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Using congruences to find the last two digits of $808^2 + 3^305times11^151$
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In the question here
Find the last two digits of the given number
The method for finding the last two digits of a single integer is fairly clear. However, when confronted with an expression like the one in this question, I am unsure of how to proceed. Ideally, I would like to avoid using any techniques asides from congruence.
elementary-number-theory modular-arithmetic
asked Jul 18 at 22:14
SolidSnackDrive
1107
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up vote
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favorite
In the question here
Find the last two digits of the given number
The method for finding the last two digits of a single integer is fairly clear. However, when confronted with an expression like the one in this question, I am unsure of how to proceed. Ideally, I would like to avoid using any techniques asides from congruence.
elementary-number-theory modular-arithmetic
asked Jul 18 at 22:14
SolidSnackDrive
1107
1
Hint: $3^2times11=99equiv-1pmod100$.
– user496634
Jul 18 at 22:19
2
"The method for finding the last two digits of a single integer is fairly clear." Good. Find the last two digits of the three integers $808^2, 3^305$ and $11^151$ separately. What do you get? What do you think you should do afterwards? There are shortcuts like the one hinted at above, and while it makes the calculations easier, it's not really necessary. At least until you feel you have the basic approach nailed down.
– Arthur
Jul 18 at 22:19
Maybe I don't understand after all. There doesn't appear to be a pattern in the last two digits for $3^n$ like there is for $7^n$. Another method shown involves $01^n$ which is very simple since it's just 1. I don't understand how doing something like $50-1$ like in the linked question, or $(10-1)$ like some of the answers below is helpful.
– SolidSnackDrive
Jul 18 at 23:31
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In the question here
Find the last two digits of the given number
The method for finding the last two digits of a single integer is fairly clear. However, when confronted with an expression like the one in this question, I am unsure of how to proceed. Ideally, I would like to avoid using any techniques asides from congruence.
elementary-number-theory modular-arithmetic
asked Jul 18 at 22:14
SolidSnackDrive
1107
In the question here
Find the last two digits of the given number
The method for finding the last two digits of a single integer is fairly clear. However, when confronted with an expression like the one in this question, I am unsure of how to proceed. Ideally, I would like to avoid using any techniques asides from congruence.
elementary-number-theory modular-arithmetic
asked Jul 18 at 22:14
SolidSnackDrive
1107
asked Jul 18 at 22:14
SolidSnackDrive
1107
asked Jul 18 at 22:14
SolidSnackDrive
1107
asked Jul 18 at 22:14
SolidSnackDrive
1107
1107
1
Hint: $3^2times11=99equiv-1pmod100$.
– user496634
Jul 18 at 22:19
2
"The method for finding the last two digits of a single integer is fairly clear." Good. Find the last two digits of the three integers $808^2, 3^305$ and $11^151$ separately. What do you get? What do you think you should do afterwards? There are shortcuts like the one hinted at above, and while it makes the calculations easier, it's not really necessary. At least until you feel you have the basic approach nailed down.
– Arthur
Jul 18 at 22:19
Maybe I don't understand after all. There doesn't appear to be a pattern in the last two digits for $3^n$ like there is for $7^n$. Another method shown involves $01^n$ which is very simple since it's just 1. I don't understand how doing something like $50-1$ like in the linked question, or $(10-1)$ like some of the answers below is helpful.
– SolidSnackDrive
Jul 18 at 23:31
add a comment |Â
1
Hint: $3^2times11=99equiv-1pmod100$.
– user496634
Jul 18 at 22:19
2
"The method for finding the last two digits of a single integer is fairly clear." Good. Find the last two digits of the three integers $808^2, 3^305$ and $11^151$ separately. What do you get? What do you think you should do afterwards? There are shortcuts like the one hinted at above, and while it makes the calculations easier, it's not really necessary. At least until you feel you have the basic approach nailed down.
– Arthur
Jul 18 at 22:19
Maybe I don't understand after all. There doesn't appear to be a pattern in the last two digits for $3^n$ like there is for $7^n$. Another method shown involves $01^n$ which is very simple since it's just 1. I don't understand how doing something like $50-1$ like in the linked question, or $(10-1)$ like some of the answers below is helpful.
– SolidSnackDrive
Jul 18 at 23:31
1
1
Hint: $3^2times11=99equiv-1pmod100$.
