2017 AMC 8 Problem 19

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To help you master 2017 AMC 8 Problem 19, this page includes:

The original AMC problem to try, plus the correct answer

Clear, step-by-step solutions using different methods

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2017 AMC8 Problem 19

Solution

Solution 1

Common Factor Method

Extract the smallest factorial from the sum to turn the addition into a single multiplication, then count the factors of 5 in each part of the resulting product.

Legendre’s Formula tells us how many times a prime number $p$ divides a factorial $n!$ . You find this by dividing $n$ by $p$ , then by $p^{2}$ , then by $p^{3}$ , and so on, rounding down to the nearest whole number each time. Adding these results together gives the total number of factors of the prime.

Step 1: Express larger factorials using the smallest factorial

To factor the sum, we first express the larger factorials in terms of the smallest factorial in the expression, which is 98!.

99! = 99 \times 98!

100! = 100 \times 99 \times 98! = 9900 \times 98!

Step 2: Factor out the smallest factorial

Now, substitute these expressions back into the original sum. Because every term contains 98!, we can factor it out.

98! + 99! + 100! = 98! + 99 \times 98! + 9900 \times 98!

98! + 99 \times 98! + 9900 \times 98! = 98! (1 + 99 + 9900)

98! (1 + 99 + 9900) = 98! \times 10000

Step 3: Count factors of 5 in the integer multiplier

Our sum is now a product of two parts: 98! and 10,000. We need to find how many times 5 is a factor of 10,000. We can find this by writing 10,000 as a power of 10, and then splitting 10 into its prime factors, 2 and 5. The exponent tells us there are exactly 4 factors of 5 here.

10000 = 1 0^{4}

1 0^{4} = (2 \times 5)^{4} = 2^{4} \times 5^{4}

Step 4: Count factors of 5 in the factorial component

To find how many times 5 divides the other part, 98!, we use Legendre’s Formula. First, we count the multiples of 5 up to 98. Next, we count the multiples of 25 (which is 5 squared), as they each contribute a second factor of 5. We drop the decimal remainder in each division. There are no multiples of 125, so we stop here.

⌊ \frac{98}{5} ⌋ = ⌊ 19.6 ⌋ = 19

⌊ \frac{98}{25} ⌋ = ⌊ 3.92 ⌋ = 3

19 + 3 = 22

Step 5: Combine the counts for the final result

The total number of factors of 5 in our product determines the largest possible exponent $n$ . We add the number of 5s found in 98! to the number of 5s found in 10,000.

22 + 4 = 26

Correct answer: D. $26$
The largest integer

n

is 26, corresponding to choice D. We can verify our work by observing that

98!

is the dominant part of the factorization. The added terms

99!

and

100!

essentially multiply

98!

by exactly

10, 000

. Since

10, 000

visibly contains exactly four factors of

5

, correctly counting the

22

factors of

5

in

98!

and adding

4

is robust and complete.

Solution 2

Largest Factorial Method

Instead of factoring out the smallest factorial, we can write the smaller factorials as fractions of the largest factorial ( $100!$ ). This provides a clever alternative perspective.

Just as we can build larger factorials by multiplying (like $100! = 100 \times 99!$ ), we can represent smaller factorials by dividing larger ones (like $99! = \frac{100 !}{100}$ ). This allows us to factor out the largest factorial.

Step 1: Express terms as fractions of the largest factorial

We rewrite 99! and 98! by dividing 100! by the missing terms.

99! = \frac{100 !}{100}

98! = \frac{100 !}{100 \times 99} = \frac{100 !}{9900}

Step 2: Factor out the largest factorial and simplify

Substitute these fractions into the original sum. By factoring out 100!, we can combine the remaining fractions over a common denominator.

98! + 99! + 100! = \frac{100 !}{9900} + \frac{100 !}{100} + 100!

\frac{100 !}{9900} + \frac{100 !}{100} + 100! = 100! (\frac{1}{9900} + \frac{99}{9900} + \frac{9900}{9900})

100! (\frac{1 + 99 + 9900}{9900}) = 100! (\frac{10000}{9900}) = 100! (\frac{100}{99})

Step 3: Count factors of 5 in the factorial component

Just as we did before, we use Legendre’s Formula to count the factors of 5, this time inside 100!.

⌊ \frac{100}{5} ⌋ = ⌊ 20 ⌋ = 20

⌊ \frac{100}{25} ⌋ = ⌊ 4 ⌋ = 4

20 + 4 = 24

Step 4: Adjust the count based on the fractional multiplier

The entire expression is equivalent to 100! multiplied by 100 and divided by 99. The multiplication by 100 provides two additional factors of 5. The division by 99 removes none, because 99 is not divisible by 5. We add these to the factors found in 100!.

100 = 25 \times 4 = 5^{2} \times 2^{2}

24 + 2 = 26

Correct answer: D. $26$
The total count of factors of 5 is 26, matching choice D. This confirms our earlier answer from a genuinely different direction. Because the original sum is roughly equal to

100!

, it makes sense that it contains just a slightly different number of prime factors than

100!

itself.

Adapted Problems to Practice

1 / 10

What Formulas Do You Need for This Problem?

Legendre's Formula:

The number of times a prime number $p$ divides $n!$ is given by $⌊ \frac{n}{p} ⌋ + ⌊ \frac{n}{p ^{2}} ⌋ + ⌊ \frac{n}{p ^{3}} ⌋ + \dots$

$n$ is the positive integer of the factorial, $p$ is a prime number, and $⌊ x ⌋$ means rounding down to the nearest integer. We use it to quickly evaluate how many times the prime $5$ divides large factorials like $98!$ and $100!$ without having to write out all the numbers.

Factoring Sums of Factorials:

To simplify a sum or difference of factorials, factor out the smallest factorial in the expression.

Applies to any expression containing the addition or subtraction of multiple factorials. Factoring converts the sum $98! + 99! + 100!$ into a product $98! \times 10000$ . Finding prime factorizations requires numbers to be written as multiplication, not addition.

Key Takeaway: Remember This Strategy for Next Time

An expression containing a sum or difference of consecutive factorials.

Try: Factor out the smallest factorial to turn the addition into multiplication.

Why: Every larger factorial contains the smallest factorial as a built-in factor. Factoring it out simplifies the expression and allows you to easily analyze its prime factors.

A problem asking for the largest power of a prime that divides a factorial.

Try: Use Legendre's formula by repeatedly dividing the factorial's number by the prime $p, p^{2}, p^{3}$ , and summing the whole-number parts of the results.

Why: The first division counts all multiples of $p$ , the second division counts the extra factors of $p$ hiding inside multiples of $p^{2}$ , and so on. This systematically accounts for every factor of the prime without missing any.

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2017 AMC 8 Problem 19

In This Page:

2017 AMC8 Problem 19