shubhamsrg
  • shubhamsrg
law of conservation of sum of digits is what i call it.. how do i prove it but ?
Mathematics
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
shubhamsrg
  • shubhamsrg
whenever 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
  • shubhamsrg
same 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
  • shubhamsrg
not 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..

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anonymous
  • anonymous
youre reducing the sum modulo 9 right? thats why you took 77 and brought it down to 5?
shubhamsrg
  • shubhamsrg
yes..
anonymous
  • anonymous
Then 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
  • anonymous
This is called, Casting Out Nines by some.
hartnn
  • hartnn
*
shubhamsrg
  • shubhamsrg
how do you apply the fact during multiplication ?
anonymous
  • anonymous
Depending 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
  • anonymous
If 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
  • shubhamsrg
hmm,,surely i'd love to research on that,,give me some time,,i'll get back to you..
shubhamsrg
  • shubhamsrg
sorry 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
  • shubhamsrg
sire ? @joemath314159
anonymous
  • anonymous
when 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
  • anonymous
i guess theres not a difference is there?
anonymous
  • anonymous
I 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
  • shubhamsrg
why 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
  • anonymous
Conserve the digits don't get thee fingers cut off. Euler 1781
shubhamsrg
  • shubhamsrg
@mukushla
shubhamsrg
  • shubhamsrg
@ganeshie8
shubhamsrg
  • shubhamsrg
ohh..got it! multiplication, is nothing but just addition ! :O on that note, everything seems so simplified.. gotcha.. hmm..
yrelhan4
  • yrelhan4
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