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

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

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

Posted from my mobile device

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!

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)

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.

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.

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

Posted from my mobile device

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.

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

OA:C

Posted from my mobile device

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

=> Choice C

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.

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

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

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

(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

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