How about 45360 = 2^4 * 3^4 * 5 * 7
# factors = (4+1)(4+1)(1+1)(1+1) = 5*5*2*2 = 100
The number of factors of the number:
x = p(1)^r(1) * p(2)^r(2) * p(3)^r(3) * ... * p(n)^r(n), where this is the prime factorization, is:
(r(1) + 1)(r(2) + 1)(r(3) + 1)...(r(n) + 1)
The question did not ask for just prime factors but all factors.
For instance, if 2 and 3 are factors then so is 6.
45360 has exactly 100 factors.
They are:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 27, 28, 30, 35, 36, 40, 42, 45, 48, 54, 56, 60, 63, 70, 72, 80, 81, 84, 90, 105, 108, 112, 120, 126, 135, 140, 144, 162, 168, 180, 189, 210, 216, 240, 252, 270, 280, 315, 324, 336, 360, 378, 405, 420, 432, 504, 540, 560, 567, 630, 648, 720, 756, 810, 840, 945, 1008, 1080, 1134, 1260, 1296, 1512, 1620, 1680, 1890, 2160, 2268, 2520, 2835, 3024, 3240, 3780, 4536, 5040, 5670, 6480, 7560, 9072, 11340, 15120, 22680, 45360
2006-11-09 07:05:08
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answer #1
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answered by Scott R 6
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1^100= 1
or the center number on step 101 of Pascals Triangle
TRIP ON THIS
Step 1 = 1
Step 2 = 1 1
Step 3 = 1 2 1 starting step 3 and only for the odd numbered steps, subtract the step number -1 to get the number of factors for the middle number 2 = 2*1 = 3-1
Step 4 = 1 3 3 1
Step 5 = 1 4 6 4 1 [6 = 1*6, 2*3 ( 4 factors ) = 5-1]
Step 6 = 1 5 10 10 5 1
Step 7 = 1 6 15 20 15 6 1 [20 = 1*20, 10*2, 4*5 (6 factors) = 7-1]
Step 8 = 1 7 21 35 35 21 7 1
Step 9 = 1 8 28 56 70 56 28 8 1 [70= 1*70, 7*2*5, 2*35, 14*5, 7*10 (8 factors 1,2,5,7,10,14,35,70) = 9-1]
or in case repeat 1's and Pascal don't count.
1 + 2^98 = x
316912650057057350374175801345 =
1, 2 (98 times) and 316912650057057350374175801345
2006-11-09 14:55:13
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answer #3
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answered by bourqueno77 4
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