The product of the first seven positive multiples of three is closest

Question

Competition Mode Question

The product of the first seven positive multiples of three is closest to which of the following powers of 10?

(A)  10^9 
(B)  10^8 
(C)  10^7 
(D)  10^6 
(E)  10^5

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freedom128 · Accepted Answer

Roughly approximate = 3*(6*9*12*15)*(18*21) = 3*~(10*10*10*10)*~(20*20) = 3*10000*400 = 12,000,000 Thus, it is closest to 10^7. FINAL ANSWER IS (C) Hit kudo

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CrackverbalGMAT · Answer

This is an excellent question which tests you on your ability to do quick approximations. Of course, you also need to know some specific numbers here if you want to be good with your approximation.

The first seven positive multiples of three are 3,6,9,12, 15, 18 and 21. When we take the product of these numbers, we get 3*6*9*12*15*18*21.

We see that 3 is common in all of them. Therefore, we already have a 3^7 in the product. 3^7 = 2187. Once 3^7 is taken out, the product also has 1*2*3*4*5*6*7 = 7!. 
7! = 5040.

Therefore, the product of the first seven positive multiples of 3 = 2187 * 5040. Here’s where you need to bring in your approximation skills instead of trying to calculate the exact answer.
I would round up 2187 to 2200 and round down 5040 to 5000. The product of 5000 and 2200 is 11,000,000. The closest power of 10 to this number is 10^7.
 
Therefore,  the correct answer option is C .

Hope that helps!

CareerGeek · Answer

First seven positive multiples of 3 = {3, 6, 9, 12, 15, 18, 21}
--> Product =  3*6*9*12*15*18*21 = 3*(2*3)*(3^2)*(3*2^2)*(3*5)*(3^2*2)*(3*7) 
=  2^4*3^9*5*7 
=  10*56*19683 
=  10*56*20000  (approx)
=  112*10^5 
=  1.12*10^7

Option C

Archit3110 · Answer

first 7 multiples of 3; 3,6,9,12,15,18,21 ; its product shall be closet to ~ 10^7
IMO C

The product of the first seven positive multiples of three is closest to which of the following powers of 10?

(A) 10^9
(B) 10^8
(C) 10^7
(D) 10^6
(E) 105

Staphyk · Answer

The product of the first seven positive multiples of three is  closest to  which of the following powers of 10?

Multiple of 3 is in the form 3x where x= any positive integer 
First seven = 3,6,9,12,15,18,21
Product = 3•(2•3)•(3^2)•(2^2•3)•(3•5)•(2•3^2)•(3•7) = 2^4•3^9•5•7 
= 2^3•3^9•7•10= (2^3•70)•3^9 = 560•19,683= 11,022,480 
10,000,000 < 11,022,480 < 12,000,000
10^7 < 11,022,480 < 12•10^6 ~ 10^8
10^7 < 11,022,480 < 10^8

.: 11,022,480 is close to 10^7 since it difference is smaller than to 10^8

Hit that C

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exc4libur · Answer

prod(3,6,9,12,15,18,21)=3^7(7!)
3^7=2187~2(10^3)
7!=5040~5(10^3)
prod(…)=10(10^6)=10^7

Ans (C)

unraveled · Answer

The product of the first seven positive multiples of three is closest to which of the following powers of 10?

(A)  10^9 
(B)  10^8 
(C)  10^7 
(D)  10^6 
(E)  10^5

(3*1)*(3*2)*(3*3)*(3*4)*(3*5)*(3*6)*(3*7) = 3^7*1*2*3*4*5*6*7 = 2187*7! = 2187*5040
~ 2200 * 5000 = 11,000,000 
~ 10^7

Answer C.

lacktutor · Answer

The product of the first seven positive multiples of three is closest to which of the following powers of 10?

3*6*9*12*15*18*21= 3^{7}* 7!= (3*3^{6})*(720*7)= 21*729*720

-->  21*729*720≈ 2*10*700*700

-->  2*10*700*700= 2*10*490000= 980000*10 = 9800000≈ 10,000,000= 10^{7}

The answer is C.

