Show that ## a^{2}-a+7 ## ends in one of the digits ## 3, 7 ,9##

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In summary: Indeed. I agree that, factoring can be easier.But which is a more exact/accurate way to express this? Is it ## a^{2}-a+7=a(a-1)+7\equiv 7, 7, 9, 3, 9, 7, 7, 9, 3 ## or ## 9\pmod {10} ##, or ## a^{2}-a+7\equiv a(a-1)+7\equiv 7, 7, 9, 3, 9, 7, 7, 9, 3 ##, or ## 9\pmod {10} ##?But which is a more exact/accurate way to express this? Is it ## a
  • #1
Math100
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Homework Statement
For any integer ## a ##, show that ## a^{2}-a+7 ## ends in one of the digits ## 3, 7 ##, or ## 9 ##.
Relevant Equations
None.
Proof:

Let ## a ## be any integer.
Then ## a\equiv 0, 1, 2, 3, 4, 5, 6, 7, 8 ## or ## 9\pmod {10} ##.
Note that ## a^{2}\equiv 0, 1, 4, 9, 6, 5, 6, 9, 4 ## or ## 1\pmod {10} ##.
Thus ## a^{2}-a+7\equiv 7, 7, 9, 9, 7, 7, 7, 9, 9 ## or ## 7\pmod {10} ##.
Therefore, ## a^{2}-a+7 ## ends in one of the digits ## 3, 7 ##, or ## 9 ## for any integer ## a ##.
 
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  • #2
Math100 said:
Homework Statement:: For any integer ## a ##, show that ## a^{2}-a+7 ## ends in one of the digits ## 3, 7 ##, or ## 9 ##.
Relevant Equations:: None.

Proof:

Let ## a ## be any integer.
Then ## a\equiv 0, 1, 2, 3, 4, 5, 6, 7, 8 ## or ## 9\pmod {10} ##.
Note that ## a^{2}\equiv 0, 1, 4, 9, 6, 5, 6, 9, 4 ## or ## 1\pmod {10} ##.
Thus ## a^{2}-a+7\equiv 7, 7, 9, 9, 7, 7, 7, 9, 9 ## or ## 7\pmod {10} ##.
I get ##7,7,9,3,9,7,7,9,3,9## from ##a^2-a+7=a(a-1)+7## which I think is easier.
Math100 said:
Therefore, ## a^{2}-a+7 ## ends in one of the digits ## 3, 7 ##, or ## 9 ## for any integer ## a ##.
 
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  • #3
fresh_42 said:
I get ##7,7,9,3,9,7,7,9,3,9## from ##a^2-a+7=a(a-1)+7## which I think is easier.
Indeed. I agree that, factoring can be easier.
 
  • #4
But which is a more exact/accurate way to express this? Is it ## a^{2}-a+7=a(a-1)+7\equiv 7, 7, 9, 3, 9, 7, 7, 9, 3 ## or ## 9\pmod {10} ##, or ## a^{2}-a+7\equiv a(a-1)+7\equiv 7, 7, 9, 3, 9, 7, 7, 9, 3 ##, or ## 9\pmod {10} ##?
 
  • #5
Math100 said:
But which is a more exact/accurate way to express this? Is it ## a^{2}-a+7=a(a-1)+7\equiv 7, 7, 9, 3, 9, 7, 7, 9, 3 ## or ## 9\pmod {10} ##, or ## a^{2}-a+7\equiv a(a-1)+7\equiv 7, 7, 9, 3, 9, 7, 7, 9, 3 ##, or ## 9\pmod {10} ##?
I would go even further.

We only need to show that ##a^2-a## always ends on ##\{0,2,6\}.## Now ##a(a-1)\equiv 0\pmod {10}## if one factor is from ##\{0,5\}##. That leaves us with ##2\cdot 1=2\, , \,3\cdot 2=6\, , \,4\cdot 3=12\, , \,7\cdot 6=42\, , \,8\cdot 7=56## and ##9\cdot 8=72, ## all ending on ##2## or ##6##.

This is a regular pattern: simplify if possible, and explain what you actually do.
 
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  • #6
fresh_42 said:
I would go even further.

We only need to show that ##a^2-a## always ends on ##\{0,2,6\}.## Now ##a(a-1)\equiv 0\pmod {10}## if one factor is from ##\{0,5\}##. That leaves us with ##2\cdot 1=2\, , \,3\cdot 2=6\, , \,4\cdot 3=12\, , \,7\cdot 6=42\, , \,8\cdot 7=56## and ##9\cdot 8=72, ## all ending on ##2## or ##6##.

This is a regular pattern: simplify if possible, and explain what you actually do.
And what about the sign? Equal sign is okay above or congruence sign?
 
  • #7
Math100 said:
And what about the sign? Equal sign is okay above or congruence sign?
I think equal is ok since it is modulo ##10## and everybody can see the last digit.

I made a mistake as I first wrote ##a(a-1)=0## for one factor in ##\{0,5\}.##
This was wrong because what I meant was ##a(a-1)\equiv 0\pmod{10}## for one factor in ##\{0,5\}.##
Here it is important since I only wrote the last digit, and not ##\{0,20,30\}.##
 
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Related to Show that ## a^{2}-a+7 ## ends in one of the digits ## 3, 7 ,9##

1. What is the significance of the expression "a^{2}-a+7" in this problem?

The expression "a^{2}-a+7" represents a quadratic function, where the value of "a" can be any real number. The goal of this problem is to show that when this function is evaluated, the result always ends in one of the digits 3, 7, or 9.

2. How do you approach this problem as a scientist?

As a scientist, I would use the scientific method to approach this problem. This involves making observations, formulating a hypothesis, conducting experiments or calculations, and analyzing the results to draw a conclusion.

3. What is the mathematical proof for this problem?

The proof for this problem involves using the properties of modular arithmetic. We can rewrite the expression "a^{2}-a+7" as (a-1)^{2}+6. By using modular arithmetic, we can show that the remainder when (a-1)^{2} is divided by 10 will always be either 3, 7, or 9. Therefore, when 6 is added to this remainder, the resulting value will always end in one of the digits 3, 7, or 9.

4. Can you provide an example to illustrate this problem?

Sure, let's take the value of a=5. When we plug this into the expression (a-1)^{2}+6, we get (5-1)^{2}+6= 16+6= 22. The remainder when 22 is divided by 10 is 2. When we add 6 to this remainder, we get 8, which ends in the digit 8. Therefore, this example shows that the expression "a^{2}-a+7" does not always end in one of the digits 3, 7, or 9.

5. What are the real-world applications of this problem?

This problem may seem abstract, but it has practical applications in fields such as cryptography and computer science. The concept of modular arithmetic is used in encryption algorithms to ensure the security of sensitive data. Additionally, this problem can help students develop critical thinking and problem-solving skills, which are valuable in many scientific and mathematical fields.

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