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What is the remainder when \(7^{548}\) is divided by 10?

A. 1
B. 3
C. 7
D. 8
E. 9

Attachment:
2024-01-25_13-51-58.png

When dividing a positive integer by 10, the remainder is always the units digit of that integer. For instance, 123 divided by 10 yields the remainder of 3. Hence, essentially we need to find the units digit of \(7^{548}\).

    \(7^{548} = (7^2)^{274} = 49^{274} = (50 - 1)^{274}\).

When expanding \((50 - 1)^{274}\), all terms but the last one will have 50 as their factors making them divisible by 10 and the last term will be \((-1)^{274}=1\), which when divided by 10 yields the remainder of 1. Here, we could also note that when expanding we'll get all the terms with 50 and the last term \((-1)^{274}=1\), hence the sum would be something with the units digit of 1, giving the remainder of 1 when divided by 10.

Alternatively, we can use the cyclicity of 7 in positive integer power, which is four, meaning that the units digit of 7 in positive integer power repeats in blocks of four {7, 9, 3, 1}{7, 9, 3, 1}... The power, 548, is a multiple of 4, so the units digit of \(7^{548}\) will be the fourth number in the cyclicity block, which is 1, giving the remainder of 1 when divided by 10.

Answer: A.
­I am still having quite some trouble understanding how to come to the answer. Is there another approach to look at this?
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What is the remainder when \(7^{548}\) is divided by 10?

A. 1
B. 3
C. 7
D. 8
E. 9

Attachment:
2024-01-25_13-51-58.png

When dividing a positive integer by 10, the remainder is always the units digit of that integer. For instance, 123 divided by 10 yields the remainder of 3. Hence, essentially we need to find the units digit of \(7^{548}\).

    \(7^{548} = (7^2)^{274} = 49^{274} = (50 - 1)^{274}\).

When expanding \((50 - 1)^{274}\), all terms but the last one will have 50 as their factors making them divisible by 10 and the last term will be \((-1)^{274}=1\), which when divided by 10 yields the remainder of 1. Here, we could also note that when expanding we'll get all the terms with 50 and the last term \((-1)^{274}=1\), hence the sum would be something with the units digit of 1, giving the remainder of 1 when divided by 10.

Alternatively, we can use the cyclicity of 7 in positive integer power, which is four, meaning that the units digit of 7 in positive integer power repeats in blocks of four {7, 9, 3, 1}{7, 9, 3, 1}... The power, 548, is a multiple of 4, so the units digit of \(7^{548}\) will be the fourth number in the cyclicity block, which is 1, giving the remainder of 1 when divided by 10.

Answer: A.
­I am still having quite some trouble understanding how to come to the answer. Is there another approach to look at this?
­
Maybe practicing similar questions could help: Units digits, exponents, remainders problems
 
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What is the remainder when \(7^{548}\) is divided by 10?

A. 1
B. 3
C. 7
D. 8
E. 9

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guddo
What is the remainder when \(7^{548}\) is divided by 10?

A. 1
B. 3
C. 7
D. 8
E. 9

Attachment:
2024-01-25_13-51-58.png
­
 First we must remember that the remainder when dividing a number by 10 equals the units digit.

So, to determine the unit digit of 7^548, we must first see that 7 raised to a power has units digits in a repeating pattern of 4. If you don’t have this memorized, we can confirm it below:

7^1 has a untis digit of 7

7^2 has a units digit of 9

7^3 has a units digit of 3

7^4 has a units digit of 1

7^5 has a units digit of 7.

So, the pattern of units digits is 7-9-3-1 and most importantly, when 7 is raised to a power that is a multiple of 4, it has a units digit of 1.

Thus, since 548 is a multiple of 4, 7^548 has a units digit of 1, and therefore 5^548 divided by 10 also has a remainder of 1.

Answer: A
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↧↧↧ Detailed Video Solution to the Problem ↧↧↧




What is the remainder when \(7^{548}\) is divided by 10

Theory: Remainder of a number by 10 is same as the unit's digit of the number

(Watch this Video to Learn How to find Remainders of Numbers by 10)

Using Above theory Remainder of \(7^{548}\) by 10 = unit's digit of \(7^{548}\)

Now to find the unit's digit of \(7^{548}\), we need to find 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.
=> We need to divided 548 by 4 and check what is the remainder
=> 548 divided by 4 gives 0 remainder

=> \(7^{548}\) will have the same unit's digit as \(7^{Cycle}\) = \(7^4\)
=> Unit's digits of \(7^{548}\) = 1

So, Answer will be A
Hope it helps!

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

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