Last visit was: 13 Jun 2024, 15:24 It is currently 13 Jun 2024, 15:24
Toolkit
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

# Around the World in 80 Questions (Day 8): The infinite sequence A is

SORT BY:
Tags:
Show Tags
Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 93696
Own Kudos [?]: 631532 [16]
Given Kudos: 82279
GMAT Club Legend
Joined: 18 Aug 2017
Status:You learn more from failure than from success.
Posts: 7909
Own Kudos [?]: 4151 [3]
Given Kudos: 242
Location: India
Concentration: Sustainability, Marketing
GMAT Focus 1:
545 Q79 V79 DI73
GPA: 4
WE:Marketing (Energy and Utilities)
Intern
Joined: 09 Jul 2023
Posts: 28
Own Kudos [?]: 52 [2]
Given Kudos: 23
Manager
Joined: 07 May 2023
Posts: 198
Own Kudos [?]: 241 [2]
Given Kudos: 46
Location: India
Re: Around the World in 80 Questions (Day 8): The infinite sequence A is [#permalink]
2
Kudos
Bunuel wrote:
The infinite sequence A is such that $$a_1=3$$ and $$a_{n} = a_{n-1} + 3*10^{n-1}$$, for all n > 1. If each term in the product $$x_1*x_2*x_3*...*x_k$$ is taken from the infinite sequence A, repetitions allowed, and the product equals 173,446,418,443,770,747, which of the following could be equal to k?

A. 16
B. 17
C. 18
D. 19
E. 20

 This question was provided by GMAT Club for the Around the World in 80 Questions Win over $20,000 in prizes: Courses, Tests & more a1 = 3 a2 = 3 +30= 33 a3 = 33 + 300 = 333 a4 = 333 + 3000 = 3333 The unit digit of each of the number is 3. Hence the unit digit of the product will be the unit digit of 3^k. Cyclicity of 3, 9 , 7, 1 So k is of the form k=4n + 3 A. 16 = 4n B. 17 = 4n+1 C. 18 = 4n+2 D. 19 = 4n+3 E. 20 = 4n IMO D Tutor Joined: 26 Jun 2014 Status:Mentor & Coach | GMAT Q51 | CAT 99.98 Posts: 450 Own Kudos [?]: 786 [1] Given Kudos: 8 Re: Around the World in 80 Questions (Day 8): The infinite sequence A is [#permalink] 1 Kudos Expert Reply Bunuel wrote: The infinite sequence A is such that $$a_1=3$$ and $$a_{n} = a_{n-1} + 3*10^{n-1}$$, for all n > 1. If each term in the product $$x_1*x_2*x_3*...*x_k$$ is taken from the infinite sequence A, repetitions allowed, and the product equals 173,446,418,443,770,747, which of the following could be equal to k? A. 16 B. 17 C. 18 D. 19 E. 20  This question was provided by GMAT Club for the Around the World in 80 Questions Win over$20,000 in prizes: Courses, Tests & more

a1 = 3
a2 = 3 + 3*10 = 33
a3 = 33 + 3*100 = 333. and so on
Thus, the n-th term = 3333.... (n times)
We have: x(1) * x(2) * ... * x(k) = 173446418443770747; where all of x(1), x(2) etc. are from the above series of a1, a2, etc.

It is important to note the units' digit of the product - that is 7
Since all the terms in a1, a2 ... have units' digit 3, we basically need to check how many 3s are need to be multiplied to have the units' digit 7
We know that the units' digit cycle for powers of 3 is:
3^1 => 3
3^2 => 9
3^3 => 7
3^4 => 1
Thus, whenever 3 is multiplied 3 times or 3+4=7 times or 3+4+4=11 times etc, the units' digit will be 7
From the options, only 19 is possible since 19 = 3 + 4*4 i.e., the cycle of 3 was completed 4 times and then 3 more times.

Alternatively, we check the remainder when 19 is divided by 4 - it is 3. So, 3^19 => 3^3 => 7 (units' digit)

VP
Joined: 03 Jul 2022
Posts: 1207
Own Kudos [?]: 819 [1]
Given Kudos: 21
GMAT 1: 680 Q49 V34
Re: Around the World in 80 Questions (Day 8): The infinite sequence A is [#permalink]
1
Kudos
The infinite sequence A is such that a1=3 and an=an−1+3∗10n−1, for all n > 1. If each term in the product x1∗x2∗x3∗...∗xk is taken from the infinite sequence A, repetitions allowed, and the product equals 173,446,418,443,770,747, which of the following could be equal to k?

A. 16
B. 17
C. 18
D. 19
E. 20

that a1=3 and an=an−1+3∗10n−1;

a1 =3
a2 =33
a3 = 333
a4 = 3333

Thus we see that all values ie, a1, a2, a3, a4 etc are multiples of 3

So, if the product equals 173,446,418,443,770,747 ie, ending with 7 as unit's digit;
then,

product x1∗x2∗x3∗...∗xk, taken from the infinite sequence A, wiil have

k is of the form 4n + 3 since xxxx3 ^ (4n + 3) will have 7 as unit's digit.

Only 19 is of the form 4n+3

(D) is the CORRECT answer
Re: Around the World in 80 Questions (Day 8): The infinite sequence A is [#permalink]
Moderator:
Math Expert
93696 posts