Max@Math Revolution wrote:[GMAT math practice question]
What is the remainder when 7^100 is divided by 50?
A. 0
B. 1
C. 7
D. 21
E. 49
If the last two digits of an integer form a value less than 50, then dividing the integer by 50 will yield a remainder equal to the last 2 digits of the integer:
1
21/50 = 2 R
21
90
44/50 = 180 R
44
250
38/50 = 500 R
38.
Examine the last two digits for small powers of 7 and look for a PATTERN:
7¹ =
07
7² =
49
7³ = 3
43
7� = 24
01
7� = 168
07.
7� has the same last two digits as 7¹, implying the last two digits for consecutive powers of 7 repeat in the following cycle:
07, 49, 43, 01...07, 49, 43, 01...07, 49, 43, 01...
Since the last two digits repeat in a CYCLE OF 4, raising 7 to a power that is a multiple of 4 will always yield 01 for the last two digits.
Since the exponent for 7¹�� is a multiple of 4, the last two digits for 7¹�� must be 01.
Thus, dividing 7¹�� by 50 will yield a remainder of 1.
The correct answer is
B.
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