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If \(n\) is a positive integer less than 10, \(k\) is a positive integer greater than 10, and \(n^{k+4} + n^{k+3} + n^{k+2} + n^{k+1}\) is divisible by 10, \(n\) could be equal to
I) 3
II) 7
III) 8
A) none
B) I only
C) II only
D) I and II only
E) I, II and III
I) 3
II) 7
III) 8
A) none
B) I only
C) II only
D) I and II only
E) I, II and III
06 Jan 2023, 08:44
Kudos
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n is a positive integer less than 10, k is a positive integer greater than 10
Is \(n^{k+4} + n^{k+3} + n^{k+2} + n^{k+1}\) divisible by 10?
--> expression should end in 10.
take common term \(n^{k+1} [1+n+n^2+n^3]\)
1. n = 3 --> 1+3+3x3+3x3x3 = 1+3+9+27 = 40 = divisible by 10.
2. n = 7 --> 1+7+7x7+7x7x7 = 1+7+49+343 = 400 = divisible by 10.
3. n = 8 --> 1+8+8x8+8x8x8 = 1+8+64+512 = ends in 5 = NOT divisible by 10.
BUT: it ends in 5! which means, if I multiple it by anything that has a factor of two, I will get the last digit as 10!
Powers of 8 = 8, 64, 512, _ _ 6...all multiplied by something that ends with 5 result in a 10.
Answer: all three
Is \(n^{k+4} + n^{k+3} + n^{k+2} + n^{k+1}\) divisible by 10?
--> expression should end in 10.
take common term \(n^{k+1} [1+n+n^2+n^3]\)
1. n = 3 --> 1+3+3x3+3x3x3 = 1+3+9+27 = 40 = divisible by 10.
2. n = 7 --> 1+7+7x7+7x7x7 = 1+7+49+343 = 400 = divisible by 10.
3. n = 8 --> 1+8+8x8+8x8x8 = 1+8+64+512 = ends in 5 = NOT divisible by 10.
BUT: it ends in 5! which means, if I multiple it by anything that has a factor of two, I will get the last digit as 10!
Powers of 8 = 8, 64, 512, _ _ 6...all multiplied by something that ends with 5 result in a 10.
Answer: all three
30 Jul 2022, 04:58
Kudos
Bookmarks
nislam
Divisibility by 10 depends on the unit’s digit, that is the unit’s digit as 0 will make that number divisible by 10.
\(n^{k+4} + n^{k+3} + n^{k+2} + n^{k+1}\) is nothing but sum of 4 consecutive powers.
We know that the consecutive powers of each digit have a cyclicity of the units digit. They could repeat after every 2 or 4 powers or could also be same.
Let us work on the two properties: units digit as 0 and cyclicity of units digit when increased to consecutive powers
Same : Numbers ending with digits 0, 1, 5, and 6 will have sum as 4*0, 4*1, 4*5 and 4*6 or 0,4,0 and4.
Repeat after 2: Numbers ending with digits 4 and 9 will have sum as 4+6+4+6 and 9+1+9+1 or 0 and 0.
Repeat after 4: Numbers ending with digits 2, 3, 7 and 8 will have sum as 2+4+6+8 and 1+3+7+9 or 0 and 0 again.
Thus, all numbers ending with 0, 2, 3, 4, 5, 7, 8 and 9 will have a sum ending with 0, a property that will make the integer divisible by 10.
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