For each positive integer k, let ak = 7k. Which of the following is

Question

For each positive integer k, let a_k= 7k . Which of the following is the greatest value of n such that 10^{n} divides (a_1)(a_2)(a_3)...(a_{28}) ? A. 7 B. 6 C. 5 D. 4 E. 3 https://gmatclub.com/forum/download/file.php?id=81443 GMAT-Club-Forum-w7zu81np.png

gmatophobia · Accepted Answer

a_k = 7k

Let's write a few terms to understand the pattern -

a_1 = 7*1
 a_2 = 7*2
 a_3 = 7*3
.
.
.
 a_{27} = 7*27
 a_{28} = 7*28

a_1*a*_2*a_3*...a_{27}*a_{28} = 7*1*7*2*7*3....7*27*7*28

a_1*a*_2*a_3*...a_{27}*a_{28} = 7^{28} * 28!

7^{28}  is not divisible by 10, hence we have to find the power of 10 in 28!

10 = 2 * 5; the power of 5 will always be smaller than the power of 2.

Target question : What is the power of 5 in 28!

We can use successive division to find the power of 5.

28 / 5 = 5
5 / 5 = 1

Total = 5 + 1 = 6

Option B

mcelroytutoring · Answer

This is a confusing new GMAT Focus question, so I am going to try to explain it in words rather than math notation whenever possible.

We can agree that the expression being asked about is equal to  7^28 x 28!

It then asks us for the largest power of 10^n for which "10^n divides" the expression: in other words, the largest power of n for which 10^n is a  factor  of the expression.

Simply put, this means that  (expression) / 10^n = integer  — not the other way around, a common misinterpretation.

The 7^28 (clearly not a multiple of 10) is not a problem, because  it's in the numerator —only the denominator needs to cancel fully for the answer to remain an integer.

Now we simply need to see how many 10s are in 28 factorial (28!).

Break 10 into its prime factorization of 5 and 2, and—since there are obviously plenty of 2s in the numerator—start counting 5s instead: they should be the limiting factor.

In 28! we find exactly  six 5s: two in 25, one in 20, one in 15, one in 10, and one in 5 . The correct answer is  Choice B (6)  because there are only six fives in 28 factorial (28!).

We will still have a bunch of numbers left in the numerator, of course—including most of 28 factorial, and a whopping 28 sevens—but that's OK, because the answer is still an integer.

JeffTargetTestPrep · Answer

y = (a₁)(a₂)…(a₂₈) = (7×1)(7×2)…(7×28)

y = (7^28)(28!)

We need to find the greatest n such that:

y/10^n = int

y/  = int

The expression above is an integer if n is not greater than the total number of 2s in the prime factored form of y and if n is not greater than the total number of 5s in the prime factored form of y.

Since in the prime factored form of y, there are fewer 5s than 2s, we need to concentrate on the total number of 5s, which we can quickly determine with the following technique:

28/5 = 5 (ignore the remainder)

5/5 = 1

The total number of 5s is 5 + 1 = 6, so we have:

n ≤ 6

Answer: B

(Note: We can ignore 7^28 because it doesn’t contain neither 2s nor 5s.)

Crater · Answer

( a_1 )( a_2 )( a_3 )...(a28) 
  (7*1)*(7*2)(7*3)...........(7*28) 
 (7^{28})*(28!)

For this number to be divisible by  10^n : 
  \frac{7^{28}*28!}{2^n*5^n}

Since 7 is a prime number, so we need to find number of 5s in 28!. 
  \frac{28}{5}=5 
 \frac{5}{5}=1

Hence, total 5s in 28!=6

Hence, Option (B)

felix1909 · Answer

Can someone explain the math behind dividing 28 by 5? Does it not matter that it is 28 factorial?

Thanks in advance
F.

Bunuel · Answer

I think the following should help:
    Everything about Factorials on the GMAT  
   Trailing Zeros  
   Power of a number in a factorial

For more check  Ultimate GMAT Quantitative Megathread

Crater · Answer

Since  28!  means  28*27*26*.......*1 . So there are going to be 5s and powers of 5s ( 5^2  in this case). So we are basically just counting multiples of  5^1, 5^2  here.

nikaro · Answer

Basically, we need to find the greatest power of 5 in 28! which is 6, hence the answer

Regor60 · Answer

7 is a red herring since it introduces no factors of either 5 or 2, so it should be ignored.

The number of instances of 5 will be fewer than those of 2, so we only need to count the instances of 5.

28/5 yields 5 instances of 5, but that undercounts the second 5 in 25, so:

n=6

Posted from my mobile device

dueconcern4024 · Answer

Hi,

Thanks for this, its a great explanation. Just one clarifying question which I didn't understand in the whole explanation :

"10 = 2 * 5; the power of 5 will always be smaller than the power of 2." Why will the power of 5 always be smaller than 2? Is it wrt '28!' As it's an even number? I'm not sure of the reasoning behind your statement.

KarishmaB · Answer

Here is a post that discusses how to find the highest exponent of a number in a factorial: https://anaprep.com/number-properties-highest-power-in-factorials/ GMAT-Club-Forum-ebodyvxu.png

vshel1990 · Answer

Hi Crater,

Can you, please, explain why are we dividing 5 by 5? Thank you.

Bunuel · Answer

I think the following should help:

Everything about Factorials on the GMAT  
  Trailing Zeros  
  Power of a number in a factorial

For each positive integer k, let ak = 7k. Which of the following is

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