Quite hard question.
The question is n=?
Provided information
- n = integers+
- n>2
- so n is basically 3,4,5,...
(1) The tens' digit of 11^n is 4- tens' digit = 4 means that the number is somemthing like this _ _ _ _ _ 4 _
- what I did is to fastly calculate the products of 11
- 11^1 =
1111^2 = 1
2111^3 = 13
3111^4 = 146
41- OK! I know that if n=4 is the only answer, then the answer is sufficient
- At first I think A is ok, but when I think closely, I see the pattern, and I believe that the number is going to be a loop, so that the tens digit is going to be like 1,2,3,4,...,9,0,1,2,3,4,...
- I try to prove myself by calculating 91*11 and the result is something end with 01, and 01*11 is end with 11! so the loop is surely happened!
- with this calculation, I know that there is going to be more than one number for n to have 11^n that has tens' digit of 4
- So it is Not sufficient
(2) The hundreds' digit of 5^n is 6- hundreds' digit = 6 means that the number is somemthing like this _ _ _ _ 6 _ _
- I approach the same way with the first question by trying to input number to see pattern
- 5^2 =
255^3 = 1
255^4 =
625 Yes! this is the one
- Then I start to ask myself about the loop again, would it has a loop? So I try to input a little more number
- 5^5 = ...125 Gotcha! we found the loop. And I know that there is going to be more than one number of n to have 5^n with the hundred digit of 6
- So it is not sufficient
(3)We try to think about both (1)&(2)- So the number should be when
11^n = _ _ _ 4 _
5^n = _ _ 6 _ _
- from the previous calculation, I know that when n=4, it is sufficient in both (1) & (2) since
11^4 = 146
41
5^4 =
625
- but then the question popup, I ask myself "
Would n=4 be the only number? or are there any matching numbers after that?"
- since I have already waste a lot of time, I have to think fast. I recognize the pattern that
for 11^n, the number for tens digit that equal to 4 is happened in the pattern n=4,14,24,34,... (since the 11^n pattern of tens digit runs from 0,1,2,3,4,..,9,0,1,... so that the number reach the loop every 10 numbers)
for 5^n, that the number for hundreds digit that equal to 6 is happened in the pattern n=4,6,8,10,12,14... (sine the 5^n pattern of hundred digits runs from 1,6,1,6,1,6,... so that the number reach the loop every 2 numbers)
- So I think my answer should be
E. This is because n could be 4,14,24,... and still suffice both (1) and (2), and that we can not find the exact number of n. (I hope I am correct
If I am wrong please give me suggestions. But to be honestly, at first, I answer A which is totally wrong)
My learning- we usually remember about the
unit digits pattern when odd/even integers expotent eg. in every 4 pattern
Odd1^4=1
3^4=1
5^4=5 (exceptional)
7^4=1
9^4=1
Even2^4=6
4^4=6
6^4=6
8^4=6
10^4=0 (exceptional)
But now we know that there are also pattern in tens, hundreds digits that could occur a loop. We dont have to remember every answer numbers. Just remember that there are loops about them.
- When we have to prove the numbers, we dont really need to calculate all of it, or we are running out of time, we just anticipate the future pattern (like for 5 the loop happens every 2 numbers, and for 11 the loop happens every 10 numbers)