Mixing
Determining $a+b$
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The sum of the digits of $x$ is $17$. When $(x^3+x^5)$ and $(x^2+7!)$ are divided by $3$ and $9$, the remainder outcomes as $a$ and $b$ respectively. Then determine $a+b$
I couldn't think of any way to solve this problem. Could I get help?
Regards!
algebra-precalculus divisibility
asked 18 hours ago
Maxwell
296
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The sum of the digits of $x$ is $17$. When $(x^3+x^5)$ and $(x^2+7!)$ are divided by $3$ and $9$, the remainder outcomes as $a$ and $b$ respectively. Then determine $a+b$
I couldn't think of any way to solve this problem. Could I get help?
Regards!
algebra-precalculus divisibility
asked 18 hours ago
Maxwell
296
@CameronBuie Sum of digits?
– TheSimpliFire
18 hours ago
@CameronBuie For instance, $a+b = 17, a = 8, b = 9$
– Maxwell
18 hours ago
@TheSimpliFire That was the right terminology I've been looking for.
– Maxwell
18 hours ago
You have 2 numbers ($x^3+x^5, x^2+7!$) and divide them by 2 numbers (3 and 9). How come you get only 2 reminders when there should be 4? I guess some of them are assumed to be equal, but which ones?
– Ingix
18 hours ago
You mean $x^3+x^5equiv amod 3$ and $(x^2+7!)equiv bmod 9$ ?
– Nicolas FRANCOIS
18 hours ago
 |Â
show 4 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The sum of the digits of $x$ is $17$. When $(x^3+x^5)$ and $(x^2+7!)$ are divided by $3$ and $9$, the remainder outcomes as $a$ and $b$ respectively. Then determine $a+b$
I couldn't think of any way to solve this problem. Could I get help?
Regards!
algebra-precalculus divisibility
asked 18 hours ago
Maxwell
296
The sum of the digits of $x$ is $17$. When $(x^3+x^5)$ and $(x^2+7!)$ are divided by $3$ and $9$, the remainder outcomes as $a$ and $b$ respectively. Then determine $a+b$
I couldn't think of any way to solve this problem. Could I get help?
Regards!
algebra-precalculus divisibility
asked 18 hours ago
Maxwell
296
edited 18 hours ago
asked 18 hours ago
Maxwell
296
asked 18 hours ago
Maxwell
296
asked 18 hours ago
Maxwell
296
296
@CameronBuie Sum of digits?
– TheSimpliFire
18 hours ago
@CameronBuie For instance, $a+b = 17, a = 8, b = 9$
– Maxwell
18 hours ago
@TheSimpliFire That was the right terminology I've been looking for.
– Maxwell
18 hours ago
You have 2 numbers ($x^3+x^5, x^2+7!$) and divide them by 2 numbers (3 and 9). How come you get only 2 reminders when there should be 4? I guess some of them are assumed to be equal, but which ones?
– Ingix
18 hours ago
You mean $x^3+x^5equiv amod 3$ and $(x^2+7!)equiv bmod 9$ ?
– Nicolas FRANCOIS
18 hours ago
 |Â
show 4 more comments
@CameronBuie Sum of digits?
– TheSimpliFire
18 hours ago
@CameronBuie For instance, $a+b = 17, a = 8, b = 9$
– Maxwell
18 hours ago
@TheSimpliFire That was the right terminology I've been looking for.
– Maxwell
18 hours ago
You have 2 numbers ($x^3+x^5, x^2+7!$) and divide them by 2 numbers (3 and 9). How come you get only 2 reminders when there should be 4? I guess some of them are assumed to be equal, but which ones?
– Ingix
18 hours ago
You mean $x^3+x^5equiv amod 3$ and $(x^2+7!)equiv bmod 9$ ?
– Nicolas FRANCOIS
18 hours ago
@CameronBuie Sum of digits?
– TheSimpliFire
18 hours ago
@CameronBuie Sum of digits?
