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28 Jan 2022, 02:30
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B
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Difficulty:
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52% (02:20) correct
48%
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If k and m are positive integers, is \(3^{2k} + 7^{4m}\) divisible by 10?
(1) km is even
(2) 39k + 5 is even
Are You Up For the Challenge: 700 Level Questions
(1) km is even
(2) 39k + 5 is even
Are You Up For the Challenge: 700 Level Questions
28 Jan 2022, 03:31
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Pre-thinking
An integer is divisible by 10 if and only if its unit digit is 0.
3^(2k)=9^k. Its unit digit is 9 when k is odd; Its unit digit is 1 when k is even.
7^(4m)=(49*49)^(m). Its unit digit is always 1.
So the original question is equivalent to "is k odd or even"?
(1)Only
k can be even.
Or k can be odd (and m is even）。
Not sufficient.
(2)Only
For 39k + 5 to be even, 39k must be odd.
So k must be odd.
Sufficient.
The answer is (B).
An integer is divisible by 10 if and only if its unit digit is 0.
3^(2k)=9^k. Its unit digit is 9 when k is odd; Its unit digit is 1 when k is even.
7^(4m)=(49*49)^(m). Its unit digit is always 1.
So the original question is equivalent to "is k odd or even"?
(1)Only
k can be even.
Or k can be odd (and m is even）。
Not sufficient.
(2)Only
For 39k + 5 to be even, 39k must be odd.
So k must be odd.
Sufficient.
The answer is (B).
30 Jan 2022, 11:54
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If k and m are positive integers, is \(3^{2k} + 7^{4m}\) divisible by 10?
We can see, for whatever the integer value of m, \(7^{4m}\) provides a cyclicity of 4 and last digit will be 1. To divide the whole equation by 10, we need last digit of \(3^{2k}\) = 9
Stat1: km is even
Either k = even or m = even or both are even. If k = even, \(3^{2k} + 7^{4m}\) is not divisible by 10; but if k = odd, then \(3^{2k} + 7^{4m}\) is divisible by 10. Not sufficient.
Stat2: 39k + 5 is even
So, k = odd, then \(3^{2k} + 7^{4m}\) is divisible by 10. Sufficient.
So, I think B.
We can see, for whatever the integer value of m, \(7^{4m}\) provides a cyclicity of 4 and last digit will be 1. To divide the whole equation by 10, we need last digit of \(3^{2k}\) = 9
Stat1: km is even
Either k = even or m = even or both are even. If k = even, \(3^{2k} + 7^{4m}\) is not divisible by 10; but if k = odd, then \(3^{2k} + 7^{4m}\) is divisible by 10. Not sufficient.
Stat2: 39k + 5 is even
So, k = odd, then \(3^{2k} + 7^{4m}\) is divisible by 10. Sufficient.
So, I think B.
