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# If x is a positive integer, what is the remainder when 7^(12

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

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Updated on: 17 Oct 2013, 02:08
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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

Originally posted by hogann on 13 Oct 2009, 06:33.
Last edited by Bunuel on 17 Oct 2013, 02:08, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: (7 ^ 12x+3) + 3 divided by 5; what is the remainder?  [#permalink]

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07 Oct 2012, 11:02
8
4
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.
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Re: (7 ^ 12x+3) + 3 divided by 5; what is the remainder?  [#permalink]

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06 Oct 2012, 10:29
8
I have put in a solution in the below format. Plz see if it helps.
Attachments

Remainder Problem.png [ 9.72 KiB | Viewed 10790 times ]

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Re: Tuesday Q5 - Remainder  [#permalink]

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13 Oct 2009, 07:53
2
1
[7^( 12x + 3 ) + 3 ] can be rewritten as -

[(7^(12x) . 7^3) + 3 ]
[ (7^(4)(3x) . 7^3 ) +3 ]

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
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Re: Tuesday Q5 - Remainder  [#permalink]

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13 Oct 2009, 11:20
5
2
7^(12x+3) + 3 is divided by 5

for x = 1 >> R = 1

7^(15) + 3 ..cyclicity of 7 is 4 hence 15/4 ... R >> 3 corresponds to 7,9,3,1
hence 3+3 = 6 /5 R = 1 OA B
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Re: Tuesday Q5 - Remainder  [#permalink]

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31 Jan 2010, 06:32
2
hogann wrote:
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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Re: Tuesday Q5 - Remainder  [#permalink]

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13 Feb 2010, 00:24
2
(7^(12x+3) + 3)/5

x is a positive integer, so we can put the value 1,2,3......

for x=1 x=2 or x=3
7^15 7^27 7^39
7 has the cycilicity of 4
15 mod 4, 27 mod 4 or 39 mod 4 leads to same result which is 7^3 will be unit digit = 343.

343+3/5 = gives 1 as remainder

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

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06 Oct 2012, 10:24
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.
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Re: (7 ^ 12x+3) + 3 divided by 5; what is the remainder?  [#permalink]

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06 Oct 2012, 10:37
1
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.

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.

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.

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

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06 Oct 2012, 10:41
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.
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Re: If x is a positive integer, what is the remainder when 7^(12  [#permalink]

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05 Sep 2013, 11:41
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!!!
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Re: If x is a positive integer, what is the remainder when 7^(12  [#permalink]

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25 Oct 2013, 19:25
I plugged numbers into the formula and did calculation to the 15th power to determine units digits
7^(12x+3)+3 using x=1

1,2 - 7's = 49
3,4 - 7's = 49 ---> 1,2,3,4 7's = 49*49 ---> units digits = 1
5,6 - 7's = 49
7,8 - 7's = 49 ---> 5,6,7,8 7's = 49*49 ---> units digits = 1
9,10 - 7's = 49
11,12 - 7's = 49 ---> 9,10,11,12 - 7's = 49*49 ---> units digits = 1
13,14 - 7's = 49
15 - 7's = 7 ---> 13,14,15 - 7's = 49*7 ---> units digits = 3

1 * 1 * 1 * 3 = 3 + 3 = 6/5= 1
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Re: If x is a positive integer, what is the remainder when 7^(12  [#permalink]

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14 Nov 2013, 05:42
1
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

Way I did it was:

Pick x=1.
7^15+3/5

Then split them

7^15/5 + 3/5

Then 7 = (5+2) ^15

So then every term will be divisible by 5 except 2^15.
The units digit on 2^15 is going to be 8. So 8/5 Remainder 3.

6/5 gives remainder 1.

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

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27 Jan 2014, 05:41
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

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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, 15:20
1
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.
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Re: If x is a positive integer, what is the remainder when 7^(12  [#permalink]

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14 Feb 2014, 15:43
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.

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

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15 Aug 2015, 00:12
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

7 cong 2 mod 5
7^2 cong 4 mod 5
7^3 cong 3 mod 5
7^4 cong 1 mod 5
7^5 cong 2 mod 5 and so on
the cycle is 2, 4, 3, 1
put x=1,2,3
we get the exponent of 7 as 15, 27 and so on
when you divide 15 or 27 or 39 by 4 you always get a remainder 3
therefore, for the first part of term, the remainder is 3
3+3=6; when divided by 5, the remainder is 1
Hence, the correct option is B
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Re: If x is a positive integer, what is the remainder when 7^(12  [#permalink]

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05 Dec 2016, 01:41
1
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

We can solve this question without plugging in numbers.

$$7^{(12x+3)} = 7^{3*(4x+1)}$$

Unit digit of 7 has a cycle of 4 (7, 9, 3, 1)

$$\frac{3*(4x+1)}{4} = \frac{3*(0+1)}{4} = \frac{3}{4}$$ ---> remainder is 3 so our power of 7 is 3 with units digit 3.

and we have $$\frac{3 + 3}{5} = \frac{6}{5}$$

Remainder 1.
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Re: If x is a positive integer, what is the remainder when 7^(12  [#permalink]

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