– TheSimpliFire
18 hours ago
@CameronBuie For instance, $a+b = 17, a = 8, b = 9$
– Maxwell
18 hours ago
@CameronBuie For instance, $a+b = 17, a = 8, b = 9$
– Maxwell
18 hours ago
@TheSimpliFire That was the right terminology I've been looking for.
– Maxwell
18 hours ago
@TheSimpliFire That was the right terminology I've been looking for.
– Maxwell
18 hours ago
You have 2 numbers ($x^3+x^5, x^2+7!$) and divide them by 2 numbers (3 and 9). How come you get only 2 reminders when there should be 4? I guess some of them are assumed to be equal, but which ones?
– Ingix
18 hours ago
You have 2 numbers ($x^3+x^5, x^2+7!$) and divide them by 2 numbers (3 and 9). How come you get only 2 reminders when there should be 4? I guess some of them are assumed to be equal, but which ones?
– Ingix
18 hours ago
You mean $x^3+x^5equiv amod 3$ and $(x^2+7!)equiv bmod 9$ ?
– Nicolas FRANCOIS
18 hours ago
You mean $x^3+x^5equiv amod 3$ and $(x^2+7!)equiv bmod 9$ ?
– Nicolas FRANCOIS
18 hours ago
 |Â
show 4 more comments
3 Answers
3
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up vote
2
down vote
accepted
The sum of the digits of $x$ has the same remainder modulo $9$ or $3$ than $x$ itself. So
$$xequiv 17equiv -1mod 3text and xequiv 17equiv -1mod 9$$
Then
$$x^3+x^5equiv (-1)^3+(-1)^5equiv 1mod 3$$
and, because $7!equiv 0mod 9$,
$$x^2+7!equiv (-1)^2+0equiv 1mod 9$$
Therefore $a=b=1$ and $a+b=2$.
answered 18 hours ago
Nicolas FRANCOIS
3,0771313
Why does $x equiv 17 equiv -1$?
– Maxwell
18 hours ago
@NicolasFRANCOIS Or you could just note that $a+b$ is the same for any $x$. When $x=89$, then you can carry on like you said.
– Toby Mak
18 hours ago
@Maxwell $17 pmod 3 equiv 18-1 pmod 3 equiv -1 pmod 3$ since $18$ is divisible by $3$.
– Toby Mak
18 hours ago
@TobyMak I still didn't get it properly.
– Maxwell
18 hours ago
@Maxwell Where did you get this problem from? If you're a beginner at modular arithmetic, this problem seems a bit challenging.
– Toby Mak
18 hours ago
 |Â
show 2 more comments
up vote
0
down vote
Since the sum of digits of x is 17. It should in the form of $3k+2$.
Now, $x^3 = 3k_1 + 8$, $x^5 = 3k'_2 +32 = 3k_2 +2$. So, when $x^3 + x^5$ is divided by $3$ we get a reminder of $1$ i.e, $a=1$.
Similarly, we can write x as $9k+8$. Which gives us $x^2 = 9k'_3+64 = 9k_3 +1$ and $7! $ is divisible by $9$. Reminder of $x^2 + 7!$ when divided by $9$ is $1$ i.e, $b=1$.
Hence $a+b = 2$.
Hope this helps!
answered 18 hours ago
zero
13
add a comment |Â
up vote
0
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THe is a well known (to some) theorem that if $s(x) = $ sum of the digits and $a equiv x mod n$ means that $a$ and $x$ have the same remainder when divided by $n$ that:
$x equiv s(x) mod 3$
and $x equiv s(x) mod 9$.
Always. For all $x$.
So if $s(x) = 17$ then $s(x) equiv 2equiv -1 mod 3$ and $s(x) equiv 8 equiv -1 mod 9$. (Note: $-1 = -1*3 + 2$ and $-1 = -1*9 + 8$ so $-1$ has remainder $2$ when divided by $3$ and $-1$ has remainder $8$ when divided by $9$
Now $7! = 1* 2*3*4*5*6*7 equiv 0 mod 3$ and $7! equiv 0 mod 9$.
