find number of zeroes and sum of digits in
3^4 x 2^17 x 5^16​

Question

find number of zeroes and sum of digits in
3^4 x 2^17 x 5^16​

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Natalia 1 year 2021-07-17T14:56:11+00:00 1 Answers 0 views 0

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    2021-07-17T14:58:02+00:00

    Answer:

    We can form a total of 4! or 24 numbers. When we add all these numbers, let us look at the contribution of the digit 2 to the sum.

    When 2 occurs in the thousands place in a particular number, its contribution to the total will be 2000. The number of numbers that can be formed with 2 in the thousands place is 3!, i.e., 6 numbers. Hence, when 2 in the thousands place its contribution to the sum is 3! x 2000.

    Similarly, when 2 occurs in the hundreds place in a particular number, its contribution to the total will be 200 and since there are 3! Numbers with 2 in the hundreds place, the contribution 2 makes to the sum when it comes in the hundreds place is 3! x 200.

    Again, when 2 occurs in the tens and units place respectively, its contribution to the sum is 3! x 20 and 3! x 2 respectively. Hence, the total contribution of 2 to the sum is 3! (2000+200+20+2) i.e. 3! x 2222. This takes care of the digit 2 completely.

    In a similar manner, the contribution of the 3,4, and 5 to the sum will respectively be 3! x 3333, 3! x 4444, 3! x 5555, i.e.,

    3! x (2+3+4+5) x 1111

    We can now generalize the above as

    “If all the possible n-digit numbers using n distinct digits are formed, the sum of all the numbers so formed is equal to (n-1)! x {sum of all the digits} x {111…….} n times.

    E.g:

    Example 1: What would be the sum of all the numbers which can be formed by using the digits 1, 3, 5, 7 all at a time and which have no digits repeated?

    Sol: Sum of the numbers formed by taking all the given n digits is ( sum of all the n digits ) x (n-1) ! x (111…..n times).

    Here n = 4. and Sum of 4 digits = 16

    The sum of all the numbers which can be formed by using the digits 1, 3, 5, 7 is

    = ( 16 ) x ( 4 – 1)! x ( 1111)

    = 16 x 3! x 1111.

    hope it’s helpful

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