IMO 4
7^1-7
^2-49
^3-343
^4-2401
^5-16807
^6-....649
.
.
.
and so on... if you see the 4 appears in the pattern shown when it is squared to 2 and its multiples.. get a calculator and see for the first 10 numbers....
tens' digit
This topic has expert replies
Source: Beat The GMAT — Problem Solving |
please elaborate some more on ur solution..shibal wrote:IMO 4
7^1-7
^2-49
^3-343
^4-2401
^5-16807
^6-....649
.
.
.
and so on... if you see the 4 appears in the pattern shown when it is squared to 2 and its multiples.. get a calculator and see for the first 10 numbers....
thanks
-
gmat_dest
- Master | Next Rank: 500 Posts
- Posts: 186
- Joined: Fri Jan 16, 2009 4:57 am
- Thanked: 7 times
- GMAT Score:720
Please check the pattern.
The tens digit follows a pattern of 0, 4, 4, 0 for first, second, third, fourth square powers of 7. This pattern is repeated.
so, 2002 = 2000/4 + 2.
So, after 2000 times, the tens digit will be 0.
after 2002 times, the tens digit will be 4.
Hope you got it.
The tens digit follows a pattern of 0, 4, 4, 0 for first, second, third, fourth square powers of 7. This pattern is repeated.
so, 2002 = 2000/4 + 2.
So, after 2000 times, the tens digit will be 0.
after 2002 times, the tens digit will be 4.
Hope you got it.
-
wanttobeat
- Junior | Next Rank: 30 Posts
- Posts: 14
- Joined: Tue Jul 14, 2009 1:28 am
- Thanked: 1 times
I have found another way of solving the problem...but aint sure if the mechanism is alright:
7^2002 = (7^2)^1001 = 49^1001.
If we look closely, we can find that the last two digits of the result for 49^1001 will be 49, because 49 is being multiplied by 01 of 1001 while deriving the result.
So the answer is 4.
7^2002 = (7^2)^1001 = 49^1001.
If we look closely, we can find that the last two digits of the result for 49^1001 will be 49, because 49 is being multiplied by 01 of 1001 while deriving the result.
So the answer is 4.
