Danou wrote:
Bunuel wrote:
guddo wrote:
What is the remainder when\(7^{548}\) is divided by 10?
A. 1
B. 3
C. 7
D. 8
E.9
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,
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Statistics : Posted by Bunuel • on 25 Jan 2024, 04:03 • Replies 4 • Views 1664
