Show that the sum or difference of $a_i$ and $a_j$ from a set of seven distinct integers is divisible by 10
$begingroup$
I know this is a duplicate of this question, but I don't understand the top answer at all. How does squaring $a$ show what we're supposed to show?
Here's my intuition on how to prove this:
$a_i$ and $a_j$ can be decomposed as a sum of multiples of powers of $10$ since we're working in base $10$. When we add or subtract $a_i$ and $a_j$ from each other we want to show that their remainder is zero using the pigeonhole principle.
Is this a correct way of looking at it?
Thanks.
elementary-number-theory pigeonhole-principle
asked Jan 13 at 23:43
jd94jd94
31917
$endgroup$
add a comment |
$begingroup$
I know this is a duplicate of this question, but I don't understand the top answer at all. How does squaring $a$ show what we're supposed to show?
Here's my intuition on how to prove this:
$a_i$ and $a_j$ can be decomposed as a sum of multiples of powers of $10$ since we're working in base $10$. When we add or subtract $a_i$ and $a_j$ from each other we want to show that their remainder is zero using the pigeonhole principle.
Is this a correct way of looking at it?
Thanks.
elementary-number-theory pigeonhole-principle
asked Jan 13 at 23:43
jd94jd94
31917
$endgroup$
$begingroup$
Note that if reduced mod 10 there can be repeats. [so the reduced numbers might not be distinct] Also does it mean there is a sum div by 10 and also a difference div by 10, or does it mean either a sum div by 10 or a difference div by 10?
$endgroup$
– coffeemath
Jan 13 at 23:50
add a comment |
$begingroup$
I know this is a duplicate of this question, but I don't understand the top answer at all. How does squaring $a$ show what we're supposed to show?
Here's my intuition on how to prove this:
$a_i$ and $a_j$ can be decomposed as a sum of multiples of powers of $10$ since we're working in base $10$. When we add or subtract $a_i$ and $a_j$ from each other we want to show that their remainder is zero using the pigeonhole principle.
Is this a correct way of looking at it?
Thanks.
elementary-number-theory pigeonhole-principle
asked Jan 13 at 23:43
jd94jd94
31917
$endgroup$
I know this is a duplicate of this question, but I don't understand the top answer at all. How does squaring $a$ show what we're supposed to show?
Here's my intuition on how to prove this:
$a_i$ and $a_j$ can be decomposed as a sum of multiples of powers of $10$ since we're working in base $10$. When we add or subtract $a_i$ and $a_j$ from each other we want to show that their remainder is zero using the pigeonhole principle.
Is this a correct way of looking at it?
Thanks.
elementary-number-theory pigeonhole-principle
elementary-number-theory pigeonhole-principle
asked Jan 13 at 23:43
jd94jd94
31917
asked Jan 13 at 23:43
jd94jd94
31917
asked Jan 13 at 23:43
jd94jd94
31917
asked Jan 13 at 23:43
jd94jd94
31917
asked Jan 13 at 23:43
jd94jd94
31917
31917
$begingroup$
Note that if reduced mod 10 there can be repeats. [so the reduced numbers might not be distinct] Also does it mean there is a sum div by 10 and also a difference div by 10, or does it mean either a sum div by 10 or a difference div by 10?
$endgroup$
– coffeemath
Jan 13 at 23:50
add a comment |
$begingroup$
Note that if reduced mod 10 there can be repeats. [so the reduced numbers might not be distinct] Also does it mean there is a sum div by 10 and also a difference div by 10, or does it mean either a sum div by 10 or a difference div by 10?
$endgroup$
– coffeemath
Jan 13 at 23:50
$begingroup$
Note that if reduced mod 10 there can be repeats. [so the reduced numbers might not be distinct] Also does it mean there is a sum div by 10 and also a difference div by 10, or does it mean either a sum div by 10 or a difference div by 10?
$endgroup$
– coffeemath
Jan 13 at 23:50
$begingroup$
Note that if reduced mod 10 there can be repeats. [so the reduced numbers might not be distinct] Also does it mean there is a sum div by 10 and also a difference div by 10, or does it mean either a sum div by 10 or a difference div by 10?
