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Re M3201
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21 May 2016, 03:21
Official Solution:How many trailing zeroes does \(49! + 50!\) have?A. 9 B. 10 C. 11 D. 12 E. 22 Factor out 49! from the expression: \(49! + 50!=49!(1+50)=49!*51\). 51 won't contribute to the number of zeros at the end of the number, therefore all zeros will come from 49!. Trailing zeros in 49!: \(\frac{49}{5}+\frac{49}{5^2}=9+1=10\) THEORY: Trailing zeros: Trailing zeros are a sequence of 0's in the decimal representation of a number, after which no other digits follow. For example 125,000 has 3 trailing zeros; The number of trailing zeros n!, the factorial of a nonnegative integer \(n\), can be determined with this formula: \(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where \(k\) must be chosen such that \(5^{(k+1)} \gt n\) It's easier if we consider an example: How many zeros are in the end (after which no other digits follow) of 32!? \(\frac{32}{5}+\frac{32}{5^2}=6+1=7\). Notice that the last denominator (\(5^2\)) must be less than 32. Also notice that we take into account only the quotient of the division, that is \(\frac{32}{5}=6\). So there are 7 zeros in the end of 32!. Another example, how many trailing zeros does 125! have? \(\frac{125}{5}+\frac{125}{5^2}+\frac{125}{5^3}=25+5+1=31\), The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero. Answer: B
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Re: M3201
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28 May 2016, 09:10
Hello, Please help me out in understanding why the answer is 10 and not 11. 49! has 10 trailing zeros OK. But 49!(10*5+1)? doesn't the 10 in the expression add an extra zero to the trailing zeros of 49! ?



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Re: M3201
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28 May 2016, 09:19
Avigano wrote: Hello, Please help me out in understanding why the answer is 10 and not 11. 49! has 10 trailing zeros OK. But 49!(10*5+1)? doesn't the 10 in the expression add an extra zero to the trailing zeros of 49! ? Hi, no, 49!(10*5+1) = 49!(50+1) = 49!*51... Now 51 does not contain any 5 in it, so the term depends on ONLY 49! and as you have said 49! has 10 5s so ans is 10
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Re: M3201
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28 May 2016, 09:38
chetan2u wrote: Avigano wrote: Hello, Please help me out in understanding why the answer is 10 and not 11. 49! has 10 trailing zeros OK. But 49!(10*5+1)? doesn't the 10 in the expression add an extra zero to the trailing zeros of 49! ? Hi, no, 49!(10*5+1) = 49!(50+1) = 49!*51... Now 51 does not contain any 5 in it, so the term depends on ONLY 49! and as you have said 49! has 10 5s so ans is 10 Ok! got it! thanks



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29 Dec 2016, 13:13
chetan2u wrote: Avigano wrote: Hello, Please help me out in understanding why the answer is 10 and not 11. 49! has 10 trailing zeros OK. But 49!(10*5+1)? doesn't the 10 in the expression add an extra zero to the trailing zeros of 49! ? Hi, no, 49!(10*5+1) = 49!(50+1) = 49!*51... Now 51 does not contain any 5 in it, so the term depends on ONLY 49! and as you have said 49! has 10 5s so ans is 10 Hi All  I don't 100% understand the rationale behind 51 not contributing to the number of trailing zeros. Does 51 not contain 10^1? Thanks.



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Re: M3201
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29 Dec 2016, 21:28
Dear Bunuel,
I know the concept and deduced the final step 49! * 51. But 49/5 = 9.8 ~ 10 and 49/25 =1.96 ~ 2 .So I marked the answer as 12 instead of 10.Please clarify,if we need to select the lowest rounded integer or do we need to consider the nearest rounded integer to answer these kind of questions.



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Re: M3201
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04 Jan 2017, 13:55
chetan2u wrote: Avigano wrote: Hello, Please help me out in understanding why the answer is 10 and not 11. 49! has 10 trailing zeros OK. But 49!(10*5+1)? doesn't the 10 in the expression add an extra zero to the trailing zeros of 49! ? Hi, no, 49!(10*5+1) = 49!(50+1) = 49!*51... Now 51 does not contain any 5 in it, so the term depends on ONLY 49! and as you have said 49! has 10 5s so ans is 10 Hi. How does 51 not contain any 5s?



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Re M3201
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20 Jan 2017, 09:16
I think this is a highquality question.



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Re: M3201
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17 Mar 2017, 21:05
Sorry i have a question. what is wrong if you factor 49 and 50 factorial and add?? So you will get 22



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Re: M3201
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15 Jul 2017, 12:08
Hi, I solved with the trailing zero method only but I calculated for 49! and 50! separately and I got 22 as answer. I want to know how that is a wrong method of solving. Thanks!
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30 Sep 2017, 09:04
This was confusing for me too, but the logic lies in the fact that 49! is a factorial. 1x2x3x4....40x41x...49. vs. 51 which is just 17x3. There are no 5s in 51. The formula allows you to count how many 5s are in factorials. If it was 51! (factorial) then yes it would have 12 fives, but the answer is 10 because 51 is NOT a factorial in this problem (after you simplify it.)



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Re M3201
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04 May 2018, 21:48
I think this is a highquality question and I agree with explanation. This about the 51 that is left out. When we multiply 51 with 10^n. The answer always is giving (n1) trailing zeros. Example, 10*51=51 (1vs0),100*51=510 (2vs1), 1000*51=5100 (3vs2). So i got 10 Trailing zeros for 49! but since it is being multiplied by 51 I assumed it would remove 1 zero from the total and give 9 trailing zeros. Please let me know where i have done a mistake.



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Re: M3201
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04 May 2018, 23:16
kittyman wrote: I think this is a highquality question and I agree with explanation. This about the 51 that is left out. When we multiply 51 with 10^n. The answer always is giving (n1) trailing zeros. Example, 10*51=51 (1vs0),100*51=510 (2vs1), 1000*51=5100 (3vs2). So i got 10 Trailing zeros for 49! but since it is being multiplied by 51 I assumed it would remove 1 zero from the total and give 9 trailing zeros. Please let me know where i have done a mistake. Not sure what are you doing there. 10*51 = 510 not 51. 100*51 = 5,100 no 510. 1,000*51 = 51,000 no 5,100. 10^n, where n is a positive integer, will have n trailing zeros.
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Hello, I solved this question in slightly different manner. To find the number of trailing zeroes in 49!we know if we write 49! as product from 49 to 1 like 49*48*47*...*1, each 10(or 2*5) in this product contributes a trailing zero So to find the number of 10's in 49! we need to find how many (2*5) pairs in 49! expansion We know since each even number contributes a 2 there are more 2's than 5's in the product, so 5's are the limiting factor. Hence we need to find number of 5's in the product. So counting the number of 5's in 49! expansion Integer  Number of 5's in that integer5 (5*1) > 1 10 (5*2) > 1 15 (5*3) > 1 20 (5*4) > 1 25 (5*5) > 2 30 (5*6) > 1 35 (5*7) > 1 40 (5*8) > 1 45 (5*9) > 1 Total 10
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Re: M3201
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18 Jul 2018, 00:18
I am really trying to understand the logic behind taking 49! common this is how i solved.' 49! : 49!/5 +49!/25 =9+1= 10 trailing zeroes 50!: 50!/5 + 50!/25 = 10+2= 12 trailing zeroes total 22. Please explain why solving 50! the way I did is incorrect. What it it was only 50! in the question? then the answer would have been 12? Also just another hypothetical question. Say it was 49!+ 60! then how would you have solved?
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