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A number when divided by 100 leaves a quotient (Q) and a
remainder (R). How many three-digit natural numbers are
there such that Q + R is divisible by 11?
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Answer / nij
rule1: Number is divisible by 11 only when difference of
sumation of numbers at odd and even place is either 11 or 0.
e.g. 352 is divisible by 11 as (3+2) - 5 = 0
so Q+R would be divisible by 11 only when it satisfies rule1
upto 100 there are 9 numbers 11,22,33,44,55,66,77,88,99.
don't consider this...since 3 digit natural numbers are asked
From 100 to 200. Q would be 1 and reminder(R)need to be 10,
21,32,43,54,65,76,87,98 again 9 numbers.
From 200 to 100. Q would be 2 and R need to be
9,20,31,42,53,64,75,86,97 again 9 numbers.....
LIKEWISE
100 to 200----9 numbers
200 to 300----9 numbers
300 to 400----9
400 to 500----9
500 to 600----9
600 to 700----9
700 to 800-----9
800 to 900-----9
900 to 1000 ---9
so total 9x9=81 numbers
| Is This Answer Correct ? | 41 Yes | 2 No |
Answer / snehal
abov ans is nt correct
cuz
121/100=q=1 r=21
q+_r=1+21=22
which is also divided by 11so
so 100 to 200=9
so similarly 9 times 9
9*9=81(tat is frm 100 to 999)
| Is This Answer Correct ? | 5 Yes | 1 No |
Answer / kartik
FROM 100 TO 200 ANY NO DIVIDED BY 100 WILL GIVE Q AS 1
ALSO Q+R SHLD BE DIVISIBLE BY 11 SOIT CAN ONLY BE 11 22 33....
NOW...
Q+R Q THEREFORE THEREFORE
R NUMBER IS(Q*100+R)
11 1 10 110
22 1 21 121
33 1 32 132
44 1 43 143
55 1 54 154
66 1 65 165
77 1 76 176
88 1 87 187
99 1 98 198
THUS THERE A TOTAL OF 9 NOS BETWEEN 100 AND 200
LIKEWISE THERE WILL BE 9 NOS BET 200 AND 300 THEN 300-400
AND SO ON
SO THERE WILL BE 81 SUCH NOS
| Is This Answer Correct ? | 2 Yes | 0 No |
Answer / arthi
Some people say its 90 numbers, and some say that its 81.
The correct answer depends on the real definition of natural numbers.
Wikipedia says:
It is either the set of positive integers {1, 2, 3, ...} according to the traditional definition or the set of non-negative integers {0, 1, 2, ...} according to a definition first appearing in the nineteenth century.
Assuming the former is correct, {1, 2, 3, ...} the answer is 81. (Checking between 100 and 999)
Whereas assuming the latter is correct {0, 1, 2, ...}, the answer is 90. (Checking between 1 and 999)
Now, for finding the answer, just know if Q+R is div by 11, and Q is the digit in the hundred's place, then the three digit number itself is div by 11. (The explanation is lengthy, so Im not including it now. I might post it later.)
so only find the number of numbers divisible by 11, between 1 and 999 or 100 and 999 as the case may be.
The answer 90 in the first case, and 81 in the second case.
| Is This Answer Correct ? | 1 Yes | 0 No |
Answer / vasanth
hi frds.. i think this is the correct ans
the no should be divided by 100..more over its only an 3
digit no so..
101/100=10(Q)+1(R)=11.which s divided by11
similarly
202/100=20+2==22
303/100=30+3=33
404,505,606,707,808,909.
so totally 9 numbers..
i think its rite.. if not excuse me..
| Is This Answer Correct ? | 4 Yes | 4 No |
Answer / jalinder
rule1: Number is divisible by 11 only when difference of
sumation of numbers at odd and even place is either 11 or 0.
e.g. 352 is divisible by 11 as (3+2) - 5 = 0
so Q+R would be divisible by 11 only when it satisfies rule1
upto 100 there are 9 numbers 11,22,33,44,55,66,77,88,99.
From 100 to 200. Q would be 1 and reminder(R)need to be 10,
21,32,43,54,65,76,87,98 again 9 numbers.
From 200 to 100. Q would be 2 and R need to be
9,20,31,42,53,64,75,86,97 again 9 numbers.....
LIKEWISE
upto 100------9 numbers
100 to 200----9 numbers
200 to 300----9 numbers
300 to 400----9
400 to 500----9
500 to 600----9
600 to 700----9
700 to 800-----9
800 to 900-----9
900 to 1000 ---9
so total 9x10=90 numbers
| Is This Answer Correct ? | 18 Yes | 25 No |
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