$endgroup$
– coffeemath
Jan 13 at 23:50
add a comment |
1 Answer
1
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$begingroup$
The idea behind that proof is that
$$0=a_i^2-a_j^2 =(a_i-a_j)(a_i+a_j) pmod{10}$$
Then at least one of $(a_i-a_j),(a_i+a_j)$ is divisible by $2$ and at least one is divisible by $5$. Now show that $(a_i-a_j)$ and $(a_i+a_j) $ have the same parity.
Your idea also works:
Split the seven numbers among the following $6$ boxes:
$${ 1,9 pmod{10} } ; { 2,8 pmod{10} } ; { 3,7 pmod{10} } \ { 4,6 pmod{10} } ; { 5 pmod{10} } ; { 0 pmod{10} }$$
Then at least one box contains two numbers. If the two numbers in the same box are congruent $pmod{10}$, then their difference is divisible by $10$. Otherwise, their sum is divisible by 10.
answered Jan 13 at 23:50
N. S.N. S.
105k7116210
$endgroup$
$begingroup$
Why is the difference of squares congruent to 0 (mod 10)?
$endgroup$
– jd94
Jan 13 at 23:56
$begingroup$
@user2965071 That's exactly what the answer you posted shows: that there exists some $i,j$ such that $$a_i^2 equiv a_j^2 pmod{10}$$
$endgroup$
– N. S.
Jan 14 at 0:01
add a comment |
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1 Answer
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$begingroup$
The idea behind that proof is that
$$0=a_i^2-a_j^2 =(a_i-a_j)(a_i+a_j) pmod{10}$$
Then at least one of $(a_i-a_j),(a_i+a_j)$ is divisible by $2$ and at least one is divisible by $5$. Now show that $(a_i-a_j)$ and $(a_i+a_j) $ have the same parity.
Your idea also works:
Split the seven numbers among the following $6$ boxes:
$${ 1,9 pmod{10} } ; { 2,8 pmod{10} } ; { 3,7 pmod{10} } \ { 4,6 pmod{10} } ; { 5 pmod{10} } ; { 0 pmod{10} }$$
Then at least one box contains two numbers. If the two numbers in the same box are congruent $pmod{10}$, then their difference is divisible by $10$. Otherwise, their sum is divisible by 10.
answered Jan 13 at 23:50
N. S.N. S.
105k7116210
$endgroup$
$begingroup$
Why is the difference of squares congruent to 0 (mod 10)?
$endgroup$
– jd94
Jan 13 at 23:56
$begingroup$
@user2965071 That's exactly what the answer you posted shows: that there exists some $i,j$ such that $$a_i^2 equiv a_j^2 pmod{10}$$
$endgroup$
– N. S.
Jan 14 at 0:01
add a comment |
$begingroup$
The idea behind that proof is that
$$0=a_i^2-a_j^2 =(a_i-a_j)(a_i+a_j) pmod{10}$$
Then at least one of $(a_i-a_j),(a_i+a_j)$ is divisible by $2$ and at least one is divisible by $5$. Now show that $(a_i-a_j)$ and $(a_i+a_j) $ have the same parity.
Your idea also works:
Split the seven numbers among the following $6$ boxes:
$${ 1,9 pmod{10} } ; { 2,8 pmod{10} } ; { 3,7 pmod{10} } \ { 4,6 pmod{10} } ; { 5 pmod{10} } ; { 0 pmod{10} }$$
Then at least one box contains two numbers. If the two numbers in the same box are congruent $pmod{10}$, then their difference is divisible by $10$. Otherwise, their sum is divisible by 10.
answered Jan 13 at 23:50
N. S.N. S.
105k7116210
$endgroup$
$begingroup$
Why is the difference of squares congruent to 0 (mod 10)?
$endgroup$
– jd94
Jan 13 at 23:56
$begingroup$
@user2965071 That's exactly what the answer you posted shows: that there exists some $i,j$ such that $$a_i^2 equiv a_j^2 pmod{10}$$
$endgroup$
– N. S.
