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19 Jul 2022, 08:00
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B
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Difficulty:
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37% (02:03) correct
63%
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wrong
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If p is a prime number and p! has more than 100 but less than 1,000 positive odd factors, how many different values can p take ?
A. 1
B. 2
C. 3
D. 5
E. 7
A. 1
B. 2
C. 3
D. 5
E. 7
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Updated on: 20 Jul 2022, 11:53
Originally posted by ThatDudeKnows on 19 Jul 2022, 09:18.
Last edited by ThatDudeKnows on 20 Jul 2022, 11:53, edited 1 time in total.
Last edited by ThatDudeKnows on 20 Jul 2022, 11:53, edited 1 time in total.
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Bunuel
Number of odd factors is the product of (power+1) of each of the prime odd factors.
\(5! = 5*4*3*2*1 = 2^3*3*5\)
There are two prime odd factors, 3 and 5. 3 is to the first power and 5 is to the first power, so we have (1+1)*(1+1) = 2*2 = 4
\(7! = 7*6*above = 2^4*3^2*5*7\)
3*2*2 = 12
Lets start omitting the 2's since they don't matter.
\(11! = 11*10*9*8*above = 3^4*5^2*7*11\)
5*3*2*2 = 60
\(13! = 13*12*above = 3^5*5^2*7*11*13\)
6*3*2*2*2 = 144
\(17! = 17*16*15*14*above = 3^6*5^3*7^2*11*13*17\)
7*4*3*2*2*2 = 672
If we add 19^1, we will need to multiply 672 by at least 2, thereby exceeding 1000.
13! and 17! are our only options.
Answer choice B.
19 Jul 2022, 12:24
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As we need to find odd factors, we cannot include any power of 2 in the calculation.
We know that p is prime, so the total factors of p! will consists of the powers of all prime number greater than 2 and less than or equal to p.
Lets assume p = 13
Power of 3 in 13 ! = 5
Power of 5 in 13 ! = 2
Power of 7 in 13 ! = 1
Power of 11 in 13 ! = 1
Power of 13 in 13 ! = 1
Total number of odd factors = 6 * 3 * 2 * 2 * 2 = 144. Hence 13 ! is vaild
Lets assume p = 17
Power of 3 in 17 ! = 6
Power of 5 in 17 ! = 3
Power of 7 in 17 ! = 2
Power of 11 in 17 ! = 1
Power of 13 in 17 ! = 1
Power of 17 in 17 ! = 1
Total number of odd factors = 7 * 4 * 3 * 2 * 2 * 2 lies in the range. Hence valid.
From 19! onwards the range exceeds. Hence only valid values are 13 & 17
IMO B
We know that p is prime, so the total factors of p! will consists of the powers of all prime number greater than 2 and less than or equal to p.
Lets assume p = 13
Power of 3 in 13 ! = 5
Power of 5 in 13 ! = 2
Power of 7 in 13 ! = 1
Power of 11 in 13 ! = 1
Power of 13 in 13 ! = 1
Total number of odd factors = 6 * 3 * 2 * 2 * 2 = 144. Hence 13 ! is vaild
Lets assume p = 17
Power of 3 in 17 ! = 6
Power of 5 in 17 ! = 3
Power of 7 in 17 ! = 2
Power of 11 in 17 ! = 1
Power of 13 in 17 ! = 1
Power of 17 in 17 ! = 1
Total number of odd factors = 7 * 4 * 3 * 2 * 2 * 2 lies in the range. Hence valid.
From 19! onwards the range exceeds. Hence only valid values are 13 & 17
IMO B
