(100), which means that 2 is not a factor of h(100) +1 (based on the above rule)
Similarly, since 3 is in the product of h(100), we know that 3 is a factor of h(100), which means that 3 is not a factor of h(100) +1 (based on the above rule)
Similarly, since 5 is in the product of h(100), we know that 5 is a factor of h(100), which means that 5 is not a factor of h(100) +1 (based on the above rule)
.
.
.
.
Similarly, since 47 is in the product of h(100), we know that 47 is a factor of h(100), which means that 47 is not a factor of h(100) +1 (based on the above rule)
So, we can see that none of the primes from 2 to 47 can be factors of h(100) +1, which means the smallest prime factor of h(100)+ 1 must be greater than 47.
Answer = E
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Since the difference between them is 1, h(100) and h(100)+1 are consecutive integers.For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, the p is
A: Between 2 & 10
B: Between 10 & 20
C: Between 20 & 30
D: Between 30 & 40
E: Greater than 40
Consecutive integers are COPRIMES: they share no factors other than 1.
Let's examine why:
If x is a multiple of 2, the next largest multiple of 2 is x+2.
If x is a multiple of 3, the next largest multiple of 3 is x+3.
Using this logic, if we go from x to x+1, we get only to the next largest multiple of 1.
So 1 is the only factor common both to x and to x+1.
In other words, x and x+1 are COPRIMES.
Thus:
h(100) and h(100)+1 are COPRIMES. They share no factors other than 1.
h(100) = 2 * 4 * 6 *....* 94 * 96 * 98 * 100
Factoring out 2, we get:
h(100) = 2^50 (1 * 2 * 3 *... * 47 * 48 * 49 * 50)
Looking at the set of parentheses on the right, we can see that every prime number between 1 and 50 is a factor of h(100).
Since h(100) and h(100)+1 are coprimes, NONE of the prime numbers between 1 and 50 can be a factor of h(100)+1.
Thus, the smallest prime factor of h(100) + 1 must be greater than 50.
The correct answer is E.
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@Brent: Got your point. Can we not generalise it further to say.
For any K being divisor of N, K will not be divisor for N , N+1, N+2....N+(k-1).
Ex: say 105 is divisible by 5, so 106,107,108,019 will also be not divisible, so a period of (divisor-1). Pls confirm it.
Thanks
For any K being divisor of N, K will not be divisor for N , N+1, N+2....N+(k-1).
Ex: say 105 is divisible by 5, so 106,107,108,019 will also be not divisible, so a period of (divisor-1). Pls confirm it.
Thanks
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That's a great (and true) generalization. Nice work.ygdrasil24 wrote:@Brent: Got your point. Can we not generalise it further to say.
For any K being divisor of N, K will not be divisor for N , N+1, N+2....N+(k-1).
Ex: say 105 is divisible by 5, so 106,107,108,019 will also be not divisible, so a period of (divisor-1). Pls confirm it.
Thanks
Cheers,
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Hi Ygdrasil!ygdrasil24 wrote:@Brent: Got your point. Can we not generalise it further to say.
For any K being divisor of N, K will not be divisor for N , N+1, N+2....N+(k-1).
Ex: say 105 is divisible by 5, so 106,107,108,019 will also be not divisible, so a period of (divisor-1). Pls confirm it.
Thanks
Trivial note: this isn't true for any divisor of N, only for divisors of N not equal to 1.
A safer and broader way of stating your (excellent!) assertion is that if K is a factor of N, then K is a factor of N + R if and only if K is also a factor of R.
One way you could write this: say N = K*m and R = K*p, where m and p are integers. Then N + R = K*m + K*p = K * (m+p), so (N + R) is divisible by K.
Now say R is NOT divisible by K, so R = K*q + z, where q is an integer and z is a nonzero integer less than K. Then (N + R) = K*m + K*q + z = K*(m+q) + z ... which isn't divisible by K, because you have a remainder, namely z.
(If all that felt like alphabet soup, ignore it and go with the more intuitive explanation above!)