1)Find such 2 Consecutive numbers whose SUM OF DIGITS are divisible by 7.
I challenge u to find such numbers.Einstein had created the riddle.
And There is an answer to this question.For hints send me an email on atulmkamat@gmail.com
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So two consecutive numbers whose sum is divisible by 7...
3,4 and 10,11 and 17,18 and 24,25 and 31,32 and 38,39 and so on... (Just add seven to each)
Am I Right?
@QuinMarie...M srry..But u didnt get the riddle..
it is 2 consecutive numbera whose Sum OF DIGITS is divisible by 7...
Eg..U took 31 and 32...
Sum of digits are 3+1=4 and 3+2=5...where 4 and 5 are nt divisible by 7
Try it..
Posted: Mar 1, 2013
@QuinMarie...M srry..But u didnt get the riddle..
it is 2 consecutive numbera whose Sum OF DIGITS is divisible by 7...
Eg..U took 31 and 32...
Sum of digits are 3+1=4 and 3+2=5...where 4 and 5 are nt divisible by 7
Try it..
Okayy 1 hint ..There is only 2 numbers which r consecutive and whose sum of digits is divisible by 7...
Its impossible...I trued it in all the ways....Plz give me a hint
This is soooo damn tough
hey what about 7 and 7
*i didn't really get it so excuse me if my answers stupid*
M sorry..But it is.. :P
So Noone??
Well...there is no buzz about this..anymore
So here is the answer
The 2 consecutive numbers are:
69999
70000
Where sum of digits of 69999 adds up to 42
And 70000 .. 7!!
So here 42 and 7 both are divisible by 7
Shelves: 7
Shelves: 7
Shelves: 13
Shelves: 7
Something filthy about Mr.Haynes
Shelves: 8
Shelves: 12
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