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If k is a positive integer and 5^k is a factor of the product 07 Nov 2023, 08:45
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If k is a positive integer and \(5^k\) is a factor of the product of the odd integers from 99 to 199, inclusive, what is the greatest value of k?

A. 10
B. 13
C. 15
D. 20
E. 23
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B
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gmatophobia
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If k is a positive integer and 5^k is a factor of the product 07 Nov 2023, 09:06
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ChandlerBong
If k is a positive integer and \(5^k\) is a factor of the product of the odd integers from 99 to 199, inclusive, what is the greatest value of k?

A. 10

B. 13

C. 15

D. 20

E. 23
Show more

As we are looking only at odd integers between 99 and 199, any number that has unit digit as \(5\) is divisible by 5.

105 → 5 * 21

115 → 5 * 23

125 → 25 * 5

135 → 5 * 27

145 → 5 * 29

155 → 5 * 31

165 → 5 * 33

175 → 25 * 7

185 → 5 * 37

195 → 5 * 39

Total number of 5s = 13

Option B
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If k is a positive integer and 5^k is a factor of the product 19 Feb 2024, 23:47
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Integers divisible by 5 but not 25: These contribute one factor of 5 each (e.g., 105, 115, 125, ..., 195). There are 10 such integers in the list.
Integers divisible by 25: These contribute two factors of 5 each (e.g., 125, 175). There are 2 such integers in the list.
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If k is a positive integer and 5^k is a factor of the product 12 Mar 2024, 03:20
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Why are the other odd #s like 1,3,7,9 not considered?
gmatophobia

ChandlerBong
If k is a positive integer and \(5^k\) is a factor of the product of the odd integers from 99 to 199, inclusive, what is the greatest value of k?

A. 10

B. 13

C. 15

D. 20

E. 23
As we are looking only at odd integers between 99 and 199, any number that has unit digit as \(5\) is divisible by 5.

105 → 5 * 21

115 → 5 * 23

125 → 25 * 5

135 → 5 * 27

145 → 5 * 29

155 → 5 * 31

165 → 5 * 33

175 → 25 * 7

185 → 5 * 37

195 → 5 * 39

Total number of 5s = 13

Option B
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­
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If k is a positive integer and 5^k is a factor of the product 20 Apr 2024, 15:15
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unicornilove
Why are the other odd #s like 1,3,7,9 not considered?
gmatophobia

ChandlerBong
If k is a positive integer and \(5^k\) is a factor of the product of the odd integers from 99 to 199, inclusive, what is the greatest value of k?

A. 10

B. 13

C. 15

D. 20

E. 23
As we are looking only at odd integers between 99 and 199, any number that has unit digit as \(5\) is divisible by 5.

105 → 5 * 21

115 → 5 * 23

125 → 25 * 5

135 → 5 * 27

145 → 5 * 29

155 → 5 * 31

165 → 5 * 33

175 → 25 * 7

185 → 5 * 37

195 → 5 * 39

Total number of 5s = 13

Option B
­
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­They arent divisible by 5
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If k is a positive integer and 5^k is a factor of the product 22 Apr 2024, 21:58
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unicornilove
Why are the other odd #s like 1,3,7,9 not considered?
gmatophobia

ChandlerBong
If k is a positive integer and \(5^k\) is a factor of the product of the odd integers from 99 to 199, inclusive, what is the greatest value of k?

A. 10

B. 13

C. 15

D. 20

E. 23
As we are looking only at odd integers between 99 and 199, any number that has unit digit as \(5\) is divisible by 5.

