**If n is a positive integer, what is the remainder
When (7^(4n+3))(6^n) is divided by 10?
F. 1
G. 2
H. 4
I. 6
J. 8
Ans E
The sides of a square region, measured to the
nearest centimeter, are 6 centimeters long. The
least possible value of the actual area of the
square region is
A.36.00 sq cm
B.35.00 sq cm
C.33.75 sq cm
D.30.25 sq cm
E.25.00 sq cm
Ans : c
This is something very strange :
Doubts
This topic has expert replies
- ssmiles08
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1) I found it best to pick numbers for this problem.
n = 1
(7^(4n+3))(6^n) = (7^7)*6
you can figure out the pattern of the powers of 7 by multiplying its units digits.
7 49...
9*7 = 63 (units digit will be 3)
3*7 = 21 (units digit will be 1)
1*7 = 7 (units digit will be 7)
so the pattern goes: 7 9 3 1...repeat the set again. 7^7 power will have a units digit of 3.
units digit of 6^1 is 6
multiply both units digits; we get 6*3 = 18 (8 is the units digit)
for a number to be divisible by 10, it should end in 0. the product of (7^7)(6^1) is 8 digits away from the last zero number.
so remainder is 8.
2) I don't know why but I get D as my answer.
for the side to be round off to it's nearest centimeter, IMO it is 5.5 since it is the smallest number where it can be rounded off to 6.0
5.5^2 = (11/2)(11/2) = 121/4 = 30.25
n = 1
(7^(4n+3))(6^n) = (7^7)*6
you can figure out the pattern of the powers of 7 by multiplying its units digits.
7 49...
9*7 = 63 (units digit will be 3)
3*7 = 21 (units digit will be 1)
1*7 = 7 (units digit will be 7)
so the pattern goes: 7 9 3 1...repeat the set again. 7^7 power will have a units digit of 3.
units digit of 6^1 is 6
multiply both units digits; we get 6*3 = 18 (8 is the units digit)
for a number to be divisible by 10, it should end in 0. the product of (7^7)(6^1) is 8 digits away from the last zero number.
so remainder is 8.
2) I don't know why but I get D as my answer.
for the side to be round off to it's nearest centimeter, IMO it is 5.5 since it is the smallest number where it can be rounded off to 6.0
5.5^2 = (11/2)(11/2) = 121/4 = 30.25