Prove that number $underbrace{11 ldots1}_{100} underbrace{22 ldots2}_{100}$ is product of two consecutive...
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Prove that number $underbrace{11 ldots1}_{100}$$underbrace{22 ldots2}_{100}$ is product of two consecutive numbers
$begin{align}underbrace{11 ldots1}_{100} underbrace{22 ldots2}_{100}&=10^{199}+10^{198}+ldots+10^{100}+2(10^{99}+10^{98}+ldots+10+1)\&=(10^{100}+2)(10^{99}+10^{98}+ldots+10+1)=(10^{100}+2)frac{10^{100}-1}{10-1}end{align}$.
Is this good path or not?
discrete-mathematics
edited Nov 16 at 16:15
Yadati Kiran
1,237417
asked Nov 16 at 15:41
Marko Škorić
69810
add a comment |
up vote
1
down vote
favorite
Prove that number $underbrace{11 ldots1}_{100}$$underbrace{22 ldots2}_{100}$ is product of two consecutive numbers
$begin{align}underbrace{11 ldots1}_{100} underbrace{22 ldots2}_{100}&=10^{199}+10^{198}+ldots+10^{100}+2(10^{99}+10^{98}+ldots+10+1)\&=(10^{100}+2)(10^{99}+10^{98}+ldots+10+1)=(10^{100}+2)frac{10^{100}-1}{10-1}end{align}$.
Is this good path or not?
discrete-mathematics
edited Nov 16 at 16:15
Yadati Kiran
1,237417
asked Nov 16 at 15:41
Marko Škorić
69810
2
You need a backslash before underbrace to get what you want.
– Ross Millikan
Nov 16 at 15:43
2
I don't see any advantage in re-writing the way you have (though of course I might be missing something). I'd start small. Note that $12=3times 4$. What about $1122$? Well, that's $33times 34$. Maybe there's a pattern....
– lulu
Nov 16 at 15:45
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Prove that number $underbrace{11 ldots1}_{100}$$underbrace{22 ldots2}_{100}$ is product of two consecutive numbers
$begin{align}underbrace{11 ldots1}_{100} underbrace{22 ldots2}_{100}&=10^{199}+10^{198}+ldots+10^{100}+2(10^{99}+10^{98}+ldots+10+1)\&=(10^{100}+2)(10^{99}+10^{98}+ldots+10+1)=(10^{100}+2)frac{10^{100}-1}{10-1}end{align}$.
Is this good path or not?
discrete-mathematics
edited Nov 16 at 16:15
Yadati Kiran
1,237417
asked Nov 16 at 15:41
Marko Škorić
69810
Prove that number $underbrace{11 ldots1}_{100}$$underbrace{22 ldots2}_{100}$ is product of two consecutive numbers
$begin{align}underbrace{11 ldots1}_{100} underbrace{22 ldots2}_{100}&=10^{199}+10^{198}+ldots+10^{100}+2(10^{99}+10^{98}+ldots+10+1)\&=(10^{100}+2)(10^{99}+10^{98}+ldots+10+1)=(10^{100}+2)frac{10^{100}-1}{10-1}end{align}$.
Is this good path or not?
discrete-mathematics
discrete-mathematics
edited Nov 16 at 16:15
Yadati Kiran
1,237417
asked Nov 16 at 15:41
Marko Škorić
69810
edited Nov 16 at 16:15
Yadati Kiran
1,237417
asked Nov 16 at 15:41
Marko Škorić
69810
edited Nov 16 at 16:15
Yadati Kiran
1,237417
edited Nov 16 at 16:15
Yadati Kiran
1,237417
edited Nov 16 at 16:15
Yadati Kiran
1,237417
1,237417
asked Nov 16 at 15:41
Marko Škorić
69810
asked Nov 16 at 15:41
Marko Škorić
69810
asked Nov 16 at 15:41
Marko Škorić
69810
69810
2
You need a backslash before underbrace to get what you want.
– Ross Millikan
Nov 16 at 15:43
2
I don't see any advantage in re-writing the way you have (though of course I might be missing something). I'd start small. Note that $12=3times 4$. What about $1122$? Well, that's $33times 34$. Maybe there's a pattern....
– lulu
Nov 16 at 15:45
add a comment |
2
You need a backslash before underbrace to get what you want.
– Ross Millikan
Nov 16 at 15:43
2
I don't see any advantage in re-writing the way you have (though of course I might be missing something). I'd start small. Note that $12=3times 4$. What about $1122$? Well, that's $33times 34$. Maybe there's a pattern....
