If you write 121! as a product of its prime factors, how many times will 11 appear? (note its 121!)?

Question

If you write 121! as a product of its prime factors, how many times will 11 appear? (note its 121!)?

Answer 1

since 121 is 11 x 11, then the number 11 can only appear

in the multiples of 11 as you go down from 121 to 1.



there are 11 multiples of 11 plus an extra 11 because 121 has two 11's in it.



so 11 + 1 = 12 times



sorry john, i can't figure out what the hell you did!

Answer 2

factor(121!)

2^116*3^58*5^28*7^19*11^12*13^9*17^7*

19^6*23^5*29^4*31^3*37^3*41^2*

43^2*47^2*53^2*59^2*61*67*71

*73*79*83*89*97*101*103*107*109*113



8.0942985252734*10^200 is the number, by the way



So 12 times.



Microsoft Math has "bignum" arithmetic and a factor verb and a factorial verb, so I just computed it.



I'm sure that there is an easy way of working it out - under 121, how many multiples of 11 are there?



Cause that seems to be the answer. Plus one for the square, of course.

Answer 3

121! = 1 * 2 * ... * 11 * ... * 22 ... * 33 ... * 44 ... * 55 ... * 66 ... * 77 ... * 88 ... * 99 ... * 110 ... * 121



121! = 1 * 2 * ... * 11 * ... * 2*11 ... * 3*11 ... * 4*11 ... * 5*11 ... * 6*11 ... * 7*11 ... * 8*11 ... * 9*11 ... * 10*11 ... * 11*11



Now we count the number of times 11 appears:

1+1+1+1+1+1+1+1+1+1+2 = 12



12 TIMES



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Here's a way to find how many times any prime factor appears in a factorial:



How many times does 3 appear:

121/3 = 40.333 ----> 40 numbers are divisible by 3

121/3² = 13.444 ---> 13 numbers are divisible by 3²

121/3³ = 4.481 -----> 4 numbers are divisible by 3³

121/3⁴ = 1.494 -----> 1 number is divisible by 3⁴



(Last number (1) is < 3 - do not proceed any further)



Number of times prime factor 3 appears: 40+13+4+1 = 58



NOTE: Do not add numbers divisible by 3² twice, since we've already counted one of the factors of 3 with numbers divisible by 3. Do not add numbers divisible by 3³ three times, since we've already counted two of the factors of 3 with numbers divisible by 3 and numbers divisible by 3², ....



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How many times does 5 appear?



121/5 = 24.2 ----> 24

121/5² = 4.84 ---> 4 (always truncate, never round)



(4.84 < 5 ---> stop)



24 + 4 = 28



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So using 11, we get:

121/11 = 11

121/11² = 1



11 + 1 = 12





-- Ματπmφm --

Answer 4

121! = 2^116 × 3^58 × 5^28 × 7^19 × 11^12 × 13^9 × 17^7 × 19^6 × 23^5 × 29^4 × 31^3 × 37^3 × 41^2 × 43^2 × 47^2 × 53^2 × 59^2 × 61 × 67 × 71 × 73 × 79 × 83 × 89 × 97 × 101 × 103 × 107 × 109 × 113



12