Question

A = 11 * 22 * 33 * 44 * 55 * ……..1010. How many zeroes will be there at the end of A ?

Answer 1

Is the last number  110 or  1001  ?      1010 is not divisible by 11.

1)    A = 11 * 22 * 33 * 44 * 55 * ..... 99 * 110
       A =  11¹⁰ * 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10
  A zero at the end of  A appear  as many as tens in the product  PLUS the number of 2 X 5 products.   So we have  one  5   and  one 10.
   ===>  So  two zeroes for this product.

2)   A = 11*22*33*44*55 ....990 * 1001
      A = 11⁹¹ * 1 * 2 * 3 * 4 * 5 * ...90 * 91

  Remove factors except for 5's and 10's. There are many 2's in the product.

     A' =  5 * 15*25*35...*85 * 10*20*30*....*80*90
    There are Nine  5's and  Nine  10's.
   So there will be 18 zeros in this product.

Answer 2

Number of zeros in last 11*22*33*...*1010  = 2

Given Expression is :

• 11 * 22 * 33 * ... * 1010

To Find:

• Number of zeroes in last

Write Each number as multiple of 2 and 5 where ever possible

• 11  = 11 x 1
• 22 = 11 x 2
• 33 = 11 x 3
• 44 = 11 x 2 x 2
• 55 = 11 x 5
• 66 = 11 x 2 x 3
• 77 = 11  x 7
• 88 = 11 x 2 x 2 x 2
• 99 = 11 x 3 x 3

• 1010 = 101 x 2 x 5

Number of 5 as factors = 2

Number of 2 as factors are much more than 2

Hence number of zeroes at end  is  2   ( as 5 x 2 = 10)

number of zeros in last 11*22*33*...*1010  = 2