Divisbility Problem

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championspunch wrote:If x is a positive integer, what is the remainder when 7^(12x+3) + 3 is divided by 5?
Note that, we only need to know the units digit of 7^(12x + 3) + 3
Now, units digit of 7^4 is 1.

So, units digit of 7^(12x + 3)
= units digit of [7^(12x)]*[7^3]
= [units digit of 7^(12x)]*[units digit of 7^]
= [units digit of (7^4)^(3x)]*[units digit of 7^3]
= [units digit of 1^(3x)]*[units digit of 7^3]
= 1*3
= 3

Hence, units digit of 7^(12x + 3) + 3 is (3 + 3) = 6
Therefore, the remainder when 7^(12x + 3) + 3 is divided by 5 = remainder when 6 is divided 5 = 1
Last edited by Anju@Gurome on Mon Apr 08, 2013 7:33 pm, edited 1 time in total.
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by GMATGuruNY » Mon Apr 08, 2013 6:42 pm
Here is the correct question, along with the answer choices:
If x is a positive integer, what is the remainder when 7^(12x+3) + 3 is divided by 5?

A. 0
B. 1
C. 2
D. 3
E. 4
We can plug in any positive integer for x.
If x=1, then 7^(12x+3) + 3 = 7^15 + 3.

Question rephrased: What is the remainder when 7^15 + 3 is divided by 5?
When an integer is divided by 5, the remainder is determined by the UNITS DIGIT.
Thus, we need to know the units digit of 7^15 + 3.

When an integer is raised to consecutive powers, the resulting units digits repeat in a CYCLE:
7¹ --> units digit of 7.
7² --> units digit of 9. (Since the product of the preceding units digit and 7 = 7*7 = 49.)
7³ --> units digit of 3. (Since the product of the preceding units digit and 7 = 9*7 = 63.)
7� --> units digit of 1. (Since the product of the preceding units digit and 7 = 3*7 = 21.)
From here, the units digits will repeat in the same pattern: 7, 9, 3, 1.
The units digit repeat in a CYCLE OF 4.
Implication:
When 7 is raised to a power that is a multiple of 4, the units digit will be 1.

Thus, 7^12 has a units digit of 1.
From here, the cycle of units digits will repeat:
7^13 --> units digit of 7.
7^14 --> units digit of 9.
7^15 --> units digit of 3.

Since 7^15 has a units digit of 3, 7^15 + 3 has a units digit of 6.
When an integer with a units digit of 6 is divided by 5, the remainder in each case is 1:
16/5 = 3 R1.
26/5 = 5 R1.
36/5 = 7 R1.
Thus, when 7^15 + 3 is divided by 5, the remainder will be 1.

The correct answer is B.
Last edited by GMATGuruNY on Mon Apr 08, 2013 8:00 pm, edited 1 time in total.
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by championspunch » Mon Apr 08, 2013 7:56 pm
Thanks GMATGuruNy!!