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

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

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.

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.

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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

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

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

\(a_1=3\\
a_2=33\\
a_3=333.\\
:\\
.\\
a_n=333...\)

\(Π x_i=173,446,418,443,770,747\) n times

Let's focus on the last digit.

\(3^3≡7\) (mod 10)
\(3^{4k+3}≡7\) (\(mod\) 10)

The ans has to be of form \(4k+3\). Left with \(19\)

D)

a1 = 3 = 3*1, a2 = 33 = 3*11, a3 = 333 = 3*111 and so on.

So each term in product \(x_1*x_2*x_3*...*x_k\) will be having 3 as a factor.
Total number of terms = k so 3^k will be factor of 173,446,418,443,770,747.

so \(x_1*x_2*x_3*...*x_k\) = 3^k * {1*11*111.....} = 3^k * {......1}= 173,446,418,443,770,747

Since unit digit of 3^k and 1 are multiplied to have results of 7.

So 3^k has to have unit digit = 7
3 has a cyclicity of 4 and 3^(4n+3) = 7 where n = integer.

Only option in 4n+3 form is (D) 19. So k = 19.

IMO D.

The infinite sequence A is such that a1=3 and an=a(n−1)+3∗10^(n−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?
The terms are
a1 = 3
a2 = 33
a3 =333 and so on
Given : any of the terms can be used and repetition is allowed to obtain the given product.
Note that the units digit of the product is 7.
since, we have 3 in the units digit of all terms in the infinite sequence - the product of each term to obtain the given product must be such that its units digit is 7.
we know that 3^1 has units digit 3
3^2 has units digit 9
3^3 has units digit 7 - desired - so the number of terms could be 3, 7, 11, 15, 19, 23 and so on
3^4 has units digit 1
Out of the given options Option D is 19 which is one possible value of k
Answer D

a1 = 3, a2 = 33, a3 = 303, a4 = 3003..........
thing to notice - unit digit of all no is 3
unit digit of no given = 7
unit digits of multiples of 3 = 3,9,7,1
checking answer choices
A- if 16 numbers with unit digit 3 will be multiplied then the unit digit of the product will be 1
B- if 17 numbers with unit digis 3 will be multiplied then the unit digit of the product will be 3
C- if 18 numbers with unit digis 3 will be multiplied then the unit digit of the product will be 9
D- if 19 numbers with unit digis 3 will be multiplied then the unit digit of the product will be 7

hence correct answer is D

Important Point
Unit digit of \(3^{4d+1}\)=3, where d is a non-negative integer.
Unit digit of \(3^{4d+2}\)=9, where d is a non-negative integer.
Unit digit of \(3^{4d+3}\)=7, where d is a non-negative integer.
Unit digit of \(3^{4d+4}\)=1, where d is a non-negative integer.

The infinite sequence A is - 3,33,333,3333,33333,.... upto infinity. We can observe the pattern is any number \(a_n\) (for n\(\geq1\)) in the infinite series will be 3....n. (i.e 3 will come n times)
For example \(a_1\)=3,\(a_2\)=33,\(a_6\)=333333.

So \(x_1*x_2*x_3*...*x_k\)=173,446,418,443,770,747,
Given that each term in the above expression is taken from series A. And repetition is allowed. We can observe one thing by taking an example say value of k is 7 and each term be 3333333 so x_1*x_2*x_3*x_4*x_5*x_6*x_7=\((3333333)^7\) can also be written as \(3^7\)*(\(111111)^7\). The unit digit will depend on the power of 3.(in the example the power of 3 is 7. so \(3^7\) can be expressed as \(3^{4k+3}\), therefore the unit digit will be 7) because the unit digit of (\(111111)^7\). will be 1. Extrapolating this finding we can conclude irrespective of the value of k and the value of each term in the expression \(x_1*x_2*x_3*...*x_k\). The expression can be expressed in term of 3^k*(11*1111*11111 or 1111111^90 or any other form whose unit digit will always be 1). The unit digit of the expression will depend on the power of 3 i.e k.

On the basis of above pointer we can easily find the value of \(x_1*x_2*x_3*...*x_k\)=173,446,418,443,770,747, Since the unit digit is 7. Only for k=19 among the given options will 3^k result in 7 as the unit digit.

Therefore option D is the correct answer

From the sequence a1 = 3, a2 = 33, a3 = 333, a4 = 3333..and so on. The last digit of the product is 7 so we raise 3 to different powers.

3^1 = 3
3^2 = 9
3^3 = 7
3^4 = 1.

From the option only 19 leaves a remainder of 3 when divided by 4 so answer is D.

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Answer:

So here if find some numbers of sequence the sequence looks like
3, 33, 333, 3333 ..........
unit digit of all the numbers will be 3
now for the product the numbers will come from this sequence and the repetition is allowed and the product is 173,446,418,443,770,747
for the product the last digit is 7
when 3*3*3 = 27 unit digit is 7
and 3*3*3*3 = 81 unit digit is 1

now lets see options
A. 16 => now we divide the 16 into 4 sets that contain multiplication of unit digit 3 4times then we get (3^4 * 3^4 * 3^4 * 3^4) which will have unit digit as 1 so not the ans
B. 17 => now same thing as above we get. 1 * 3 =. 3 as unit digit
C. 18 => same as above 1* 3* 3 =9 as unit digit
D. 19=> 1 * 3*3*3 = 7 as unit digit matches with the unit digit of the product this is our ans
E. 20 =>1* 3* 3*3*3 = 1 as unit digit

So the ans is D

so ans is D

Given, a1 = 3,
\(a2 = 3 + 3*10^(2-1) = 3 + 30 = 33\)
a3 = 333, a4 = 3333, ..... and so on. So All the terms of this sequence ends with unit digit "3".

In the product x1∗x2∗x3∗...∗xk:- All terms ends with unit digit "3" and so for the unit digit of final product, we are basically multiplying "3" k times.
So we need to find among the given options, for what value of k, 3^k ends with unit digit "7".

We get unit digit "3" for every k = 1, 5, 9, 13, 17, 21.....
We get unit digit "9" for every k = 2, 6, 10, 14, 18, 22.....
We get unit digit "7" for every k = 3, 7, 11, 15, 19, 23.....
We get unit digit "1" for every k = 4, 8, 12, 16, 20, 24.....

So, among the given options, k = 19 could be a possible value of "k".
Hence answer is option D

The following sequence will be formed.
a1=3
a2=33
a3=333
a4=3333
As we can see all the terms have 3 in the unit's place.
Hence the product of all the terms of the sequence will result in 3, 9, 7 or 1 in the units place depending on the number of 3's in the unit's place.
Since we have 7 in the unit's place, the total number of terms will be of the form 4k+3.

The only answer that matches this condition is 19.

Answer Option D.