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

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

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