Question:
Find the number of zeros at the end of
P = 35! + 36! + 37! + 38! + 39! + 40!
Find the number of zeros at the end of
P = 35! + 36! + 37! + 38! + 39! + 40!
P = 35! + 36! + 37! + 38! + 39! + 40!
Updated On: Sep 10, 2024
- 8
- 9
- 10
- 11
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The Correct Option is A
Solution and Explanation
35! + 36! + 37! + 38! + 39! + 40! = 35! (1 + 36 + 36 × 37 + 36 × 37 × 38 + 36 × 37 × 38 × 39 + 36 × 37 × 38 × 39 × 40)
Here 35! has \(\frac{35}{5}+\frac{35}{25}=7+1\) = 8 zeroes
Now, the sum in brackets gives an odd number.
For the value in the bracket to be a multiple of 5, the last digit must be 5.
So the last digit = 1 + 6 + 2 + 6 + 4 + 0 = 19
As the last digit is 9, the value in the bracket does not have any multiple of 5.
So, there are only 8 zeros at the end of the 35! + 36! + 37! + 38! + 39! + 40!
So, the correct option is (A) : 8.
Here 35! has \(\frac{35}{5}+\frac{35}{25}=7+1\) = 8 zeroes
Now, the sum in brackets gives an odd number.
For the value in the bracket to be a multiple of 5, the last digit must be 5.
So the last digit = 1 + 6 + 2 + 6 + 4 + 0 = 19
As the last digit is 9, the value in the bracket does not have any multiple of 5.
So, there are only 8 zeros at the end of the 35! + 36! + 37! + 38! + 39! + 40!
So, the correct option is (A) : 8.
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