105 → 5 * 21

115 → 5 * 23

125 → 25 * 5

135 → 5 * 27

145 → 5 * 29

155 → 5 * 31

165 → 5 * 33

175 → 25 * 7

185 → 5 * 37

195 → 5 * 39

Total number of 5s = 13

Option B
­
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­We need to find k which is the power of 5, hence we're looking for odd numbers divisible by 5. So odd numbers ending with 1,3,7,9 are not considered.
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If k is a positive integer and 5^k is a factor of the product 07 Jul 2024, 00:05
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johnnywatts
Integers divisible by 5 but not 25: These contribute one factor of 5 each (e.g., 105, 115, 125, ..., 195). There are 10 such integers in the list.
Integers divisible by 25: These contribute two factors of 5 each (e.g., 125, 175). There are 2 such integers in the list.
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­and one more - integer divisible by 125, which is 125, so in total --> 10+2+1=13­
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If k is a positive integer and 5^k is a factor of the product 16 Mar 2025, 04:16
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@Bunel could you please check?
Step 1: Count the number of factors of 5 in 199!: 47
Step 2: Count the number of factors of 5 in 99!: 22
Total factors of 5=47−22=25
Divide by 2 (since we're only considering odd numbers): 12.5
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If k is a positive integer and 5^k is a factor of the product 02 Apr 2025, 22:27
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bbb1923
@Bunel could you please check?
Step 1: Count the number of factors of 5 in 199!: 47
Step 2: Count the number of factors of 5 in 99!: 22
Total factors of 5=47−22=25
Divide by 2 (since we're only considering odd numbers): 12.5
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yes please check if this is valid
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If k is a positive integer and 5^k is a factor of the product 02 Apr 2025, 22:56
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bbb1923
If k is a positive integer and \(5^k\) is a factor of the product of the odd integers from 99 to 199, inclusive, what is the greatest value of k?
A. 10
B. 13
C. 15
D. 20
E. 23

@Bunel could you please check?
Step 1: Count the number of factors of 5 in 199!: 47
Step 2: Count the number of factors of 5 in 99!: 22
Total factors of 5=47−22=25
Divide by 2 (since we're only considering odd numbers): 12.5
yes please check if this is valid
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Almost. It should be 199!/98! to get the product of all integers from 99 to 199 inclusive. Yes, 199! = x * 5^47 and 98! = y * 5^22, so 199!/98! = z * 5^25. Dividing 25 by 2 gives only an estimate, but it's good enough here.
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If k is a positive integer and 5^k is a factor of the product 03 Apr 2025, 05:56
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ChandlerBong
If k is a positive integer and \(5^k\) is a factor of the product of the odd integers from 99 to 199, inclusive, what is the greatest value of k?

A. 10
B. 13
C. 15
D. 20
E. 23
Show more
odd multiples of 5 from 99 to 199 = 105*115*125*135*145*155*165*175*185*195

Power of 5 in them respectively are = 1+1+3+1+1+2+1+1+1+1 = 13

Answer: Option B
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If k is a positive integer and 5^k is a factor of the product 03 Apr 2025, 08:07
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For anyone like me struggling with this question, this Chatgpt answer made it very clear:


We have all the odd integers from 99 to 199, inclusive. In other words, we’re looking at the numbers: 99, 101, 103, 105, 107, ..., 197, 199.

We multiply all of these odd integers together to form one gigantic product: 99 × 101 × 103 × 105 × ... × 197 × 199.

We want to find the largest exponent k such that 5^k still divides this big product.

Saying “5^k is a factor of that product” means that if you fully factor the product into primes (2, 3, 5, 7, 11, 13, ...), you can pull out at least k copies of the prime number 5.


To figure out how many times 5 divides a product, you only need to look at the factors that contribute 5s. Any integer that isn’t a multiple of 5 contributes no factor of 5 at all.

In the range 99 to 199 (odd integers only), list out the ones that are actually multiples of 5.

A number that’s a multiple of 5 ends with 0 or 5 in decimal notation. Among odd numbers, it’ll end with 5.

So let’s list all the odd multiples of 5 between 99 and 199:

Start just above 99: the first odd multiple of 5 ≥ 99 is 105.

Then keep adding 10 (since every other multiple of 5 is 10 apart if we stick to odd multiples): 105, 115, 125, 135, 145, 155, 165, 175, 185, 195.

That’s it—once you reach 195, the next would be 205 (which is > 199).

Count them. We have 10 numbers: 105, 115, 125, 135, 145, 155, 165, 175, 185, 195.

