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Re: Tuesday Q5 - Remainder [#permalink]
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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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[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
so the answer is B
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Re: Tuesday Q5 - Remainder [#permalink]
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(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

B is the answer
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Re: Tuesday Q5 - Remainder [#permalink]
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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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If x is a positive integer, what is the remainder when 7^(12 [#permalink]
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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


For remainder upon division by 5, we just need to find the units digit of the number.

7 has a cyclicity of 7, 9, 3, 1
Since we have 7^(12x + 3), 12x gives us full cycles so it will end on a 3.

So 7^(12x + 3) + 3 has the units digit of 6. So when divided by 5, remainder will be 1.

Answer (B)

Originally posted by KarishmaB on 06 Mar 2019, 03:31.
Last edited by KarishmaB on 04 Oct 2022, 00:27, edited 1 time in total.
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Re: (7 ^ 12x+3) + 3 divided by 5; what is the remainder? [#permalink]
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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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Re: If x is a positive integer, what is the remainder when 7^(12 [#permalink]
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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.

Now, addind 3+3 = 6.

6/5 gives remainder 1.

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

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

Answer: B
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Re: If x is a positive integer, what is the remainder when 7^(12 [#permalink]
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]
I made a careless mistake here marking it as remainder = 0 because I subtracted +3 from the RHS at the end of my calculations

7^(12x+3) + 3 = 5q + ?/5
Cyclicity of 7 is 7,8,3,1
when x=1, exponent is 15, divided by 4 gives remainder 3

Should have written: 7^(12x+3) = 5q + (3+?)/5
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