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:
121/50 = 2 R21
9044/50 = 180 R44
25038/50 = 500 R38.

Examine the last two digits for small powers of 7 and look for a PATTERN:
7¹ = 07
7² = 49
7³ = 343
7� = 2401
7� = 16807.

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.

=>

The remainder when 7^{100} is divided by 50 depends only on the units and tens digits.

The units digits of 7^n cycle through the four values 7, 9, 3, and 1.
The tens digits of 7^n cycle through the four values 0, 4, 4, and 0.

We have the following sequence of units and tens digits for 7^n:

7^1 = 07 ~ 07
7^2 = 49 ~ 49
7^3 = 343 ~ 43
7^4 = 2401 ~ 01
7^5 = 16807~ 07
...

So, 7^{100} = (7^4)^{25} has the same units and tens digits as 7^4, that is, 01.
Thus, the remainder when 7^{100} is divided by 50 is 1.

Therefore, B is the answer.

Answer : B