1hmmm...we can justify it :
n2=9k ...possible (for all n=3m)
n2=9k+1
n2-1=9k
factoring it
(n-1)(n+1)=9k..again possible for any (n-1)orn+1 to be 9m
similarly for 9
for 7
n2-1=9k+6
(n-1)(n+1)=3(3k+2)...again possible
\texttt{The number of natural numbers between 1 to 2000 such that sum of digits of their sqaures is eqaul to 21}
-
UP 0 DOWN 0 0 14
14 Answers
1sum of digits of squares is 21
for any perfect square the digits when added iteratively results into 1, 4, 7 or 9
example : 2342 = 54756 : 5+4+7+5+6=27, 2+7=9
sum of digits of 21 is 3, so there is no natural number whose square's sum of digits is 21
24@sir
"for any perfect square the digits when added iteratively results into 1, 4, 7 or 9"
any proof for this ?or we have to remeber it as it is. ? [12]
1@eureka123 : its just an observation
hope prophet sir will help me on this
1
1there can be 3 kinds of numbers
3k, 3k+1 , 3k+2 .
squaring the first gives a no. divisible by nine.
so the *iterative sum of digits is obviously 9
the rest of the two leave remainder of 1 when divided by 3
and any no. that leaves a rem of 1 when divided by 3 must have *iterative sum of digits as 1 4 or 7.
this simply proves that any perfect square may have only 1 ,4 , 7 ,9 as the *iterative sum of their digits.
* -- edits
1The particular question can be easily solved by saying that the sum of 21 implies an iterative sum 3.
which means that the number is divisible by 3 only once which is impossible for a perfect square.
1give a proper proof philip
"so the *iterative sum of digits is obviously 9"
"that leaves a rem of 1 when divided by 3 must have *iterative sum of digits as 1 4 or 7."
how ?
1hmmm ...
check out the criteria for divisibility by 3 and 9..
if you already know that then i think i dont have anything much to explain..
49yeah i agree with CHE!!
divisibility criteria for 3 and 9 gives *iterative sum to be in form 3m but that does not mean 3k+1 will have iterative sum in form 3m+1...yes observation proves it but is there any algebraical proof??? i guess everything in maths can be proved...so lets try this out!![1][1]
11Let me clarify what Philip wanted to say.
In general, any number when divided by 3 leaves the same remainder as the sum of its digits would leave on being divided by 3.
So all nos. whose sum of digits is 1 or 4 or 7, shall leave a rem of 1, when divided by 3.
As a proof of this, consider the number -
N=10^na_0+10^{n-1}a_1+....a_{n}
Observing 10\equiv 1(mod3), we say N\equiv \left(\sum_{i=0}^{i=n}{a_i} \right)(mod10) - as already said.
1kya yaar thats simple, just like soumik says
especially after you see this one :
http://targetiit.com/iit-jee-forum/posts/prove-divisibilty-by-9-149.html
1and I think the solution in #7 is much simpler for this question
11And on very similar - infact exactly this idea helps u to get the divisbility test for 11.