– lulu
Nov 16 at 15:45
2
2
You need a backslash before underbrace to get what you want.
– Ross Millikan
Nov 16 at 15:43
You need a backslash before underbrace to get what you want.
– Ross Millikan
Nov 16 at 15:43
2
2
I don't see any advantage in re-writing the way you have (though of course I might be missing something). I'd start small. Note that $12=3times 4$. What about $1122$? Well, that's $33times 34$. Maybe there's a pattern....
– lulu
Nov 16 at 15:45
I don't see any advantage in re-writing the way you have (though of course I might be missing something). I'd start small. Note that $12=3times 4$. What about $1122$? Well, that's $33times 34$. Maybe there's a pattern....
– lulu
Nov 16 at 15:45
add a comment |
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
Hint $ $ Apply the result below, with $,a = 10^{100}!-1, n = 3,,$ using $,n^2mid a$
$$begin{align}
&dfrac{a+n}{1} dfrac{a}{n^2}\[.3em]
= &dfrac{a+n}{n} dfrac{a}{n}\[.3em]
= &(b+1), b
end{align}$$
answered Nov 16 at 16:25
Bill Dubuque
207k29189624
add a comment |
up vote
2
down vote
You are not done because the two factors you exhibited are not consecutive. For three digits, you have shown $111222=1002cdot 111$. You can, however, make one more step and be there. lulu has given a good hint.
answered Nov 16 at 15:46
Ross Millikan
288k23195365
2
$111222=333times334$
– Yadati Kiran
Nov 16 at 15:47
2
@YadatiKiran: that is correct, but OP has not demonstrated that yet.
– Ross Millikan
Nov 16 at 15:48
add a comment |
up vote
0
down vote
Insert $1=frac{1}{3}*3$
$$
(10^{100}+2)frac{1}{3} 3frac{10^{100}-1}{10-1}\
x= frac{10^{100}+2}{3}\
y=3frac{10^{100}-1}{10-1} = frac{10^{100}-1}{3}\\
$$
answered Nov 16 at 15:51
AHusain
2,7602816
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Hint $ $ Apply the result below, with $,a = 10^{100}!-1, n = 3,,$ using $,n^2mid a$
$$begin{align}
&dfrac{a+n}{1} dfrac{a}{n^2}\[.3em]
= &dfrac{a+n}{n} dfrac{a}{n}\[.3em]
= &(b+1), b
end{align}$$
answered Nov 16 at 16:25
Bill Dubuque
207k29189624
add a comment |
up vote
1
down vote
accepted
Hint $ $ Apply the result below, with $,a = 10^{100}!-1, n = 3,,$ using $,n^2mid a$
$$begin{align}
&dfrac{a+n}{1} dfrac{a}{n^2}\[.3em]
= &dfrac{a+n}{n} dfrac{a}{n}\[.3em]
= &(b+1), b
end{align}$$
answered Nov 16 at 16:25
Bill Dubuque
207k29189624
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Hint $ $ Apply the result below, with $,a = 10^{100}!-1, n = 3,,$ using $,n^2mid a$
$$begin{align}
&dfrac{a+n}{1} dfrac{a}{n^2}\[.3em]
= &dfrac{a+n}{n} dfrac{a}{n}\[.3em]
= &(b+1), b
end{align}$$
answered Nov 16 at 16:25
Bill Dubuque
207k29189624
Hint $ $ Apply the result below, with $,a = 10^{100}!-1, n = 3,,$ using $,n^2mid a$
$$begin{align}
&dfrac{a+n}{1} dfrac{a}{n^2}\[.3em]
= &dfrac{a+n}{n} dfrac{a}{n}\[.3em]
= &(b+1), b
end{align}$$
answered Nov 16 at 16:25
Bill Dubuque
207k29189624
answered Nov 16 at 16:25
Bill Dubuque
207k29189624
answered Nov 16 at 16:25
Bill Dubuque
207k29189624
answered Nov 16 at 16:25
Bill Dubuque
207k29189624
207k29189624
add a comment |
add a comment |
up vote
2
down vote
You are not done because the two factors you exhibited are not consecutive. For three digits, you have shown $111222=1002cdot 111$. You can, however, make one more step and be there. lulu has given a good hint.
answered Nov 16 at 15:46
Ross Millikan
288k23195365
2
$111222=333times334$
– Yadati Kiran
Nov 16 at 15:47
2
@YadatiKiran: that is correct, but OP has not demonstrated that yet.