And "remainder arithmetic" is consistent under addition. If $a$ has remainder $a'$ and $b$ has remainder $b'$ then $a + b$ will have the same remainder as $a' + b'$. So "remainder arithmetic" is consistent under subtraction (the inverse of addition) and multiplication (addition multiple times), NOT division, and exponential powers (multplication multiple times).
.....
So $s(x) =17equiv -1 mod 3,9$ then $x^3 + x^5 equiv (-1)^3 + (-1)^5 = -1 -1 equiv -2 mod 3,9$ and $x^3 + x^5 equiv -2equiv 1 mod 3$ and $x^3 + x^5 equiv -2equiv 7 mod 9$.
And $x^2 + 7! equiv (-1)^2 + 0 equiv 1 mod 3,9$.
Now I do not know what "When (x3+x5) and (x2+7!) are divided by 3 and 9, the remainder outcomes as a and b respectively" is supposed to mean.
$x^3 + x^5$ when divided by $3$ has the remainder $1$ and $x^3 + x^5$ when divide by $9$ has remainder $7$. And $x^2 + 7!$ when divided by $3$ or $9$ has remainder $1$. So I guess $a$ and $b$ are supposed to be $1$ and $7$ and $a+b=8$.
I guess. The question was not very clear.
answered 15 hours ago
fleablood
60k22575
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The sum of the digits of $x$ has the same remainder modulo $9$ or $3$ than $x$ itself. So
$$xequiv 17equiv -1mod 3text and xequiv 17equiv -1mod 9$$
Then
$$x^3+x^5equiv (-1)^3+(-1)^5equiv 1mod 3$$
and, because $7!equiv 0mod 9$,
$$x^2+7!equiv (-1)^2+0equiv 1mod 9$$
Therefore $a=b=1$ and $a+b=2$.
answered 18 hours ago
Nicolas FRANCOIS
3,0771313
Why does $x equiv 17 equiv -1$?
– Maxwell
18 hours ago
@NicolasFRANCOIS Or you could just note that $a+b$ is the same for any $x$. When $x=89$, then you can carry on like you said.
– Toby Mak
18 hours ago
@Maxwell $17 pmod 3 equiv 18-1 pmod 3 equiv -1 pmod 3$ since $18$ is divisible by $3$.
– Toby Mak
18 hours ago
@TobyMak I still didn't get it properly.
– Maxwell
18 hours ago
@Maxwell Where did you get this problem from? If you're a beginner at modular arithmetic, this problem seems a bit challenging.
– Toby Mak
18 hours ago
 |Â
show 2 more comments
up vote
2
down vote
accepted
The sum of the digits of $x$ has the same remainder modulo $9$ or $3$ than $x$ itself. So
$$xequiv 17equiv -1mod 3text and xequiv 17equiv -1mod 9$$
Then
$$x^3+x^5equiv (-1)^3+(-1)^5equiv 1mod 3$$
and, because $7!equiv 0mod 9$,
$$x^2+7!equiv (-1)^2+0equiv 1mod 9$$
Therefore $a=b=1$ and $a+b=2$.
answered 18 hours ago
Nicolas FRANCOIS
3,0771313
Why does $x equiv 17 equiv -1$?
– Maxwell
18 hours ago
@NicolasFRANCOIS Or you could just note that $a+b$ is the same for any $x$. When $x=89$, then you can carry on like you said.
– Toby Mak
18 hours ago
@Maxwell $17 pmod 3 equiv 18-1 pmod 3 equiv -1 pmod 3$ since $18$ is divisible by $3$.
– Toby Mak
18 hours ago
@TobyMak I still didn't get it properly.
– Maxwell
18 hours ago
@Maxwell Where did you get this problem from? If you're a beginner at modular arithmetic, this problem seems a bit challenging.