Kinshook · Answer

Product of first seven positive multiples of 3 = 3*6*9*12*15*18*21 
= 3^7*7! = 2187*5040 = 11022480 which is closest to 10^7.

IMO C

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eakabuah · Answer

3*6*9*12*15*18*21
We can approximate 18 to 20, and 21 to 20. In addition, we can approximate 12 to 10 and 9 to 10.
So, 3*6*9*12*15*18*21≈3*6*10*10*15*20*20 = 3*6*15*4*10^4 = 3*6*60*10^4 = 3*36*10^5 = 108*10^5, but we can approximate 108 to 100, hence 108*10^5 ≈ 10^7.

The right answer is C.

madgmat2019 · Answer

3*6*9*12*15*18*21
=162*180*378
~3*10^4*4*10^2
~12*10^6
~10^7

OA:C

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ostrick5465 · Answer

3*6*9*12*15*18*21 = 11022480 = 1.1 * 10^7 
=> Closest to  10^7

=> Choice C

Bunuel · Answer

Similar questions to practice: https://gmatclub.com/forum/the-product- ... 35192.html (OG) https://gmatclub.com/forum/the-product- ... 28135.html https://gmatclub.com/forum/the-product- ... 49050.html https://gmatclub.com/forum/the-product-o ... 37076.html Hope it helps.

sujoykrdatta · Answer

Question: 
The product of the first seven positive multiples of three is closest to which of the following powers of 10?

Solution:

The product of the first seven positive multiples of three 
= (3*1) * (3*2) * (3*3) * (3*4) * (3*5) * (3*6) * (3*7)
=  3^7 * (1 * 2 * 3 * 4 * 5 * 6 * 7) 
=  (3^4 * 3^3 )* (2 * 5) * (3 * 4) * (6 * 7) 
=  (81 * 27 )* (10) * (12) * (42) 
=~  (80 * 30 )* (10) * (10) * (40) 
=  (96 * 10^5) 
=~  (100 * 10^5)  
=  10^7

Answer C

ScottTargetTestPrep · Answer

We need the approximate value of:

3 x 6 x 9 x 12 x 15 x 18 x 21

We can group the numbers as follows:

9 x 12 = 108 ≈ 100 = 10^2

6 x 18 = 108 ≈ 100 = 10^2

3 x 15 x 21 = 45 x 21 ≈ 50 x 20 = 1000 = 10^3

Therefore, the product of the first seven positive multiples of 3 is approximately:

10^2 x 10^2 x 10^3 = 10^7

Answer: C

RS81 · Answer

you need to find the closest 10^ for
- 3*6*9*12*15*18*21
= (3^7)*(1*2*3*4*5*6*7)
= (3^7)*(7!) || because i know 7! is 5040
= (3^7)*5040 ~(3^7)*5*10^3
= 3*(3^6)*5*10^3 || because i know 3^6 is 729
= 3*729*5*10^3
= 3*(4000)*10^3
= 12*10^6
= ~10^7

So Answer is C

BrentGMATPrepNow · Answer

(3)(6) (9)(12) (15)(18)(21) ≈ (3)(6) (100) (15)(18)(21)

(3) (6) (100) (15) (18)(21) ≈ (3) (100) (100)(18)(21)

(3) (100)(100) (18)(21)  ≈  (1000) (100)(100)

(1000)(100)(100) = (10³)(10²)(10²) = 10⁷

Answer: C

coolspgmba · Answer

First seven multiple of 3 are 3,6,9,12,15,18,21.

= (3x6) x (9x12) x(15x18) x 21
=(1.8x10^1) x (1.08x10^2) x (2.7x10^2) x (2.1 x 10^1)
=(1.8 x 1.08 x 2.7 x 2.1 ) x 10^6
~ (2x1x3x2) x 10^6 = 12 x 10^6 = 1.2 X 10^7

In this approach, multiplications are smaller and more manageable.

Hence option C.

The product of the first seven positive multiples of three is closest

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The product of the first seven positive multiples of three is closest

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