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Let N = 60! + 55! + 50! The unit digit of N and a number of digits to  [#permalink]

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Question Stats: 48% (01:30) correct 52% (01:56) wrong based on 44 sessions

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Competition Mode 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

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Re: Let N = 60! + 55! + 50! The unit digit of N and a number of digits to  [#permalink]

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Quote:
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

N = 60! + 55! + 50!

Number of zeros in any number = Power of 10 in that Number

but 10 = 2*5
i.e. Power of 10 in any number = Pairs of 2 and 5 in that number

i.e. To find the number of zeros in 60!, 55! and 50! we need to find the powers of 2 and 5 in those numbers

Also, Power of 2 in any factorial function > Power of 5 in that factorial function
i.e. Total possible pairs of 2 and 5 in any factorial = Power of 5 in that factorial

CONCEPT: Power of any Prime Number in any factorial can be calculated by following understanding  Power of prime x in n! = [n/x] + [n/x^2] + [n/x^3] + [n/x^4] + ... and so on
Where,
[n/x] = No. of Integers that are multiple of x from 1 to n
[n/x^2] = No. of Integers that are multiple of x^2 from 1 to n whose first power has been counted in previous step and second is being counted at this step
[n/x^3] = No. of Integers that are multiple of x^3 from 1 to n whose first two powers have been counted in previous two step and third power is counted at this step
And so on.....

For more clarity on this concept please refer to the following post
https://gmatclub.com/forum/how-many-tra ... l#p2446819

i.e. Power of 5 (Number of zeros) in 60! = [60/5] + [60/5^2] + [60/5^3] + ... and so on  = 12 + 2 = 14
i.e. Power of 5 (Number of zeros) in 55! = [55/5] + [55/5^2] + [55/5^3] + ... and so on  = 11 + 2 = 13
i.e. Power of 5 (Number of zeros) in 50! = [50/5] + [50/5^2] + [50/5^3] + ... and so on  = 10 + 2 = 12

i.e. Number of Zeros in 60!+55!+50! = Number with 14 zeros + Number with 13 zeros + number with 12 zeros
i.e. Number of Zeros in 60!+55!+50! = 12 zeros

CONCEPT VIDEO TO FIND NUMBER OF ZEROS IN ANY FACTORIAL

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Re: Let N = 60! + 55! + 50! The unit digit of N and a number of digits to  [#permalink]

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Let N = 60! + 55! + 50!
—> How many such consecutive zeros are there until the first non-zero digit?

50!— smallest one among them.
$$[\frac{50}{5}] +[ \frac{50}{5^{2}}] + ... = 10 + 2 = 12$$

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Re: Let N = 60! + 55! + 50! The unit digit of N and a number of digits to  [#permalink]

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Quote:
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

The number of consecutive zeros to the left of the units digit is the number of trailing zeros;
The number of trailing zeros of a factorial is the sum of the quotients of the factorial's integer divided by 5:
We must find trailing zeros for 50!, because its smaller than 60! and 55!, thus the first non-zero digit will appear first.

50/5=10; 50/25=2; trailing zeros=12

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Re: Let N = 60! + 55! + 50! The unit digit of N and a number of digits to  [#permalink]

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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: $$[50/5] + [50/25] = 10 + 2 = 12$$

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Re: Let N = 60! + 55! + 50! The unit digit of N and a number of digits to  [#permalink]

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N = 50!*(60*59*58...*51+ 55*54*...*51+1)
50! ; 50/5 + 50/25 ; 12 zeroes
IMO D ; 12

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
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Re: Let N = 60! + 55! + 50! The unit digit of N and a number of digits to  [#permalink]

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60!, 55!, and 50! have 14, 13, and 12 factors of 5, respectively. Of course, there is more than sufficient number of factor of 2 for each factor of 5 in any of 60!, 55!, and 50!.

The lowest number of factor of 5 (which is 12) will govern the number of consecutive zeros (which is also 12) until the first non-zero digit.

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Re: Let N = 60! + 55! + 50! The unit digit of N and a number of digits to  [#permalink]

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Individually,
50! has 50/5 + 50/25 = 12 trailing zeros.
55! has 55/5 + 55/25 = 13 trailing zeros.
60! has 60/5 + 60/25 = 14 trailing zeros.
We can assume that the non-zero digit before the trailing zeros for 50! is a, while that for 55! is b and that for 60! is c.
Then N can be written in terms of the first non-zero digit before the trailing zeros as a*10^12 + b*10^13 + c*10^14 = a*10^12 + b0*10^12 + c00*10^12 = 10^12(a + b0 + c00)= xya * 10^12
where x and y are digits from 0 to 9.
So clearly, the number if trailing zeros for N=60!+55!+50! is 12.

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Re: Let N = 60! + 55! + 50! The unit digit of N and a number of digits to  [#permalink]

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N = 60! + 55! + 50!
We can take 60! = 50!*a, for some positive integer a
& 55! = 50!*b for some positive integer b

--> N = 50!*a + 50!*b + 50!
--> N = 50!(a + b + 1) = 50!*c, for some positive integer c [Note that unit digit of c is a non zero digit as c = a+b+1]

So, number of consecutive zeros from unit digit till a non-zero digit in N is same number of consecutive zeros from unit digit till a non-zero digit of 50!
--> Number of zero digits in 50! = [50/5] + [50/5^2] =  +  = 10 + 2 = 12

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Re: Let N = 60! + 55! + 50! The unit digit of N and a number of digits to  [#permalink]

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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
In each term of N = 60! + 55! + 50!
zeros are obtained from 2*5, so the number of 5(always lesser than that of 2) decides the number of consecutive zeros before which first nonzero digit is arrived at.

So,
Number of zeros = minimum(14, 13, 12) = 12 as clearly can be seen from the expression below where the first non zero digit is 1 which comes after 12 zeros from expression 50!
N = 50!(60* ...* 51 + 55*...*51 + 1)

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Re: Let N = 60! + 55! + 50! The unit digit of N and a number of digits to  [#permalink]

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The least of the sum is 50!....

No.of zeros in 50! Is ....50/5= 10
10/5=2

Total=12

OA:D

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Re: Let N = 60! + 55! + 50! The unit digit of N and a number of digits to  [#permalink]

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Bunuel wrote:

Competition Mode 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

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We can rewrite N as 50!(60P10 + 55P5 + 1). We see that the units digit of 60P10 + 55P5 + 1 is 1 since the first two addends have units digits of 0 (notice that 60P10 has 60 as a factor and 55P5 has 55 and 54 as factors). Therefore, the number of trailing zeros of N is the same as the number of trailing zeros of 50!. Recall that the number of trailing zeros in a factorial is determined by the number of 5-and-2 pairs that occur in the factorial. Since the number of 5s is less than the number of 2s in 50!, we see that the number of 5-and-2 pairs is limited by the number of 5s that occur in 50!.

From 10, 20, 30, 40, and 50, we obtain 6 factors of 5. From 5, 15, 25, 35, and 45, we obtain 6 additional factors of 5. The total number of factors of 5 in 50! is, therefore, 6 + 6 = 12. Thus, we have twelve 5-and-2 pairs, and this indicates that there are 12 trailing zeros until the first non-zero digit.

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