(a) Consider the integers (1,..., 10). How many times does 2 occur as a factor o

Question

(a) Consider the integers (1,..., 10). How many times does 2 occur as a factor of one of these numbers? (b) Consider the integers (1,... , 10). How many times does 5 occur as a factor of one of these numbers? How many zeroes appear at the end of 10!? onsider the integers (1,.. . 100). How many times does 5 occur as a factor of of these numbers? sed on the intuition gained from parts (a)-(c) and your answer to part (d), can you predict how many zeroes will appear at the end of 100! without computing the answer directly? (f) Compute or look up the value of 100! (You may be able to compute this on a computer, or look it up on the internet or in a reference book.) Was yourT prediction in part (e) correct? If not, can you describe your error?

Explanation / Answer

[a) The simplest measure to calculate that how many times 2 appears in integers {1,,,,,10}

is to divide 10 by 2 i.e 10/2 = 5 Times (we took 10 as it is the last multiple of 2 in series ,if the series was 1 to 9 we would have taken 8 as it is the next last multiple of 2 )

lets check manually 2,4,6,8,10 are all having 2 as a factor

b) Similarly we can divide 10 (being a nearest multiple of 5 ) by 5 to arrive at number with factors of 5 = (10/2)=2 in this case

or 5 & 10

c) Trailing Zeroes appearing in 10! can be calculated by using above two explanations

a zero appears when it has a factor of 10 that is any time a number is multiplied by 10 it gets a trailing 0

for example 10 *2 =20 there is one zero multiplied and we have one zero in trail in result

10 * 10 = 100 or two zeros in trail.

and by a ) and b) above we know 10 is a multiple of 5 * 2

But we have more numbers that are multiples of 2 than there are multiples of 5. So if we take numbers with 5 as a factor, we will have sufficient even numbers to pair with them to get factors of 10 . So to find the number of times 10 is a factor, we need to find how many times 5 is a factor in all integers between 1 to 10.

we have 5 and 10 as factor so we will have 2 zeroes trailling in10!

(proof 10! = 36,28,800 has 2 zeroes)

d) Similarly to find trailling zeros in 100!

we will calculate multipules of 5 in 1 to 100 = 20 (i.e 100/5 by using a and b )

additionally 25 (5 x 5 ) has an extra factor of 5 that we need to account for

so we have 100/25 = 4 multiples of 25 (5 X 5)

total factors = 20 + 4 = 24

so 100! will have 24 trailing zeroes in the end

e) on using computer for calculating 100! we found our answer of 24 zeroes to be correct

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