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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
A. 0
B. 1
C. 2
D. 3
E. 4
rjacobsMGMAT
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Expert reply
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.
_________________
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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Pansi
Joined: 04 Jul 2011
Last visit: 01 Dec 2019
Posts: 42
Status:Fighting hard
Schools: Kellogg '15 ISB '14 (S)
GMAT Date: 10-01-2012
Schools: Kellogg '15 ISB '14 (S)
Posts: 42
I have put in a solution in the below format. Plz see if it helps.
Attachments
Remainder Problem.png [ 9.72 KiB | Viewed 35934 times ]
General Discussion
[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
[(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
Posts: 70
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
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
Posts: 175
hogann
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!
(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
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
Posts: 297
hogann
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.
hogann
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
_________________
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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Posts: 154
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).
_________________
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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ablatt4
Joined: 18 Dec 2024
Last visit: 24 Sep 2025
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Concentration: Finance
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Posts: 87
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
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
Posts: 21
Akshara777
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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