What is the units digit of the solution to 177^28 - 133^23 ?
(A) 1
(B) 3
(C) 4
(D) 6
(E) 9
UNITS DIGIT
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177^28 ends in 1 (units digit of powers of 7 follow the cycle 7, 9, 3, 1)shrey2287 wrote:What is the units digit of the solution to 177^28 - 133^23 ?
(A) 1
(B) 3
(C) 4
(D) 6
(E) 9
and 133^23 ends in 7 (units digit of powers of 3 follow the cycle 3,9,7,1)
So, a number ending in 1 - a number ending in 7 ends in 4.
The correct answer is 4, (C).
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Hey- the key to solve these type of questions is to learn the cyclicity of numbershrey2287 wrote:What is the units digit of the solution to 177^28 - 133^23 ?
(A) 1
(B) 3
(C) 4
(D) 6
(E) 9
The cyclicity of 7 is 4 ie after 7^4 the cycle of power repeats itself. Starting from 7,9,3,1 and then again 7.
Similarly the cyclicity of 3 is also 4.
Divide the power by 4 and see the remainder.
For the first the units digit will be 1 and for second number it is 7
1-7 or 11-7=4 (carry 1 makes 1 equal to 11)
Answer =4
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Solution:
To solve this problem, we need to use the following two facts:
1) If u is the units digit of a positive integer n, then n^m and u^m have the same units digit.
2) Let m and n be the units digits of M and N. If M - N is positive but m - n is negative, then the units digit of M - N is m - n + 10.
Using the first fact, we see that 177^28 has the same units digit as 7^28, and 133^23 has the same units digit as 3^23.
Since the units digit pattern of powers of 7 is 7-9-3-1,we see that 7^28 has units digit of 1. Likewise, since the units digit pattern of powers of 3 is 3-9-7-1, then 3^23 has a units digit of 7.
Using the second fact, we see that the units digit of 177^28 - 133^23 is 1 - 7 + 10 = 4.
Answer: C
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