weird question (1 Viewer)

[haboozin]

this was in 2003 trail hsc from our school...
i dont really know where to start...


if a 3 digit number contains the letter a for its hunderds and letter b for its tens and letter c for its units and a + b + c is divisible by 3 .. then prove that the 3 digit number is also divisible by 3

for those who have trouble understanding:

eg.
321
abc

a + b + c = 6
6/3 = 2
therefore 321 is visible by 3

 

[who_loves_maths]

hi haboozin,

this is one of those 'classic' questions you'd get in things like maths comps. which question was is it in in your trial? just curious to know...

but anyhow, back to the question:

if a 3 digit number contains the letter a for its hunderds and letter b for its tens and letter c for its units and a + b + c is divisible by 3 .. then prove that the 3 digit number is also divisible by 3

the three digit number 'abc' in base 10 = 100a + 10b +c ; where 0<= a,b,c <= 9 , of course 0 < a ;
now, a +b +c = 3k , where 'k' is integral, since (a+b+c) is divisible by three.
also, let the number in question = x ; ie. x = 100a +10b+c -----> x = 99a + 9b + 3k
-----> x = 3(33a +3b +k) ------> therefore, the original number 'x' is also divisible by 3.

hope that helps

 

[Templar]

Well, you can show pretty easily that amod3=100amod3, as 100=1mod3. Similarly with bmod3=10bmod3.

abc mod3=100amod3+10bmod3+cmod3
=amod3+bmod3+cmod3
=(a+b+c)mod3
=0 as a+b+c=0mod3

 

[Love child]

what a random question.. which question no. was it??

 

[haboozin]

hi haboozin,

this is one of those 'classic' questions you'd get in things like maths comps. which question was is it in in your trial? just curious to know...

but anyhow, back to the question:



the three digit number 'abc' in base 10 = 100a + 10b +c ; where 0<= a,b,c <= 9 , of course 0 < a ;
now, a +b +c = 3k , where 'k' is integral, since (a+b+c) is divisible by three.
also, let the number in question = x ; ie. x = 100a +10b+c -----> x = 99a + 9b + 3k
-----> x = 3(33a +3b +k) ------> therefore, the original number 'x' is also divisible by 3.

hope that helps

question 6 i think...
its weird cause Q6 had a question like this..
and Q8 had 5 mark 2u differentiation and liek 10 mark easy inverse function question..
I dont think they really went with Question numbers.


yea i thought of what u did...omg i just realized what i was doing wrong...
i did the same thing as u but i thought that it didnt prove it cause i dont know i was thinking weirdly..
but yea that explains it great..

Thanks,.

 

[haboozin]

Well, you can show pretty easily that amod3=100amod3, as 100=1mod3. Similarly with bmod3=10bmod3.

abc mod3=100amod3+10bmod3+cmod3
=amod3+bmod3+cmod3
=(a+b+c)mod3
=0 as a+b+c=0mod3

can u use mod in maths?

i thought it was just used in programming.. :s

 

[who_loves_maths]

Originally Posted by haboozin
can u use mod in maths?

i thought it was just used in programming.. :s

of course you can use mod in mathematics, it's part of mathematics itself...


just for fun, i got another question here for keen ppl out there:

Find an integer(s), with an arbitrary number of constituent digits, that is the product of 17 and the sum of its digits. How many such integers are there?


have fun