Difference between revisions of "2017 AMC 8 Problems/Problem 19"
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− | ==Problem | + | ==Problem== |
For any positive integer <math>M</math>, the notation <math>M!</math> denotes the product of the integers <math>1</math> through <math>M</math>. What is the largest integer <math>n</math> for which <math>5^n</math> is a factor of the sum <math>98!+99!+100!</math> ? | For any positive integer <math>M</math>, the notation <math>M!</math> denotes the product of the integers <math>1</math> through <math>M</math>. What is the largest integer <math>n</math> for which <math>5^n</math> is a factor of the sum <math>98!+99!+100!</math> ? | ||
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==Solution 1== | ==Solution 1== | ||
− | Factoring out <math>98!</math>, we have <math>98!( | + | Factoring out <math>98!+99!+100!</math>, we have <math>98!(1+99+99*100)</math> which is <math>98!(10000)</math> Next, <math>98!</math> has <math>\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{98}{25}\right\rfloor = 19 + 3 = 22</math> factors of <math>5</math>. The <math>19</math> is because of all the multiples of <math>5</math>. Now <math>10,000</math> has <math>4</math> factors of <math>5</math>, so there are a total of <math>22 + 4 = \boxed{\textbf{(D)}\ 26}</math> factors of <math>5</math>. |
==Solution 2== | ==Solution 2== | ||
The number of <math>5</math>'s in the factorization of <math>98! + 99! + 100!</math> is the same as the number of trailing zeroes. The number of zeroes is taken by the floor value of each number divided by <math>5</math>, until you can't divide by <math>5</math> anymore. Factorizing <math>98! + 99! + 100!</math>, you get <math>98!(1+99+9900)=98!(10000)</math>. To find the number of trailing zeroes in 98!, we do <math>\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{19}{5}\right\rfloor= 19 + 3=22</math>. Now since <math>10000</math> has 4 zeroes, we add <math>22 + 4</math> to get <math>\boxed{\textbf{(D)}\ 26}</math> factors of <math>5</math>. | The number of <math>5</math>'s in the factorization of <math>98! + 99! + 100!</math> is the same as the number of trailing zeroes. The number of zeroes is taken by the floor value of each number divided by <math>5</math>, until you can't divide by <math>5</math> anymore. Factorizing <math>98! + 99! + 100!</math>, you get <math>98!(1+99+9900)=98!(10000)</math>. To find the number of trailing zeroes in 98!, we do <math>\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{19}{5}\right\rfloor= 19 + 3=22</math>. Now since <math>10000</math> has 4 zeroes, we add <math>22 + 4</math> to get <math>\boxed{\textbf{(D)}\ 26}</math> factors of <math>5</math>. | ||
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+ | ==Solution 3== | ||
+ | We can rewrite the expression as <math>98!+99!+100!=98!(1+99+99\cdot100)=98!(100+99\cdot100)=98!\cdot10,000</math>. | ||
+ | The exponent of <math>5</math> in <math>10,000</math> is <math>4</math>. Onto the <math>98!</math> part. | ||
+ | |||
+ | Remember that <math>98!</math> is the product of the integers from 1 to 98. Among these, there are multiples of <math>5</math> and multiples of <math>25</math>. | ||
+ | The number of multiples of <math>5</math> below or equal to <math>98</math> is <math>\left\lfloor\frac{98}{5}\right\rfloor</math> (try to see why), which is <math>19</math>. Every such number has a factor of <math>5</math>, so they contribute <math>1</math> to the total each. So these numbers contribute <math>19</math> to the exponent of <math>5</math> in <math>98!</math>. | ||
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+ | However, we forgot about multiples of <math>25</math>. Using similar logic, we have <math>\left\lfloor\frac{98}{25}\right\rfloor=3</math>, so there are 3 multiples of 25 in this range. Multiples of 25 contribute 2 to the total each. We already counted 1 of 2 contributions while counting multiples of 5. So we need to add another 1 to the exponent of 5 for every multiple of 25. So these numbers contribute 3 to the exponent of 5 in <math>98!</math>. | ||
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+ | We thus have that the exponent of <math>5</math> in <math>98!</math> is <math>19 + 3 = 22</math>, and so our answer is <math>22 + 4 = 26 \longrightarrow \boxed{(D)26}</math>. | ||
+ | |||
+ | ~Math4Life2020 | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/HISL2-N5NVg?t=817 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==See Also== | ==See Also== |
Latest revision as of 22:18, 27 January 2021
Problem
For any positive integer , the notation denotes the product of the integers through . What is the largest integer for which is a factor of the sum ?
Solution 1
Factoring out , we have which is Next, has factors of . The is because of all the multiples of . Now has factors of , so there are a total of factors of .
Solution 2
The number of 's in the factorization of is the same as the number of trailing zeroes. The number of zeroes is taken by the floor value of each number divided by , until you can't divide by anymore. Factorizing , you get . To find the number of trailing zeroes in 98!, we do . Now since has 4 zeroes, we add to get factors of .
Solution 3
We can rewrite the expression as . The exponent of in is . Onto the part.
Remember that is the product of the integers from 1 to 98. Among these, there are multiples of and multiples of . The number of multiples of below or equal to is (try to see why), which is . Every such number has a factor of , so they contribute to the total each. So these numbers contribute to the exponent of in .
However, we forgot about multiples of . Using similar logic, we have , so there are 3 multiples of 25 in this range. Multiples of 25 contribute 2 to the total each. We already counted 1 of 2 contributions while counting multiples of 5. So we need to add another 1 to the exponent of 5 for every multiple of 25. So these numbers contribute 3 to the exponent of 5 in .
We thus have that the exponent of in is , and so our answer is .
~Math4Life2020
Video Solution
https://youtu.be/HISL2-N5NVg?t=817
~ pi_is_3.14
See Also
2017 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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All AJHSME/AMC 8 Problems and Solutions |
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