– Toby Mak
18 hours ago
 |Â
show 2 more comments
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The sum of the digits of $x$ has the same remainder modulo $9$ or $3$ than $x$ itself. So
$$xequiv 17equiv -1mod 3text and xequiv 17equiv -1mod 9$$
Then
$$x^3+x^5equiv (-1)^3+(-1)^5equiv 1mod 3$$
and, because $7!equiv 0mod 9$,
$$x^2+7!equiv (-1)^2+0equiv 1mod 9$$
Therefore $a=b=1$ and $a+b=2$.
answered 18 hours ago
Nicolas FRANCOIS
3,0771313
The sum of the digits of $x$ has the same remainder modulo $9$ or $3$ than $x$ itself. So
$$xequiv 17equiv -1mod 3text and xequiv 17equiv -1mod 9$$
Then
$$x^3+x^5equiv (-1)^3+(-1)^5equiv 1mod 3$$
and, because $7!equiv 0mod 9$,
$$x^2+7!equiv (-1)^2+0equiv 1mod 9$$
Therefore $a=b=1$ and $a+b=2$.
answered 18 hours ago
Nicolas FRANCOIS
3,0771313
answered 18 hours ago
Nicolas FRANCOIS
3,0771313
answered 18 hours ago
Nicolas FRANCOIS
3,0771313
answered 18 hours ago
Nicolas FRANCOIS
3,0771313
3,0771313
Why does $x equiv 17 equiv -1$?
– Maxwell
18 hours ago
@NicolasFRANCOIS Or you could just note that $a+b$ is the same for any $x$. When $x=89$, then you can carry on like you said.
– Toby Mak
18 hours ago
@Maxwell $17 pmod 3 equiv 18-1 pmod 3 equiv -1 pmod 3$ since $18$ is divisible by $3$.
– Toby Mak
18 hours ago
@TobyMak I still didn't get it properly.
– Maxwell
18 hours ago
@Maxwell Where did you get this problem from? If you're a beginner at modular arithmetic, this problem seems a bit challenging.
– Toby Mak
18 hours ago
 |Â
show 2 more comments
Why does $x equiv 17 equiv -1$?
– Maxwell
18 hours ago
@NicolasFRANCOIS Or you could just note that $a+b$ is the same for any $x$. When $x=89$, then you can carry on like you said.
– Toby Mak
18 hours ago
@Maxwell $17 pmod 3 equiv 18-1 pmod 3 equiv -1 pmod 3$ since $18$ is divisible by $3$.
– Toby Mak
18 hours ago
@TobyMak I still didn't get it properly.
– Maxwell
18 hours ago
@Maxwell Where did you get this problem from? If you're a beginner at modular arithmetic, this problem seems a bit challenging.
– Toby Mak
18 hours ago
Why does $x equiv 17 equiv -1$?
– Maxwell
18 hours ago
Why does $x equiv 17 equiv -1$?
– Maxwell
18 hours ago
@NicolasFRANCOIS Or you could just note that $a+b$ is the same for any $x$. When $x=89$, then you can carry on like you said.
– Toby Mak
18 hours ago
@NicolasFRANCOIS Or you could just note that $a+b$ is the same for any $x$. When $x=89$, then you can carry on like you said.
– Toby Mak
18 hours ago
@Maxwell $17 pmod 3 equiv 18-1 pmod 3 equiv -1 pmod 3$ since $18$ is divisible by $3$.
– Toby Mak
18 hours ago
@Maxwell $17 pmod 3 equiv 18-1 pmod 3 equiv -1 pmod 3$ since $18$ is divisible by $3$.
– Toby Mak
18 hours ago
@TobyMak I still didn't get it properly.
– Maxwell
18 hours ago
@TobyMak I still didn't get it properly.
– Maxwell
18 hours ago
@Maxwell Where did you get this problem from? If you're a beginner at modular arithmetic, this problem seems a bit challenging.
– Toby Mak
18 hours ago
@Maxwell Where did you get this problem from? If you're a beginner at modular arithmetic, this problem seems a bit challenging.
– Toby Mak
18 hours ago
 |Â
show 2 more comments
up vote
0
down vote
Since the sum of digits of x is 17. It should in the form of $3k+2$.
