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

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

Given: The infinite sequence A is such that \(a_1=3\) and \(a_{n} = a_{n-1} + 3*10^{n-1}\), for all n > 1.
Asked: 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_1 = 3
a_2 = 3 + 3*10 = 33
a_3 = 33 + 3*10^2 = 333
a_k = 333.....k digits

\(173,446,418,443,770,747 = 3^20 * 49743947\)
k = 20

IMO E

a1 = 3
a2 = 33
a3 = 333
a4 = 3333 and so on.
The units digit of the given product is 7.
All numbers in sequence have units digit of 3.

If we observe units digits of powers of 3,
3^1-----3
3^2-----9
3^3-----7
3^4-----1
and repeat.

Only the powers which are multiples of 3 will have units digit as 7.
18 is the only multiple of 3 in the options.
Hence, C is the right choice.

My Answer: D

a1=3, a2=33, a3=333 and so on. So we can conclude product will be cyclical and last digit will be: 3,9,7,1.

Now, only when k=19 will the last digit be 7.

The answer is C
We get a2=33, a3=33 and so on..
The numbers 173,446,418,443,770,747 can be expresses as factors of 3.
Amongst the answer choices 18 is a factor of 3 so answer is c

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

The sequence A is defined as:
a1 = 3
an = an-1 + 3 * 10n-1 for n > 1
This means:
a2 = a1 + 3 * 101 = 6
a3 = a2 + 3 * 102 = 30
a4 = a3 + 3 * 103 = 300
a5 = a4 + 3 * 104 = 3,003
etc

Given x1 * x2 * ... * xk, where product equals 173,446,418,443,770,747.
Hence plugging in values for k:
If k = 16, the product would be too small, since it would only contain terms up to a16.
If k = 17, the product would contain terms up to a17 = 3,003,030,003.
Therefore, option B.

a1 = 3
a2 = a1 + 3*10^1 = 33
a3 = a2 + 3*10^2 = 333
.
.
.
an = 3333…n times

Since last digit for all terms are 3, cyclic nature means product of the numbers would be 3, 9, 7, 1, 3, 9, 7, 1…so 7 is 3rd, 7th, 11th, 15th, 19th… etc.

Correct answer is 19th.

Posted from my mobile device

try to substitute
a1=3
a2=13
a3=313
a4=3313
a5=33313
so we can see the pattern.
The question ask 173,446,418,443,770,747 is the result of the multiplication of a1 to which a
and we find out that the digit of the power are as followings
3^1 = 3
3^2 = 9
3^3 =7
3^4 = 1
Hence the given question is ended which 7 so we know that it must be power until 19 times

A1 = 3
An = A(n-1) + [3* 10^(n-1)]

A1 = 3
A2 = 3 + (3*10) = 33
A3 = 33 + (3*100) = 333
A4 = 333 + (3*1000) = 3333

Unit digit of the product in question is 7
Now, we check the unit digit pattern: 3 - (3*33 - unit = 9) - 7 - 1 - 3 - 9 - 7 - 1
After every 4 nos. the unit digit pattern repeats - we need a number with unit digit as 7,
=> Answer = 19 [D]

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

Solution:
a1=3
a2=33
a3=333
a4=3333
and so on.
So, an=multiple of 3.

Now if each term of the product, let B= x1∗x2∗x3∗...∗xk is taken from the infinite sequence A, then B is surely a multiple of 3,
i.e. the product 173,446,418,443,770,747 is a multiple of 3.
Given, the product 173,446,418,443,770,747 ends in 7, it should be a multiple of a triplet of 3 as 3^3 ends in 7 wherein 3 has a cyclicity of 4. This implies that k should be a of the form 4n-1. Among the given options, 19 is the only one of the form 4n-1. ANSWER D

First of all, a standing oviation to the amazing person who came up with this question! It is not evry hard, but sure is very tricky!

The pattern is: \(a_{n} = a_{n-1} + 3*10^{n-1}\)
Starting from \(a_1=3\), and calculating for subsequent numbers, we get the following sequence:
3, 33, 333, 3333...
Now, the product value has a '7' at the unit place. It means we need to take numbers in multiple of three in order to get a 7 at the end of the multiple. Hence, our answer will be a multiple of 3. Only C matches.

Given : \(a_1=3\)
\(a_{n} = a_{n-1} + 3*10^{n-1}\), for all n > 1

To find : value of k such that \(x_1*x_2*x_3*...*x_k\) = 173,446,418,443,770,747

Lets analyze the sequence first:
\(a_{n} = a_{n-1} + 3*10^{n-1}\)
\(a_2 = a_1 + 3*10^1= 3 + 30 = 33 \)
\(a_3 = a_2 + 3*10^2= 33 + 300 = 333 \)
\(a_4 = a_3 + 3*10^3= 333 + 3000 = 3333 \)

We can now predict the future numbers .
Also note that the last digit in each number is 3 .
Consider \(x_1*x_2*x_3*...*x_k\) = 173,446,418,443,770,747
So the multiplication of all the 3s in the unit digit in k numbers will give us the last digit of product.
We do not need to worry about the big number . We can focus on the last digit ,i.e 7 .
The last digit of powers of 3 repeats in a cyclic fashion.
Last digit 3^1 = 3
Last digit 3^2 = 9
Last digit 3^3 = 7
Last digit 3^4 = 1
Last digit 3^5 = 3

The cyclicity is 4, i.e, after four powers the last digits repeat again in cyclic order. So we need to find the number when the last digit will appear as 7 again.
We can calculate the next number with 7 as last digit by incrementing the powers by 4 starting with 3^3.
We will have 3^7, 3^11, 3^15, 3^19,3^23

Out of the given answer choices only 3^19 has a unit digit as 7.
Thus k =19

Answer : D. 19

IMO C

173,446,418,443,770,747 - Sum of digits total to 3 and 9

Hence the number is divisible by both 3 and 9. Hence only option that has both 3 and 9 as factors is 18

If we write out the sequence mentioned in the question, it comes to
3, 33, 333, 3333...

Since the units digit of the product of x1, x2, x3... xk is 7, we can conclude that the number of numbers that are taken from the sequence is of the form 4n + 3 (where n is an integer greater than or equal to 0) because we know that,

Units digit of 3^4n = 1
Units digit of 3^(4n+1) = 3
Units digit of 3^(4n+2) = 9
Units digit of 3^(4n+3) = 7

So, k should be of the form 4n + 3. Looking at the options, only 19 is of this form.

Hence the answer is option D

I think the correct answer is B.

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?
______
It is a truly arithmetic question, we need experts reply on that . Please, help.