– user496634
Jul 18 at 22:19
Hint: $3^2times11=99equiv-1pmod100$.
– user496634
Jul 18 at 22:19
2
2
"The method for finding the last two digits of a single integer is fairly clear." Good. Find the last two digits of the three integers $808^2, 3^305$ and $11^151$ separately. What do you get? What do you think you should do afterwards? There are shortcuts like the one hinted at above, and while it makes the calculations easier, it's not really necessary. At least until you feel you have the basic approach nailed down.
– Arthur
Jul 18 at 22:19
"The method for finding the last two digits of a single integer is fairly clear." Good. Find the last two digits of the three integers $808^2, 3^305$ and $11^151$ separately. What do you get? What do you think you should do afterwards? There are shortcuts like the one hinted at above, and while it makes the calculations easier, it's not really necessary. At least until you feel you have the basic approach nailed down.
– Arthur
Jul 18 at 22:19
Maybe I don't understand after all. There doesn't appear to be a pattern in the last two digits for $3^n$ like there is for $7^n$. Another method shown involves $01^n$ which is very simple since it's just 1. I don't understand how doing something like $50-1$ like in the linked question, or $(10-1)$ like some of the answers below is helpful.
– SolidSnackDrive
Jul 18 at 23:31
Maybe I don't understand after all. There doesn't appear to be a pattern in the last two digits for $3^n$ like there is for $7^n$. Another method shown involves $01^n$ which is very simple since it's just 1. I don't understand how doing something like $50-1$ like in the linked question, or $(10-1)$ like some of the answers below is helpful.
– SolidSnackDrive
Jul 18 at 23:31
add a comment |Â
1 Answer
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Use binomial theorem along with $3^2 = 10 -1$ and $11 = 10 + 1$.
So $808 = (800 + 8) = 800^2 + 2*8*800 + 8^2equiv 8^2=64 mod 100$
And $3^305 = 3*(3^2)^152 = 3*(10 - 1)^152 = 3(...... + 152*10*(-1)^151 + (-1)^152)equiv 3(-20 + 1) =-57equiv 43mod 100$.
And $11^151 = (10 + 1)^151 = ...... + 151choose 210^2 + 151*10 + 1equiv 11mod 100$.
So $808^2 + 3^305 cdot 11^151 equiv 64 + 43 cdot 11 = 64 + 473equiv 37 mod 100$.
=====
Oh, for goodness sake!
$3^2*11 = (10-1)(10+1) =100-1equiv -1$
So $3^2k11^k =(99)^k equiv (-1)^kmod 100$
so
$3^305cdot 11^151= 3^3*3^302cdot 11^151=$
$27*(-1)^151 equiv -27mod 100$.
So $808^2 + 3^30511^151equiv 64-27= 37mod 100$.
answered Jul 18 at 22:59
fleablood
60.4k22575
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Use binomial theorem along with $3^2 = 10 -1$ and $11 = 10 + 1$.
So $808 = (800 + 8) = 800^2 + 2*8*800 + 8^2equiv 8^2=64 mod 100$
And $3^305 = 3*(3^2)^152 = 3*(10 - 1)^152 = 3(...... + 152*10*(-1)^151 + (-1)^152)equiv 3(-20 + 1) =-57equiv 43mod 100$.
And $11^151 = (10 + 1)^151 = ...... + 151choose 210^2 + 151*10 + 1equiv 11mod 100$.
So $808^2 + 3^305 cdot 11^151 equiv 64 + 43 cdot 11 = 64 + 473equiv 37 mod 100$.
=====
Oh, for goodness sake!
$3^2*11 = (10-1)(10+1) =100-1equiv -1$
So $3^2k11^k =(99)^k equiv (-1)^kmod 100$
so
$3^305cdot 11^151= 3^3*3^302cdot 11^151=$
$27*(-1)^151 equiv -27mod 100$.
So $808^2 + 3^30511^151equiv 64-27= 37mod 100$.
answered Jul 18 at 22:59
fleablood
60.4k22575
add a comment |Â
up vote
0
down vote
Use binomial theorem along with $3^2 = 10 -1$ and $11 = 10 + 1$.
So $808 = (800 + 8) = 800^2 + 2*8*800 + 8^2equiv 8^2=64 mod 100$
And $3^305 = 3*(3^2)^152 = 3*(10 - 1)^152 = 3(...... + 152*10*(-1)^151 + (-1)^152)equiv 3(-20 + 1) =-57equiv 43mod 100$.