Now, $x^3 = 3k_1 + 8$, $x^5 = 3k'_2 +32 = 3k_2 +2$. So, when $x^3 + x^5$ is divided by $3$ we get a reminder of $1$ i.e, $a=1$.
Similarly, we can write x as $9k+8$. Which gives us $x^2 = 9k'_3+64 = 9k_3 +1$ and $7! $ is divisible by $9$. Reminder of $x^2 + 7!$ when divided by $9$ is $1$ i.e, $b=1$.
Hence $a+b = 2$.
Hope this helps!
answered 18 hours ago
zero
13
add a comment |Â
up vote
0
down vote
Since the sum of digits of x is 17. It should in the form of $3k+2$.
Now, $x^3 = 3k_1 + 8$, $x^5 = 3k'_2 +32 = 3k_2 +2$. So, when $x^3 + x^5$ is divided by $3$ we get a reminder of $1$ i.e, $a=1$.
Similarly, we can write x as $9k+8$. Which gives us $x^2 = 9k'_3+64 = 9k_3 +1$ and $7! $ is divisible by $9$. Reminder of $x^2 + 7!$ when divided by $9$ is $1$ i.e, $b=1$.
Hence $a+b = 2$.
Hope this helps!
answered 18 hours ago
zero
13
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Since the sum of digits of x is 17. It should in the form of $3k+2$.
Now, $x^3 = 3k_1 + 8$, $x^5 = 3k'_2 +32 = 3k_2 +2$. So, when $x^3 + x^5$ is divided by $3$ we get a reminder of $1$ i.e, $a=1$.
Similarly, we can write x as $9k+8$. Which gives us $x^2 = 9k'_3+64 = 9k_3 +1$ and $7! $ is divisible by $9$. Reminder of $x^2 + 7!$ when divided by $9$ is $1$ i.e, $b=1$.
Hence $a+b = 2$.
Hope this helps!
answered 18 hours ago
zero
13
Since the sum of digits of x is 17. It should in the form of $3k+2$.
Now, $x^3 = 3k_1 + 8$, $x^5 = 3k'_2 +32 = 3k_2 +2$. So, when $x^3 + x^5$ is divided by $3$ we get a reminder of $1$ i.e, $a=1$.
Similarly, we can write x as $9k+8$. Which gives us $x^2 = 9k'_3+64 = 9k_3 +1$ and $7! $ is divisible by $9$. Reminder of $x^2 + 7!$ when divided by $9$ is $1$ i.e, $b=1$.
Hence $a+b = 2$.
Hope this helps!
answered 18 hours ago
zero
13
answered 18 hours ago
zero
13
answered 18 hours ago
zero
13
answered 18 hours ago
zero
13
13
add a comment |Â
add a comment |Â
up vote
0
down vote
THe is a well known (to some) theorem that if $s(x) = $ sum of the digits and $a equiv x mod n$ means that $a$ and $x$ have the same remainder when divided by $n$ that:
$x equiv s(x) mod 3$
and $x equiv s(x) mod 9$.
Always. For all $x$.
So if $s(x) = 17$ then $s(x) equiv 2equiv -1 mod 3$ and $s(x) equiv 8 equiv -1 mod 9$. (Note: $-1 = -1*3 + 2$ and $-1 = -1*9 + 8$ so $-1$ has remainder $2$ when divided by $3$ and $-1$ has remainder $8$ when divided by $9$
Now $7! = 1* 2*3*4*5*6*7 equiv 0 mod 3$ and $7! equiv 0 mod 9$.
And "remainder arithmetic" is consistent under addition. If $a$ has remainder $a'$ and $b$ has remainder $b'$ then $a + b$ will have the same remainder as $a' + b'$. So "remainder arithmetic" is consistent under subtraction (the inverse of addition) and multiplication (addition multiple times), NOT division, and exponential powers (multplication multiple times).
.....
