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

Question

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

Archit3110 · Accepted Answer

given that 
a1= 3
a2 will be 33 
a3 = 333
a4=3333
product of values is 173,446,418,443,770,747,
last digit is  7 
cyclicity of 3 is (4)
3^1 =3
3^2=9
3^3=7
3^4=1
value of k must be 19 to get the product value ending with digit 7

option D is correct

NoobGmatTaker · Answer

Let's calculate the first few terms of the sequence A to get a better understanding:

a1 = 3
a2 = a1 + 3*10^1 = 3 + 30 = 33
a3 = a2 + 3*10^2 = 33 + 300 = 333
a4 = a3 + 3*10^3 = 333 + 3000 = 3333
and so on...

We can see that each term in the sequence A is a string of 3's, the length of which is equal to the term's position in the sequence.

Now, we are given that the product of some terms from the sequence A equals 173,446,418,443,770,747. We are asked to find the number of terms in this product, which is represented by k.

Since unit digit is 7 and we know unit digit in power series of 3 is : 3,9,7,1 
To have unit digit as 7, we need k of form 4p+3
So 19 is the correct answer

Answer is D

missionmba2025 · Answer

a1 =  3

a2 = 3 +30= 3 3

a3 = 33 + 300 = 33 3

a4 = 333 + 3000 = 333 3

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

sujoykrdatta · Answer

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)

Answer D

pintukr · Answer

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

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

bumpbot · Answer

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

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