Each of these 10 numbers contributes at least one factor of 5 because they’re multiples of 5.

Accounting for Higher Powers (25, 125, etc.)

But we’re not done: some numbers might be multiples of 25 = 5^2, which means they don’t just contribute one factor of 5—they contribute at least two. Likewise, if any is a multiple of 125 = 5^3, that number contributes at least three factors of 5, and so on.

Multiples of 25 among those odd multiples of 5:

Which of the 10 numbers above are divisible by 25? 25 goes into 125 and 175. So each of these two numbers (125 and 175) contributes one extra factor of 5 (beyond the first one).

Multiples of 125 among them:

125 itself is 5^3. That means 125 contributes yet another factor of 5 beyond the “multiple of 25” count. 175 is not a multiple of 125—only 125 is. So 125 gives us one more factor of 5 on top of everything else.

No multiples of 625 (5^4) here because 625 > 199. So we stop.

Tallying Up the 5-Factors

Let’s sum the contributions:

One factor of 5 from each of the 10 odd multiples of 5: 10 (for 105, 115, 125, 135, 145, 155, 165, 175, 185, 195).

An extra factor of 5 from each multiple of 25. That’s 125 and 175, so that’s 2 additional.

An extra factor of 5 from each multiple of 125. That’s just 125, so that’s 1 additional.

Add them up: 10 + 2 + 1 = 13.

So altogether, the product has 13 copies of the prime 5 in its factorization. That is, 5^13 divides the product, but 5^14 does not.

Hence the largest k is 13.
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If k is a positive integer and 5^k is a factor of the product 19 Apr 2025, 01:57
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ChandlerBong
If k is a positive integer and \(5^k\) is a factor of the product of the odd integers from 99 to 199, inclusive, what is the greatest value of k?

A. 10
B. 13
C. 15
D. 20
E. 23
Show more
Problem interpretation: Determine the number of times the prime factor 5 appears in the product of odd integers from 99 to 199

(1) Identify the qualified numbers:
  • Condition 1: has 5 as a factor -> divisible by 5 -> numbers ending with 0 or 5
  • Condition 2: odd numbers
=> Only the case of numbers ending with 5 is qualified

(2) List qualifying numbers: The sequence of numbers from 99 to 195 ending in 5, is 105, 115, 125, ..., 195. This is an arithmetic sequence with a common difference of 10.
  • Number of terms: [(195 - 105) / 10] + 1 = 10 numbers.
  • Each number contributes one digit 5 (e.g., 105 has one 5).
  • Total 5s from these numbers: 10.

(3) Check for additional 5s: Removed 5s that already count in step 2 (e.g., 105 = 5 × 21, 115 = 5 × 23, ..., 195 = 5 × 39); we have new set of odd integers (21, 23, ..., 39) include 25 = 5 × 5 and 35 = 5 × 7, which are divisible by 5.
  • 25 = 5 × 5, contributes 2 additional 5s.
  • 35 = 5 × 7, contributes 1 additional 5.
  • Additional 5s: 2 (from 25) + 1 (from 35) = 3.

Total count: 10 + 3 = 13 => k=13
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If k is a positive integer and 5^k is a factor of the product 19 May 2025, 22:52
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ChandlerBong
If k is a positive integer and \(5^k\) is a factor of the product of the odd integers from 99 to 199, inclusive, what is the greatest value of k?

A. 10
B. 13
C. 15
D. 20
E. 23
Show more
Responding to a pm:

All multiplies of 5 end with a 0 or a 5. So from 99 to 199, we will have
105, 115, 125, 135.... 195 - these 10 odd multiples only. Multiples ending with 0 will be even. Each odd multiple given here will provide a 5 and we will be left with
21, 23, 25, 27, 29, .... 39 - there are 2 more multiples of 5 here - 25 and 35 so we get another 3 5s (2 from 25 and 1 from 35)

Hence total 13 5s.

Here is a blog post discussing power in factorials:
https://anaprep.com/number-properties-h ... actorials/
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