Without computing the value of 100 determine how many zeros

Without computing the value of 100!, determine how many zeros are at the end of this number when it is written in decimal form. Justify your answer. Find all solutions to m^2 - n^2 = 105, for which both m and n are integers.

Solution

1)

as given we need to find the number of zeros at the end of 100! when written n decimal form.

First find the number of multiples of 5, 5² , 5³ , 5⁴ , 5⁵ , . . . . between number 1 to 100 and and them which gives the number of 0 in 100!

100/5 = 20

100/5² = 100/25 = 4

100/5³ = 100/125 = 0

so we can say that the number of zeros at the end of 100! is 20 + 4 + 0 = 24

2)

we need to find all the solution of m² - n² = 105 for which both m and n are integers.

we know that we can write m² - n² = 105 as (m-n)(m+n) = 105

now factorized 105 we can write

105 = 105 *1 = 21*5 = 35*3 = 15*7

so we can say that,

take the first case 105 = 105 *1

(m+n)(m-n) = 105 * 1

hence m+n = 105 and m-n =1

put n = -1+m in m+n=105 we have,m+m-1 = 105 so m =106 and m = 53

n = -1+m = -1 + 53 = 52

we can say that m = 53 and n = 52 is one solution as 53² - 52² = 105

take the second case 105 = 21*5

(m+n)(m-n) = 21*5

hence m+n = 21 and m-n =5

put n = -5+m in m+n=21 we have,m+m-5 = 21 so 2m =26 and m = 13

n = -5+m = -5 + 13 = 8

we can say that m = 13 and n = 8 is other a solution as 13² - 8² = 105

take the third case 105 = 15*7

(m+n)(m-n) = 15*7

hence m+n = 15 and m-n =7

put n = -7+m in m+n=15 we have,m+m-7 = 15 so 2m =22 and m = 11

n = -7+m = -7 + 11 = 4

we can say that m = 11 and n = 4 is other solution as 11² - 4² = 105

take the fourth case 105 = 35*3

(m+n)(m-n) = 35*3

hence m+n = 35 and m-n =3

put n = -3+m in m+n=35 we have,m+m-3 = 35 so 2m =38 and m = 19

n = -3+m = -3 + 19 = 16

we can say that m = 19 and n = 16 is other a solution as 19² - 16² = 105