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shubhamsrg
 3 years ago
law of conservation of sum of digits is what i call it..
how do i prove it but ?
shubhamsrg
 3 years ago
law of conservation of sum of digits is what i call it.. how do i prove it but ?

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shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0whenever you multiply or add ANY 2 numbers, sum of digits on LHS always = sum of digits on RHS you may verify.. lets say 119 * 322 = 38318 sum of respective digits on LHS = (11) * (7) => 77 => 14 => 5 and on RHS , 3+8+3+1+8 = 23 => 5 note : for (11)*(7) , we can also treat it as (2)*(7) = 14=>5 sum of digits do not change ..

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0same is true for addition 119 + 322 = 441 11 + 7 = 18 =>9 on LHS 4+4+1 =9 on RHS how do i prove that sum of digits is always conserved ?

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0not 2 numbers infact,,any quantity of numbers you add or multiply, SUM remains conserved. by any quantity means you may take as many numbers you want on LHS..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0youre reducing the sum modulo 9 right? thats why you took 77 and brought it down to 5?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Then the reason is because we usually have our numbers in base 10. Say you have the 5 digit number 54321. This is really:\[5(10)^4+4(10)^3+3(10)^2+2(10)+1\]If you reduce 10 modulo 9, they become ones. So mod 9 we have:\[5(10)^4+4(10)^3+3(10)^2+2(10)+1\equiv 5(1)^4+4(1)^3+3(1)^2+2(1)+1\] \[=5+4+3+2+1\]So remainders when dividing by 9 are preserved in the digits.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This is called, Casting Out Nines by some.

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0how do you apply the fact during multiplication ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Depending on what lvl of math you have seen, that can be a rough question or an easy one. The short answer is that the act of taking an integer an looking at its remainder when you divide by nine is a ring homomorphism from\[\mathbb{Z}\rightarrow \mathbb{Z}_9\]the integers modulo 9. A ring homomorphism is a function that preserves the operations of the ring, the addition and multiplication. So for all integers x, y:\[f(x+y)=f(x)+f(y), f(x\cdot y)=f(x)\cdot f(y)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If some of those terms are a little fuzzy, to understand why, you would have to multiply two numbers in the base 10 expanded form and check out whats happening.

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0hmm,,surely i'd love to research on that,,give me some time,,i'll get back to you..

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0sorry for my late reply..but OS was down for quite a long time today.. as you told , and rightly so, i worked on reducing to modulo 9. here is my progress: a * b = c let a = (9m + n) b = (9s + t) c =(9p + q) on that substitution , we are reduced to 9h + mn = q for some number h now what do i do next.. m,n,q are all single digits

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0sire ? @joemath314159

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0when you multiply two numbers, the sum of the digits on the lhs is equal to the sum of the digits on the rhs? that does not seem intuitive but always seems to work. is this a property of integers, or any rational numbers with finite decimal representations ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i guess theres not a difference is there?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I guess at first we need to prove that a number is divisible by 3 only if its sum of digits is divisible by 3

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0why you take 3? we rightly took 9 and this fact can be easily proven suppose we have 100a+ 10b + c =>9 q + a+b+c i.e. whole no. is divisible by 9 if sum of digits is divisible by 9 am just stuck at proving sum of digits remain conserved in 9h + mn = q for some number h now what do i do next.. m,n,q are all single digits as in my last comment..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Conserve the digits don't get thee fingers cut off. Euler 1781

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0ohh..got it! multiplication, is nothing but just addition ! :O on that note, everything seems so simplified.. gotcha.. hmm..
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