– Ross Millikan
Nov 16 at 15:48
add a comment |
up vote
2
down vote
You are not done because the two factors you exhibited are not consecutive. For three digits, you have shown $111222=1002cdot 111$. You can, however, make one more step and be there. lulu has given a good hint.
answered Nov 16 at 15:46
Ross Millikan
288k23195365
2
$111222=333times334$
– Yadati Kiran
Nov 16 at 15:47
2
@YadatiKiran: that is correct, but OP has not demonstrated that yet.
– Ross Millikan
Nov 16 at 15:48
add a comment |
up vote
2
down vote
up vote
2
down vote
You are not done because the two factors you exhibited are not consecutive. For three digits, you have shown $111222=1002cdot 111$. You can, however, make one more step and be there. lulu has given a good hint.
answered Nov 16 at 15:46
Ross Millikan
288k23195365
You are not done because the two factors you exhibited are not consecutive. For three digits, you have shown $111222=1002cdot 111$. You can, however, make one more step and be there. lulu has given a good hint.
answered Nov 16 at 15:46
Ross Millikan
288k23195365
edited Nov 16 at 15:48
answered Nov 16 at 15:46
Ross Millikan
288k23195365
answered Nov 16 at 15:46
Ross Millikan
288k23195365
answered Nov 16 at 15:46
Ross Millikan
288k23195365
288k23195365
2
$111222=333times334$
– Yadati Kiran
Nov 16 at 15:47
2
@YadatiKiran: that is correct, but OP has not demonstrated that yet.
– Ross Millikan
Nov 16 at 15:48
add a comment |
2
$111222=333times334$
– Yadati Kiran
Nov 16 at 15:47
2
@YadatiKiran: that is correct, but OP has not demonstrated that yet.
– Ross Millikan
Nov 16 at 15:48
2
2
$111222=333times334$
– Yadati Kiran
Nov 16 at 15:47
$111222=333times334$
– Yadati Kiran
Nov 16 at 15:47
2
2
@YadatiKiran: that is correct, but OP has not demonstrated that yet.
– Ross Millikan
Nov 16 at 15:48
@YadatiKiran: that is correct, but OP has not demonstrated that yet.
– Ross Millikan
Nov 16 at 15:48
add a comment |
up vote
0
down vote
Insert $1=frac{1}{3}*3$
$$
(10^{100}+2)frac{1}{3} 3frac{10^{100}-1}{10-1}\
x= frac{10^{100}+2}{3}\
y=3frac{10^{100}-1}{10-1} = frac{10^{100}-1}{3}\\
$$
answered Nov 16 at 15:51
AHusain
2,7602816
add a comment |
up vote
0
down vote
Insert $1=frac{1}{3}*3$
$$
(10^{100}+2)frac{1}{3} 3frac{10^{100}-1}{10-1}\
x= frac{10^{100}+2}{3}\
y=3frac{10^{100}-1}{10-1} = frac{10^{100}-1}{3}\\
$$
answered Nov 16 at 15:51
AHusain
2,7602816
add a comment |
up vote
0
down vote
up vote
0
down vote
Insert $1=frac{1}{3}*3$
$$
(10^{100}+2)frac{1}{3} 3frac{10^{100}-1}{10-1}\
x= frac{10^{100}+2}{3}\
y=3frac{10^{100}-1}{10-1} = frac{10^{100}-1}{3}\\
$$
answered Nov 16 at 15:51
AHusain
2,7602816
Insert $1=frac{1}{3}*3$
$$
(10^{100}+2)frac{1}{3} 3frac{10^{100}-1}{10-1}\
x= frac{10^{100}+2}{3}\
y=3frac{10^{100}-1}{10-1} = frac{10^{100}-1}{3}\\
$$
answered Nov 16 at 15:51
AHusain
2,7602816
answered Nov 16 at 15:51
AHusain
2,7602816
answered Nov 16 at 15:51
AHusain
2,7602816
answered Nov 16 at 15:51
AHusain
2,7602816
2,7602816
add a comment |
add a comment |
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You need a backslash before underbrace to get what you want.
– Ross Millikan
Nov 16 at 15:43
2
I don't see any advantage in re-writing the way you have (though of course I might be missing something). I'd start small. Note that $12=3times 4$. What about $1122$? Well, that's $33times 34$. Maybe there's a pattern....
– lulu
Nov 16 at 15:45