So $s(x) =17equiv -1 mod 3,9$ then $x^3 + x^5 equiv (-1)^3 + (-1)^5 = -1 -1 equiv -2 mod 3,9$ and $x^3 + x^5 equiv -2equiv 1 mod 3$ and $x^3 + x^5 equiv -2equiv 7 mod 9$.
And $x^2 + 7! equiv (-1)^2 + 0 equiv 1 mod 3,9$.
Now I do not know what "When (x3+x5) and (x2+7!) are divided by 3 and 9, the remainder outcomes as a and b respectively" is supposed to mean.
$x^3 + x^5$ when divided by $3$ has the remainder $1$ and $x^3 + x^5$ when divide by $9$ has remainder $7$. And $x^2 + 7!$ when divided by $3$ or $9$ has remainder $1$. So I guess $a$ and $b$ are supposed to be $1$ and $7$ and $a+b=8$.
I guess. The question was not very clear.
answered 15 hours ago
fleablood
60k22575
add a comment |Â
up vote
0
down vote
THe is a well known (to some) theorem that if $s(x) = $ sum of the digits and $a equiv x mod n$ means that $a$ and $x$ have the same remainder when divided by $n$ that:
$x equiv s(x) mod 3$
and $x equiv s(x) mod 9$.
Always. For all $x$.
So if $s(x) = 17$ then $s(x) equiv 2equiv -1 mod 3$ and $s(x) equiv 8 equiv -1 mod 9$. (Note: $-1 = -1*3 + 2$ and $-1 = -1*9 + 8$ so $-1$ has remainder $2$ when divided by $3$ and $-1$ has remainder $8$ when divided by $9$
Now $7! = 1* 2*3*4*5*6*7 equiv 0 mod 3$ and $7! equiv 0 mod 9$.
And "remainder arithmetic" is consistent under addition. If $a$ has remainder $a'$ and $b$ has remainder $b'$ then $a + b$ will have the same remainder as $a' + b'$. So "remainder arithmetic" is consistent under subtraction (the inverse of addition) and multiplication (addition multiple times), NOT division, and exponential powers (multplication multiple times).
.....
So $s(x) =17equiv -1 mod 3,9$ then $x^3 + x^5 equiv (-1)^3 + (-1)^5 = -1 -1 equiv -2 mod 3,9$ and $x^3 + x^5 equiv -2equiv 1 mod 3$ and $x^3 + x^5 equiv -2equiv 7 mod 9$.
And $x^2 + 7! equiv (-1)^2 + 0 equiv 1 mod 3,9$.
Now I do not know what "When (x3+x5) and (x2+7!) are divided by 3 and 9, the remainder outcomes as a and b respectively" is supposed to mean.
$x^3 + x^5$ when divided by $3$ has the remainder $1$ and $x^3 + x^5$ when divide by $9$ has remainder $7$. And $x^2 + 7!$ when divided by $3$ or $9$ has remainder $1$. So I guess $a$ and $b$ are supposed to be $1$ and $7$ and $a+b=8$.
I guess. The question was not very clear.
answered 15 hours ago
fleablood
60k22575
add a comment |Â
up vote
0
down vote
up vote
0
down vote
THe is a well known (to some) theorem that if $s(x) = $ sum of the digits and $a equiv x mod n$ means that $a$ and $x$ have the same remainder when divided by $n$ that:
$x equiv s(x) mod 3$
and $x equiv s(x) mod 9$.
Always. For all $x$.
So if $s(x) = 17$ then $s(x) equiv 2equiv -1 mod 3$ and $s(x) equiv 8 equiv -1 mod 9$. (Note: $-1 = -1*3 + 2$ and $-1 = -1*9 + 8$ so $-1$ has remainder $2$ when divided by $3$ and $-1$ has remainder $8$ when divided by $9$
Now $7! = 1* 2*3*4*5*6*7 equiv 0 mod 3$ and $7! equiv 0 mod 9$.
And "remainder arithmetic" is consistent under addition. If $a$ has remainder $a'$ and $b$ has remainder $b'$ then $a + b$ will have the same remainder as $a' + b'$. So "remainder arithmetic" is consistent under subtraction (the inverse of addition) and multiplication (addition multiple times), NOT division, and exponential powers (multplication multiple times).
