If 5^k is a factor of the product of odd integers from 99 to 199, what

Question

If 5^k is a factor of the product of odd integers from 99 to 199, what is the greatest possible integer value of k?

A. 12
B. 13
C. 22
D. 23
E. 24

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TestPrepUnlimited · Accepted Answer

We want to count the number of multiples of 5 from 105, 115, 125, 135, ..., 195. We only include odd integers since the question said odd integers only.

Then note some of these numbers have more than one multiple of 5 in them, these numbers are 125 and 175 as they are multiples of 25.

So in total, we have 1 copy from each number, an extra 2 copies from 125, an extra copy from 175.

10 + 2 + 1 = 13.

Ans: B

Kinshook · Answer

Asked: If 5^k is a factor of the product of odd integers from 99 to 199, what is the greatest possible integer value of k?

x = 99*101*103*105*.......*199 = (200!/2^100*100!)/(98!/2^49*49!)

Highest power of 5 in 200! = 40 + 8 + 1 = 49
Highest power of 5 in 100! = 20 + 4 = 24
Highest power of 5 in 98! = 19 + 3 = 22
Highest power of 5 in 49! = 9+ 1 = 10

Highest power of 5 in (200!/2^100*100!)/(98!/2^49*49!) = 49 - 24 - 22 +10 = 13

IMO B

sahibr · Answer

Easiest way is to use AP

(Don't be intimidated by the long answer given below. Ive overexplained the solution)

The question states that odd no.s between 99 and 199 are multiplied ie. 99 x 101 x 103......197x199

It also states that 5^k is a factor ie. completely divisible to the above product. Hence all 5s in the denominator must be eliminated.

Now any multiple of 5 must end in either 5 or 0.

But the above product has only odd no.s ie there are no no.s ending with 0.

Hence, the no.s we are looking for are 105,115,125.....,195

last term (l) = 195
First term(a) = 105
Common difference(d)=10

Now, l=a+(n-1)d
195=105+(n-1)10
n=10

Hence, k must also equal to 10. If 5^10 was the denominator, it would get eliminated completely.

But a few no.s in the product can be divided by 5 even further.

125/5=25
25/5=5
5/5=1

Since we had already counted the first 5, we get two more 5s.
Now, k=12.

Again, 175 can be divided more than once.
175/5=35
35/5=7

Since we already counted the first 5, we get another 5.

Hence, we get k=10 + 2 + 1 = 13.

Hence, the answer B

samarpan.g28 · Answer

Odd multiples of 5 - 105,115,125,135,....,195. Total=10 (I have used the formula for AP).
Odd multiples of 25 - 125,175. Total=2. These two terms are also involved in the above series.
Remove 125 and 175 from series 1 for ease. So total terms in series 1=8.
Now in 1st series, each term has one 5. In the second series, 125 has 3 5s and 175 has 2 5s. Therefore total 5s are 8+3+2=13(B).

pratiksha1998 · Answer

Here's my approach:-

Find max power of 5 in 199! = 47
Find max power of 5 in 99! = 22

We need to consider only the 5(s) from 100 to 199, therefore number of 5(s) to consider = 47 - 22 = 25

Even numbers in the range of 100 to 199 with 5(s) in it are = 100,110,120,130,140,150,160,170,180,190. Out of these 10 numbers, 100 and 150 has 2 5(s) and others have 1 5(s). The number of 5s to be removed = 2 + 1 + 1 + 1 + 1 + 2 + 1 + 1 + 1 + 1 = 12

Final answer = 25 - 12 = 13

Globethrotter · Answer

Kinshook  Can you explain " x = 99*101*103*105*.......*199 = (200!/2^100*100!)/(98!/2^49*49!)" a bit?
Use of 200! and 49! are clear, but the rest of it I am not able to understand.

Kinshook · Answer

Globethrotter

x = 99*101*103*105*..*199 = (1*3*5*...*199)/(1*3*5....97)
1*3*5*..*199 = 200!/(2*4*.....198*200) = 200!/2^100*100!
1*3*5*...97 = 98!/(2*4*...96*98) = 98!/2^49*49!

If 5^k is a factor of the product of odd integers from 99 to 199, what

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If 5^k is a factor of the product of odd integers from 99 to 199, what

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