Question: Let N = 60! + 55! + 50! The unit digit of N and a number of digits to the left of the units digits are consecutive zeros before we come to the first non-zero digit. How many such consecutive zeros are there until the first non-zero digit?
A. 16
B. 15
C. 14
D. 12
E. 10
Solution:The number of consecutive zeros till the first non-zero digit is basically the number of trailing zeroes in a number
The number of trailing zeroes will be decided by the exponent of 10
For example, in the number 10100, there are 2 trailing zeroes since \(10100 = 101 * 10^2\)
Thus, in the given number, we need to determine the highest exponent of 10
Since 10 = 2 * 5 we need to determine the highest exponent of 2 and that of 5 and then choose the smaller of the two values. For example, if we have \(2^8 * 5^3\), the exponent of 10 will be 3 (i.e. \(10^3\)) => 3 trailing zeroes
We have:
\(60! + 55! + 50!\)
\(= 50!*51*52*53*54*...*60 + 50!*51*52*53*54*55 + 50!\)
\(= 50! * [51*52*...*60 + 51*52*...*55 + 1]\)
Let us check the part within [ ]:
\((51*52*...*60)\) is a multiple of 5; \((51*52*...*55)\) is a multiple of 5; but 1 is not a multiple of 5
Hence, the term within [ ] is NOT a multiple of 5
Also: (51*52*...*60) is even; (51*52*...*55) is even; but 1 is not even
Hence, the term within [ ] is NOT even
Thus, the required 2s and 5s are present only in 50!
Since the highest power of 2 in 50! will be much higher than that of 5 in 50!, we simply need to determine the highest power of 5 in 50!
The required power is: \([50/5] + [50/25] = 10 + 2 = 12\)
Answer D
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