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If n is a positive integer less than 10, k is a positive integer great

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nislam
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If n is a positive integer less than 10, k is a positive integer great [#permalink] If n is a positive integer less than 10, k is a positive integer great   Updated on: 30 Jul 2022, 02:02
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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
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E

Originally posted by nislam on 28 Jul 2022, 22:43.
Last edited by nislam on 30 Jul 2022, 02:02, edited 2 times in total.
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thisisit25
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If n is a positive integer less than 10, k is a positive integer great [#permalink] If n is a positive integer less than 10, k is a positive integer great   06 Jan 2023, 08:44
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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
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If n is a positive integer less than 10, k is a positive integer great [#permalink] If n is a positive integer less than 10, k is a positive integer great   30 Jul 2022, 04:58
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nislam wrote:
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


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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adityakaregamba
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Re: If n is a positive integer less than 10, k is a positive integer great [#permalink] If n is a positive integer less than 10, k is a positive integer great   05 Jun 2024, 06:09
thisisit25 wrote:
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

­Hi Bunuel And KarishmaB can you help me explain how n= 8 here is divisisble by 10 ?
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Re: If n is a positive integer less than 10, k is a positive integer great [#permalink] If n is a positive integer less than 10, k is a positive integer great   05 Jun 2024, 06:51
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adityakaregamba wrote:
thisisit25 wrote:
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

­Hi Bunuel And KarishmaB can you help me explain how n= 8 here is divisisble by 10 ?

­
We have \(n^{k+1}(1+n+n^2+n^3)\)­

For n = 8, the units digit of 1 + n + n^2 + n^3 is 5 and when this is multiplied by \(8^{k+1}\), the units digit becomes 0.
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Re: If n is a positive integer less than 10, k is a positive integer great [#permalink]
05 Jun 2024, 06:51
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