Jan 14 at 0:01
add a comment |
$begingroup$
The idea behind that proof is that
$$0=a_i^2-a_j^2 =(a_i-a_j)(a_i+a_j) pmod{10}$$
Then at least one of $(a_i-a_j),(a_i+a_j)$ is divisible by $2$ and at least one is divisible by $5$. Now show that $(a_i-a_j)$ and $(a_i+a_j) $ have the same parity.
Your idea also works:
Split the seven numbers among the following $6$ boxes:
$${ 1,9 pmod{10} } ; { 2,8 pmod{10} } ; { 3,7 pmod{10} } \ { 4,6 pmod{10} } ; { 5 pmod{10} } ; { 0 pmod{10} }$$
Then at least one box contains two numbers. If the two numbers in the same box are congruent $pmod{10}$, then their difference is divisible by $10$. Otherwise, their sum is divisible by 10.
answered Jan 13 at 23:50
N. S.N. S.
105k7116210
$endgroup$
The idea behind that proof is that
$$0=a_i^2-a_j^2 =(a_i-a_j)(a_i+a_j) pmod{10}$$
Then at least one of $(a_i-a_j),(a_i+a_j)$ is divisible by $2$ and at least one is divisible by $5$. Now show that $(a_i-a_j)$ and $(a_i+a_j) $ have the same parity.
Your idea also works:
Split the seven numbers among the following $6$ boxes:
$${ 1,9 pmod{10} } ; { 2,8 pmod{10} } ; { 3,7 pmod{10} } \ { 4,6 pmod{10} } ; { 5 pmod{10} } ; { 0 pmod{10} }$$
Then at least one box contains two numbers. If the two numbers in the same box are congruent $pmod{10}$, then their difference is divisible by $10$. Otherwise, their sum is divisible by 10.
answered Jan 13 at 23:50
N. S.N. S.
105k7116210
answered Jan 13 at 23:50
N. S.N. S.
105k7116210
answered Jan 13 at 23:50
N. S.N. S.
105k7116210
answered Jan 13 at 23:50
N. S.N. S.
105k7116210
105k7116210
$begingroup$
Why is the difference of squares congruent to 0 (mod 10)?
$endgroup$
– jd94
Jan 13 at 23:56
$begingroup$
@user2965071 That's exactly what the answer you posted shows: that there exists some $i,j$ such that $$a_i^2 equiv a_j^2 pmod{10}$$
$endgroup$
– N. S.
Jan 14 at 0:01
add a comment |
$begingroup$
Why is the difference of squares congruent to 0 (mod 10)?
$endgroup$
– jd94
Jan 13 at 23:56
$begingroup$
@user2965071 That's exactly what the answer you posted shows: that there exists some $i,j$ such that $$a_i^2 equiv a_j^2 pmod{10}$$
$endgroup$
– N. S.
Jan 14 at 0:01
$begingroup$
Why is the difference of squares congruent to 0 (mod 10)?
$endgroup$
– jd94
Jan 13 at 23:56
$begingroup$
Why is the difference of squares congruent to 0 (mod 10)?
$endgroup$
– jd94
Jan 13 at 23:56
$begingroup$
@user2965071 That's exactly what the answer you posted shows: that there exists some $i,j$ such that $$a_i^2 equiv a_j^2 pmod{10}$$
$endgroup$
– N. S.
Jan 14 at 0:01
$begingroup$
@user2965071 That's exactly what the answer you posted shows: that there exists some $i,j$ such that $$a_i^2 equiv a_j^2 pmod{10}$$
$endgroup$
– N. S.
Jan 14 at 0:01
add a comment |
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$begingroup$
Note that if reduced mod 10 there can be repeats. [so the reduced numbers might not be distinct] Also does it mean there is a sum div by 10 and also a difference div by 10, or does it mean either a sum div by 10 or a difference div by 10?
$endgroup$
– coffeemath
Jan 13 at 23:50