And $11^151 = (10 + 1)^151 = ...... + 151choose 210^2 + 151*10 + 1equiv 11mod 100$.
So $808^2 + 3^305 cdot 11^151 equiv 64 + 43 cdot 11 = 64 + 473equiv 37 mod 100$.
=====
Oh, for goodness sake!
$3^2*11 = (10-1)(10+1) =100-1equiv -1$
So $3^2k11^k =(99)^k equiv (-1)^kmod 100$
so
$3^305cdot 11^151= 3^3*3^302cdot 11^151=$
$27*(-1)^151 equiv -27mod 100$.
So $808^2 + 3^30511^151equiv 64-27= 37mod 100$.
answered Jul 18 at 22:59
fleablood
60.4k22575
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Use binomial theorem along with $3^2 = 10 -1$ and $11 = 10 + 1$.
So $808 = (800 + 8) = 800^2 + 2*8*800 + 8^2equiv 8^2=64 mod 100$
And $3^305 = 3*(3^2)^152 = 3*(10 - 1)^152 = 3(...... + 152*10*(-1)^151 + (-1)^152)equiv 3(-20 + 1) =-57equiv 43mod 100$.
And $11^151 = (10 + 1)^151 = ...... + 151choose 210^2 + 151*10 + 1equiv 11mod 100$.
So $808^2 + 3^305 cdot 11^151 equiv 64 + 43 cdot 11 = 64 + 473equiv 37 mod 100$.
=====
Oh, for goodness sake!
$3^2*11 = (10-1)(10+1) =100-1equiv -1$
So $3^2k11^k =(99)^k equiv (-1)^kmod 100$
so
$3^305cdot 11^151= 3^3*3^302cdot 11^151=$
$27*(-1)^151 equiv -27mod 100$.
So $808^2 + 3^30511^151equiv 64-27= 37mod 100$.
answered Jul 18 at 22:59
fleablood
60.4k22575
Use binomial theorem along with $3^2 = 10 -1$ and $11 = 10 + 1$.
So $808 = (800 + 8) = 800^2 + 2*8*800 + 8^2equiv 8^2=64 mod 100$
And $3^305 = 3*(3^2)^152 = 3*(10 - 1)^152 = 3(...... + 152*10*(-1)^151 + (-1)^152)equiv 3(-20 + 1) =-57equiv 43mod 100$.
And $11^151 = (10 + 1)^151 = ...... + 151choose 210^2 + 151*10 + 1equiv 11mod 100$.
So $808^2 + 3^305 cdot 11^151 equiv 64 + 43 cdot 11 = 64 + 473equiv 37 mod 100$.
=====
Oh, for goodness sake!
$3^2*11 = (10-1)(10+1) =100-1equiv -1$
So $3^2k11^k =(99)^k equiv (-1)^kmod 100$
so
$3^305cdot 11^151= 3^3*3^302cdot 11^151=$
$27*(-1)^151 equiv -27mod 100$.
So $808^2 + 3^30511^151equiv 64-27= 37mod 100$.
answered Jul 18 at 22:59
fleablood
60.4k22575
edited Jul 18 at 23:09
answered Jul 18 at 22:59
fleablood
60.4k22575
answered Jul 18 at 22:59
fleablood
60.4k22575
answered Jul 18 at 22:59
fleablood
60.4k22575
60.4k22575
add a comment |Â
add a comment |Â
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1
Hint: $3^2times11=99equiv-1pmod100$.
– user496634
Jul 18 at 22:19
2
"The method for finding the last two digits of a single integer is fairly clear." Good. Find the last two digits of the three integers $808^2, 3^305$ and $11^151$ separately. What do you get? What do you think you should do afterwards? There are shortcuts like the one hinted at above, and while it makes the calculations easier, it's not really necessary. At least until you feel you have the basic approach nailed down.
– Arthur
Jul 18 at 22:19
Maybe I don't understand after all. There doesn't appear to be a pattern in the last two digits for $3^n$ like there is for $7^n$. Another method shown involves $01^n$ which is very simple since it's just 1. I don't understand how doing something like $50-1$ like in the linked question, or $(10-1)$ like some of the answers below is helpful.
– SolidSnackDrive
Jul 18 at 23:31