TheFerg wrote:
If x is a positive integer, what is the remainder when (7 ^ 12x+3) + 3 is divided by 5?
Can someone explain how to solve this in a different way than the
MGMAT CAT Test does?
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To find the remainder when a number is divided by 5, all we need to know is the units digit, since every number that ends in a zero or a five is divisible by 5.
For example, 23457 has a remainder of 2 when divided by 5 since 23455 would be a multiple of 5, and 23457 = 23455 + 2.
Since we know that x is an integer, we can determine the units digit of the number 712x+3 + 3. The first thing to realize is that this expression is based on a power of 7. The units digit of any integer exponent of seven can be predicted since the units digit of base 7 values follows a patterned sequence:
Units Digit = 7
Units Digit = 9
Units Digit = 3
Units Digit = 1
71
72
73
74
75
76
77
78
712x
712x+1
712x+2
712x+3
We can see that the pattern repeats itself every 4 integer exponents.
The question is asking us about the 12x+3 power of 7. We can use our understanding of multiples of four (since the pattern repeats every four) to analyze the 12x+3 power.
12x is a multiple of 4 since x is an integer, so 712x would end in a 1, just like 74 or 78.
712x+3 would then correspond to 73 or 77 (multiple of 4 plus 3), and would therefore end in a 3.
However, the question asks about 712x+3 + 3.
If 712x+3 ends in a three, 712x+3 + 3 would end in a 3 + 3 = 6.
If a number ends in a 6, there is a remainder of 1 when that number is divided by 5.
The correct answer is B.
I guess it should be \(7^{12x+3}+3\). The solution explains that the last digits of the powers of \(7\) repeat in a cyclical manner: \(7, 9, 3, 1, 7, 9, 3, 1,...\)
Meaning, if the exponent is a multiple of 4 + 1 (\(M4+1\) like 1, 5, 9,...), the last digit is \(7,\) for a \(M4+2\) (like 2, 6, 8,...) the last digit is \(9,\) etc.
\(1\) is also a \(M4 + 1\), because \(1 = 4\cdot0 + 1.\)
In this question, it is obvious the answer does not depend on the value of \(x\). But, I am going to tell you a secret: even if they say \(x\) is positive, you can still plug in \(0\) for \(x,\) the answer will be the correct one. So, just look at \(7^3+3\), and find the last digit, which is \(6,\) so the remainder when dividing by \(5\) is \(1.\)
They just state that \(x\) is positive, meaning at least \(1,\) so that you cannot raise \(7\) to the \(12\cdot1+3=15\)th power.
Since \(12x\) is a multiple of \(4,\) \(\,\,12x + 3\) is a \(M4+3.\) In other words, only the remainder is important, you can ignore \(12x\) or just take it as 0.
Answer B.
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