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Many good insights above. I'll add: there is actually a whole class of questions on the GMAT that ask either "what is the remainder when x is divided by 5" or "what is the remainder when x is divided by 10"? Both of these questions can be answered just by knowing the units digit of x. The reason why that works should be fairly clear if you do a few division problems (try finding the remainder for 7÷5, 37÷5, and 127÷5).

Once you understand that part of the problem, the question becomes "what is the units digit of (7 ^ 12x+3) + 3?" To figure that out, we first need to know the units digit of (7 ^ 12x+3), since the final +3 will be easy to deal with.

If you have no idea what to do at this point, I think the best thing is just to figure out what the units digit of 7^1 is, then of 7^2, then of 7^3, then of 7^4, etc. It's easy for the first two (7 and 9), but remember that to find the units digit of 7^3, you don't actually need to figure out 49*7. You just need to figure out 9*7. 9*7 is 63 (and by the way, 40*7 is 280 - since that ends in a 0, it won't affect the units digit when we add it to 63 to figure out what 7^3 is). So the units digit of 7^3 is 3. And the units digit of 7^4 is therefore the same as the one of 3*7, or 1. The units digit of 7^5 is 7 again, because 1*7 is 7. Note that once you get to a units digit of 1, that's when the pattern starts repeating. Check out Pansi's chart above.

So assume x=1. The units digit of 7^15 is the same as 7^11, 7^7, and 7^3. That would be 3. Add 3 to that and you get 6. Divide by 5 and you get a remainder of 1.

Final note: 7^x isn't the only base with an interesting units digit pattern: 2^x goes 2-4-8-6 3^x goes 3-9-7-1 4^x goes 4-6 5^x is always 5 6^x is always 6 8^x goes 8-4-2-6 9^x goes 9-1 and by the way, 17^x goes 7-9-3-1 just like 7^x does. So does 27^x. You get the idea. No need to memorize these patterns, just understand how to figure them out.
_________________

Ryan Jacobs | Manhattan GMAT Instructor | San Francisco

now 7 cube is 343 the unit digit is '3', and 7 raised to the power of 4 is 2401, unit digit is 1 and inturn raised to the power 3x will result with unit digit as 1,

so the equations boils down to [( 1 . 3 ) + 3]/5 which is 1 so the answer is B

If x is a positive integer, what is the remainder when 7^(12x+3) + 3 is divided by 5? 1. 0 2. 1 3. 2 4. 3 5. 4

Edit - Fixed Exponent

7^(12x+3)Mod 5 + 3 Mod 5 = (7 mod 5)^12x * (7 mod 5)^3 + 3 = (2 mod 5)^12x * (2^3 Mod 5) + 3 = (4 mod 5)^6x * (8 Mod 5) + 3 = (-1)^6x * (-2) + 3 = 1 * (-2) + 3 = 1

Therefore B!
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|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

Can someone explain how to solve this in a different way than the MGMAT CAT Test does?

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.

Re: (7 ^ 12x+3) + 3 divided by 5; what is the remainder? [#permalink]

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06 Oct 2012, 09:37

1

This post received KUDOS

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?

0 1 2 3 4

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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PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: (7 ^ 12x+3) + 3 divided by 5; what is the remainder? [#permalink]

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06 Oct 2012, 09:24

Honestly, I don't see another way.

With this specific question, you can also assume x=0 and get the correct result, although the question states that this is not allowed. But since this is a dangerous approach, where you have to know that it will lead to the same result as a positive integer, I don't recommend it.

But anyways: if x=0, then 7^3 + 3. 49*7+3=346, so remainder is 1.

Re: (7 ^ 12x+3) + 3 divided by 5; what is the remainder? [#permalink]

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06 Oct 2012, 09:41

Zinsch123 wrote:

Honestly, I don't see another way.

With this specific question, you can also assume x=0 and get the correct result, although the question states that this is not allowed. But since this is a dangerous approach, where you have to know that it will lead to the same result as a positive integer, I don't recommend it.

But anyways: if x=0, then 7^3 + 3. 49*7+3=346, so remainder is 1.

In this case, no danger to consider x = 0. The only restriction should be 12x + 3 non-negative. The powers of 7 have their last digit repeating cyclically, and for x = 0 the exponent is positive, so no problem.
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: If x is a positive integer, what is the remainder when 7^(12 [#permalink]

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05 Sep 2013, 10:41

Without having to keep writing out beyond the basic pattern of the unit digits of 7, which is 7,9,3,1..... how do you determine that 7^12X is multiple of 4 instead of 2 or 3 and use that respective digit? How do you determine that 7^15 unit digit is based on 7^3 instead of 7^5. I opted to solve by plugging in 1 for X and thought that the unit digit for 7^15 was 7 because I based it on 7^1 and 7^5. Thanks!!!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If x is a positive integer, what is the remainder when 7^(12 [#permalink]

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03 Feb 2014, 14:20

The way i did it : 7^(12x+3) -- > 7^(12x) * 7^3 7 has a cyclicity of 4 so - 12x/4 will always give us a remainder of 0 thus last digit is 1. - 3/4 remainder 3 so the last digit is 3 . So we know the last digit of 7^(12x+3) is 3*1 = 3 That last digit will be added to 3 since we have 7^(12x+3) + 3 so 3 +3 = 6 6/5 will give us a remainder of 1 so B.

Re: If x is a positive integer, what is the remainder when 7^(12 [#permalink]

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14 Feb 2014, 14:43

Different approach:

Since x is a positive integer, assume x = 1.

7^(12x+3) + 3 = 7^(15) + 3 = (5+2)^15 + 3

The remainder of 5^15 / 5 will be 0, so that leaves us with 2^15 + 3

Cycle of 2^15 becomes easy to work with: 2^1 = 2 2^2 = 4 2^3 = 8 2^4 = 16 [units digit of 2^x repeats] ... Working through the cycle, we know that 2^15 = xx8. So, the final number will be xx8 + 3 = xx1. xx1/5 will leave a remainder of 1.

Re: If x is a positive integer, what is the remainder when 7^(12 [#permalink]

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21 Mar 2014, 04:50

jlgdr wrote:

hogann wrote:

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

Binomial method works well here. First lets assume x=1

so 7^12x+3 + 3 / 5 what is the remainder?

7^15 + 3 /5

(5+2)^15 / 5, all terms will be divisible by 5 except 2^15

What is remainder of 2^15 / 5?

2^15 = 2* 4^7 = 2(5-1)^7

All terms will again be divisible by 5 except -1^7 Then -1*2 = -2 + 3 = 1

Hence we have a remainder of 1 when divided by 7

Answer is thus B

Cheers! J

Just wondering is it possible to multiply both numerator and denominator by 2 in order to get 10 as denominator? That way we would only need to worry about the units digit of the expression

Bunuel, could you clarify whether this approach is valid?