BMAD Posted August 18 Report Share Posted August 18 Assume that n is a natural number, prove that n and n^{5} will always have the same one's digit. e.g. 13 and 13^{5}=371,293 both end in 3. Quote Link to comment Share on other sites More sharing options...

0 EventHorizon Posted August 26 Report Share Posted August 26 Proof 1: Modulus 10 Spoiler The last digit is just the number mod 10. (x * y) mod z = ((x mod z) * (y mod z)) mod z, so we just need to check x^5 mod 10 for each of 0-9. x,^2,^3,^4,^5 0,0,0,0,0 1,1,1,1,1 2,4,8,6,2 3,9,7,1,3 4,6,4,6,4 5,5,5,5,5 6,6,6,6,6 7,9,3,1,7 8,4,2,6,8 9,1,9,1,9 Done. Proof 2: Modulus 5 Spoiler Let n be a natural number. n can be represented as 5x + a, where x is an integer and a is one of {-2,-1,0,1,2} . Case 1: a = 0 Any power of an odd multiple of 5 remains an odd multiple of 5, and must have a 5 in the ones digit. Any power of an even multiple of 5 remains an even multiple of 5, and must have a 0 in the evens digit. So n^5 has the same ones digit as n. Case 2: a = 1 or -1 (5x +/- 1) * (5x +/- 1) = 25x^2 +/- 10x + 1 = 1 mod 5 Squaring again it remains 1 mod 5. Case 3: a = 2 or -2 (5x +/- 2) * (5x +/- 2) = 25x^2 +/- 20x + 4 = 4 mod 5 From case 2, we see if we square something 4 mod 5 (a=-1), it becomes 1 mod 5. Finishing cases 2 and 3: n^4 = 1 mod 5, so it can be represented as 5y+1 for an integer y. (5y+1)*(5x+a) = 25xy + 5x + 5ay + a = a mod 5. So n^5 mod 5 = a mod 5 = n mod 5. n^5 mod 2 = n mod 2 since any power of an even number remains even and any power of an odd number remains odd. n^5 mod 10 = n mod 10, because 5 and 2 are relatively prime. Same mod 10 = same last digit. Done. It's also interesting from this proof... n^4 ends in 1 for odd non multiples of 5, n^4 ends in 6 for even non multiples of 5. Also, no fourth power of any integer ends in 2,3,4,7,8, or 9. I have a third proof idea... I'll see if I can flush it out later. Are you a math professor/researcher? If not, where do you find / get inspiration for these? Quote Link to comment Share on other sites More sharing options...

0 EventHorizon Posted August 28 Report Share Posted August 28 Proof 3: Factors of x^5-x Spoiler If x differs by x^5 by a multiple of 10, they must have the same one's digit. x^5-x = x * (x^4-1) = x * (x^2-1) * (x^2+1) = x * (x-1) * (x+1) * (x^2+1) Since x and x-1 are factors, x^5-x is even and we only need to show it is also a multiple of 5. If x is a multiple of 5 or one away from a multiple of 5, x^5-x is a multiple of 5 since x, x-1, and x+1 are all factors. The remaining possibilities to check differ from multiples of 5 by 2, so lets substitute them into the final factor. x^2+1 = (5a +/- 2) * (5a +/- 2) + 1 = (25a^2 +/- 20a + 4) + 1 = 25a^2 +/- 20a +5 = 5 * (5a^2 +/- 4a + 1) Since x^5-x will always be a multiple of 10, x and x^5 must have the same one's digit. Quote Link to comment Share on other sites More sharing options...

0 BMAD Posted September 10 Author Report Share Posted September 10 On 8/25/2023 at 8:58 PM, EventHorizon said: Proof 1: Modulus 10 Reveal hidden contents The last digit is just the number mod 10. (x * y) mod z = ((x mod z) * (y mod z)) mod z, so we just need to check x^5 mod 10 for each of 0-9. x,^2,^3,^4,^5 0,0,0,0,0 1,1,1,1,1 2,4,8,6,2 3,9,7,1,3 4,6,4,6,4 5,5,5,5,5 6,6,6,6,6 7,9,3,1,7 8,4,2,6,8 9,1,9,1,9 Done. Proof 2: Modulus 5 Reveal hidden contents Let n be a natural number. n can be represented as 5x + a, where x is an integer and a is one of {-2,-1,0,1,2} . Case 1: a = 0 Any power of an odd multiple of 5 remains an odd multiple of 5, and must have a 5 in the ones digit. Any power of an even multiple of 5 remains an even multiple of 5, and must have a 0 in the evens digit. So n^5 has the same ones digit as n. Case 2: a = 1 or -1 (5x +/- 1) * (5x +/- 1) = 25x^2 +/- 10x + 1 = 1 mod 5 Squaring again it remains 1 mod 5. Case 3: a = 2 or -2 (5x +/- 2) * (5x +/- 2) = 25x^2 +/- 20x + 4 = 4 mod 5 From case 2, we see if we square something 4 mod 5 (a=-1), it becomes 1 mod 5. Finishing cases 2 and 3: n^4 = 1 mod 5, so it can be represented as 5y+1 for an integer y. (5y+1)*(5x+a) = 25xy + 5x + 5ay + a = a mod 5. So n^5 mod 5 = a mod 5 = n mod 5. n^5 mod 2 = n mod 2 since any power of an even number remains even and any power of an odd number remains odd. n^5 mod 10 = n mod 10, because 5 and 2 are relatively prime. Same mod 10 = same last digit. Done. It's also interesting from this proof... n^4 ends in 1 for odd non multiples of 5, n^4 ends in 6 for even non multiples of 5. Also, no fourth power of any integer ends in 2,3,4,7,8, or 9. I have a third proof idea... I'll see if I can flush it out later. Are you a math professor/researcher? If not, where do you find / get inspiration for these? I am a math professor. Quote Link to comment Share on other sites More sharing options...

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Assume that n is a natural number, prove that n and n

^{5}will always have the same one's digit. e.g. 13 and 13^{5}=371,293 both end in 3.## Link to comment

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