.....
So $s(x) =17equiv -1 mod 3,9$ then $x^3 + x^5 equiv (-1)^3 + (-1)^5 = -1 -1 equiv -2 mod 3,9$ and $x^3 + x^5 equiv -2equiv 1 mod 3$ and $x^3 + x^5 equiv -2equiv 7 mod 9$.
And $x^2 + 7! equiv (-1)^2 + 0 equiv 1 mod 3,9$.
Now I do not know what "When (x3+x5) and (x2+7!) are divided by 3 and 9, the remainder outcomes as a and b respectively" is supposed to mean.
$x^3 + x^5$ when divided by $3$ has the remainder $1$ and $x^3 + x^5$ when divide by $9$ has remainder $7$. And $x^2 + 7!$ when divided by $3$ or $9$ has remainder $1$. So I guess $a$ and $b$ are supposed to be $1$ and $7$ and $a+b=8$.
I guess. The question was not very clear.
answered 15 hours ago
fleablood
60k22575
THe is a well known (to some) theorem that if $s(x) = $ sum of the digits and $a equiv x mod n$ means that $a$ and $x$ have the same remainder when divided by $n$ that:
$x equiv s(x) mod 3$
and $x equiv s(x) mod 9$.
Always. For all $x$.
So if $s(x) = 17$ then $s(x) equiv 2equiv -1 mod 3$ and $s(x) equiv 8 equiv -1 mod 9$. (Note: $-1 = -1*3 + 2$ and $-1 = -1*9 + 8$ so $-1$ has remainder $2$ when divided by $3$ and $-1$ has remainder $8$ when divided by $9$
Now $7! = 1* 2*3*4*5*6*7 equiv 0 mod 3$ and $7! equiv 0 mod 9$.
And "remainder arithmetic" is consistent under addition. If $a$ has remainder $a'$ and $b$ has remainder $b'$ then $a + b$ will have the same remainder as $a' + b'$. So "remainder arithmetic" is consistent under subtraction (the inverse of addition) and multiplication (addition multiple times), NOT division, and exponential powers (multplication multiple times).
.....
So $s(x) =17equiv -1 mod 3,9$ then $x^3 + x^5 equiv (-1)^3 + (-1)^5 = -1 -1 equiv -2 mod 3,9$ and $x^3 + x^5 equiv -2equiv 1 mod 3$ and $x^3 + x^5 equiv -2equiv 7 mod 9$.
And $x^2 + 7! equiv (-1)^2 + 0 equiv 1 mod 3,9$.
Now I do not know what "When (x3+x5) and (x2+7!) are divided by 3 and 9, the remainder outcomes as a and b respectively" is supposed to mean.
$x^3 + x^5$ when divided by $3$ has the remainder $1$ and $x^3 + x^5$ when divide by $9$ has remainder $7$. And $x^2 + 7!$ when divided by $3$ or $9$ has remainder $1$. So I guess $a$ and $b$ are supposed to be $1$ and $7$ and $a+b=8$.
I guess. The question was not very clear.
answered 15 hours ago
fleablood
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answered 15 hours ago
fleablood
60k22575
answered 15 hours ago
fleablood
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answered 15 hours ago
fleablood
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60k22575
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@CameronBuie Sum of digits?
– TheSimpliFire
18 hours ago
@CameronBuie For instance, $a+b = 17, a = 8, b = 9$
– Maxwell
18 hours ago
@TheSimpliFire That was the right terminology I've been looking for.
– Maxwell
18 hours ago
You have 2 numbers ($x^3+x^5, x^2+7!$) and divide them by 2 numbers (3 and 9). How come you get only 2 reminders when there should be 4? I guess some of them are assumed to be equal, but which ones?
– Ingix
18 hours ago
You mean $x^3+x^5equiv amod 3$ and $(x^2+7!)equiv bmod 9$ ?
– Nicolas FRANCOIS
18 hours ago