Prove that the number $\underbrace{111\ldots 11}_{1997}\underbrace{22\ldots 22}_{1998}5$ (which has 1997 of 1-s and 1998 of 2-s) is a perfect square.
Problem
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Tags: induction, Perfect Square
tµtµ
30.10.2005 12:03
A solution is to prove by induction that :
$\underbrace{111\ldots 11}_{n}\underbrace{22\ldots 22}_{n+1}5 = (\underbrace{333\ldots 33}_{n}5)^2$
darij grinberg
30.10.2005 13:21
Looool... http://www.mathlinks.ro/Forum/viewtopic.php?t=5797 . Although, thinking about it once again, probably knowing the initial problem was not that much of an advantage in solving the ISL problem. Darij
Valentin Vornicu
30.10.2005 13:22
darij grinberg wrote: Looool... http://www.mathlinks.ro/Forum/viewtopic.php?t=5797 Darij You got to be kidding me
shyong
30.10.2005 14:59
It is same as $\frac{10^{1997}-1}{9}\cdot 10^{1999}+20\left(\frac{10^{1998}-1}{9}\right)+5$
Letting $\frac{10^{1997}-1}{9}=a$ , we get $\frac{10^{1998}-1}{9}=10\cdot \frac{10^{1997}-1}{9}+1=10a+1$
Also $10^{1999}=10^{1997}\cdot 100=100(9a+1)$
So the expression become $100a(9a+1)+20(10a+1)+5=25(6a+1)^2$ which is obviously a perfect square since $a$ is integer .
chess64
30.10.2005 18:53
$m = \underbrace{111\ldots 11}_{1997} \underbrace{222\ldots 22}_{1997} 25$
$n =\underbrace{111\ldots 11}_{1997} \underbrace{222\ldots 22}_{1997}$
If $n$ can be expressed as $k(k+1)$ for some positive integral $k$, then $m$ is a perfect square.
$n=10^{1997}(10^{1996} + 10^{1995} + \cdots + 10^{1} + 10^{0})+2(10^{1996} + 10^{1995} + \cdots + 10^{1} + 10^{0})\\ =\left( \frac{1}{9}\right)(10^{1997} + 2)(10^{1997}-1)$
nvm, i don't think this approach works
devarshruparelia
19.02.2013 04:52
tµtµ wrote:
A solution is to prove by induction that :
$\underbrace{111\ldots 11}_{n}\underbrace{22\ldots 22}_{n+1}5 = (\underbrace{333\ldots 33}_{n}5)^2$
Your solution is the most easier to picture out and I appreciate your innovative approach.Do you know some more kinds of ways in which induction works
DanDumitrescu
26.07.2014 18:56
1111...1(1997 of 1-s)222...2(1998 2-s)5 =x10^1998+x*11+4=x*(9*x+1)+x*11+4=(3*x+2)(3*x+2) X=11…1(1998 of 1-s)
Doc.AK
30.07.2015 19:14
N=111…1222…25 = 〖10〗^1999∙((〖10〗^1997-1)/9)+2∙((〖10〗^1998-1)/9)∙10+5=〖10〗^1999∙((〖10〗^1997-1)/9)+20∙((〖10〗^1998-1)/9)+5= (〖10〗^1999 (〖10〗^1997-1)+20(〖10〗^1998-1)+45)/9=(〖10〗^3996-〖10〗^1999+20∙〖10〗^1998-20+45)/9= (〖10〗^3996-〖10〗^1999+2∙〖10〗^1999+25 )/9= (〖10〗^3996+〖10〗^1999+25)/9= (〖10〗^3996+2∙5∙〖10〗^1998+5^2)/3^2 = 〖((〖10〗^1998+5)/3)〗^2 Now we will show that 3/ 〖10〗^1998+5 〖10〗^1998+5=〖10〗^1998+3+2=〖10〗^1998+3+3-1≡〖10〗^1998-1 (*)=9k≡0(mod3) *a^n-b^n=(a-b)k Thus N is a perfect square
azerbaijan2015
29.11.2015 20:44
It equals square of 33....3
gghx
25.07.2019 12:02
@above its a square that ends in 5 it should be 33...3335