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E
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Difficulty:
95%
(hard)
Question Stats:
53% (02:52) correct
47%
(02:23)
wrong
based on 169
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If \(k\) is an odd positive integer greater than 10, which of the following must be divisible by 10?
I) \(2^k + 3^k + 7^k+ 8^k\)
II) \(9^k + 5^k − 4^k\)
III) \(2^k + 3^k + 5^k\)
A) none
B) I only
C) II only
D) I and II only
E) I, II and III
I) \(2^k + 3^k + 7^k+ 8^k\)
II) \(9^k + 5^k − 4^k\)
III) \(2^k + 3^k + 5^k\)
A) none
B) I only
C) II only
D) I and II only
E) I, II and III
20 Jan 2022, 23:38
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nislam
Divisibility by 10 depends on the units digit of the expression or the term.
So, we look for whether the units digit is 0.
There is a cyclicity of 4 when it comes to units digit of successive powers. Here, we are looking for odd powers.
11, 15, 19 and so on are of type \(a^{4k+3}\), so they will have same units digit as \(a^3.\)
13, 17, 21 and so on are of type \(a^{4k+1}\), so they will have same units digit as \(a^1.\)
Hence, if the expressions with k as both 1 and 3 gives 0 as units digit, the answer is ‘yes’, otherwise ‘no’.
I) \(2^k + 3^k + 7^k+ 8^k\)
k=1….. \(2^1 + 3^1 + 7^1+ 8^1=2+3+7+8=20\)…YES
k=3……. \(2^3 + 3^3 + 7^3+ 8^3=8+7+3+2=20\)…YES
Answer is YES
II) \(9^k + 5^k − 4^k\)
k=1…… \(9^1 + 5^1 − 4^1=9+5-4=10\)….YES
k=3……. \(9^3 + 5^3 − 4^3=9+5-4=10\)….YES
Answer is YES
III) \(2^k + 3^k + 5^k\)
k=1….. \(2^1 + 3^1 + 5^1=2+3+5=10\)…YES
k=3……. \(2^3 + 3^3 + 5^3=8+7+5=20\)…YES
Answer is YES
All three are divisible by 10.
E
General Discussion
09 Apr 2024, 10:50
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Cyclicity of the exponents is the key here for all the numbers the cyclicity is matching. Given that k is odd.
For option i
Cyclicity of 2,3,7,8 is same i.e. 4.
Last digit of exponents of 2 is.
2, 4, 8, 6
Last digit of exponents of 3 is. 3, 9, 7, 1
Last digit of exponents of 7 is. 7, 9, 3, 1
Last digit of exponents of 8 is. 8, 4, 2, 6
As k is odd we have to consider only 1st and third digits of each cyclicity. Here we can see their addition obtain multiples of 10. Like this we can check for rest.
Answer E
Posted from my mobile device
For option i
Cyclicity of 2,3,7,8 is same i.e. 4.
Last digit of exponents of 2 is.
2, 4, 8, 6
Last digit of exponents of 3 is. 3, 9, 7, 1
Last digit of exponents of 7 is. 7, 9, 3, 1
Last digit of exponents of 8 is. 8, 4, 2, 6
As k is odd we have to consider only 1st and third digits of each cyclicity. Here we can see their addition obtain multiples of 10. Like this we can check for rest.
Answer E
Posted from my mobile device
