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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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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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hogann
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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(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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hogann
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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hogann
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)
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We know to find what is the remainder when \( 7^{(12x+3)} + 3 \) is divided by 5

Theory: Remainder of a number by 5 is same as the Remainder of the unit's digit of the number by 5

(Watch this Video to Learn How to find Remainders of Numbers by 5)

We can do this by finding the pattern / cycle of unit's digit of power of 7 and then generalizing it.

Unit's digit of \(7^1\) = 7
Unit's digit of \(7^2\) = 9
Unit's digit of \(7^3\) = 3
Unit's digit of \(7^4\) = 1
Unit's digit of \(7^5\) = 7

So, unit's digit of power of 7 repeats after every \(4^{th}\) number.
=> Remainder of 12x + 3 by 4 = 3
=> Units' digit of \( 7^{(12x+3)} + 3 \) = Units' digit of \(7^3\) + 3 = 3 + 3 = 6

=> Remainder = Remainder of 6 by 5 = 1

So, Answer will be B
Hope it helps!

MASTER How to Find Remainders with 2, 3, 5, 9, 10 and Binomial Theorem

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I will break down [7^(12x+3) + 3]/5 as 7^(12x+3) /5 + 3/5
Since we are dealing only with remainders, I can rewrite 7^(12x+3) /5 as 2^(12x+3) /5 because 7/5 leaves a remainder of 2.

Now, find out the pattern of remainders when different powers of 2 are divided by 5:
2^1/5 --> R = 2
2^2/5 --> R = 4
2^3/5 --> R = 3
2^4/5 --> R = 1
2^5/5 --> R = 2 again.

So we get a repeating pattern of (2, 4, 3, 1), which is a cycle of 4 terms.

Now, look at the exponent in 7^(12x+3) /5 and notice that 12x is a multiple of 4. Regardless of what x is, the exponent represents 12x complete cycles of remainders + 3, which means that the remainder will be the 3rd term in our pattern --> R = 3 when 7^(12x+3) is divided by 5.

We know that 3/5 also leaves a remainder of 3. Finally, add both remainders and divide by 5 --> (3+3)/5 = 6/5 --> R = 1.

The answer is (B).
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7^(12x+3) +3 / 5 =?

the pattern with 7 to any power is 7,9,3,1
using this pattern we can plug in integers for x, and we will find that no matter what we plug in (on account of 12 being a multiple of 4) we will get the 3 term of 3 in our sequence.

If we know we will get 3 as the units digit, and we add 3 we get 6, which means when we divide by 5 we will geta remainder of 1
hogann
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
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can we substiture x as any positive integar in this sum? I am not understanding where the question means all values and where it doesent. Please help Bunuel KarishmaB
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Akshara777
can we substiture x as any positive integar in this sum? I am not understanding where the question means all values and where it doesent. Please help Bunuel KarishmaB
Yes, you can substitute any positive integer. Since this is a Problem Solving question with a single correct answer and x is restricted to positive integers, the remainder must be the same for every positive integer value of x.
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