Quote:
Bunuel wrote:
Let n be a 5-digit number, and let q and r be the quotient and the remainder, respectively, when n is divided by 100. For how many values of n is q + r divisible by 11?
(A) 8180
(B) 8181
(C) 8182
(D) 9000
(E) 9090
Let "
abcde" is the 5 digit number. the minimum value of abcde shall be 10000 & maximum 99999
as per the question q shall be "abc" and r shall be "de".
ex: 10000/100 = 100.00 (q = 100 & r = 100 * 0.00 = 00); 99999/100 = 999.99 (q = 999 & r = 0.99*100 = 99)
now the q+r will be equal to abc+de (q = 100a+10b+c, r = 10d+e)
q+r = abc+de
given to find the number of 5 digit numbers "abcde", such that q+r i.e.,
abc+de divisible by 11.
we know that the divisibility rule of 11 is that "
the given number can only be completely divided by 11 if the difference of the sum of digits at odd position and sum of digits at even position in a number is 0 or 11". now we need to figure out the sum of digits at odd and even positions in q+r i.e.,
abc+de.
first let us find how many digits does the number
abc+de will have. we know that the minimum "
abcde" is 10000 and maximum is 99999. so the minimum abc+de shall contain 3 digits i.e., 100+00 = 100 and maximum
abc+de shall contain 4 digits i.e., 999+99 = 1098.
let us find the difference of the sum of digits at odd position and sum of digits at even position of three digit
abc+de
abc+de = (100a+10b+c) + (10d+e)When we add abc+de, there will be "c+e" in 1's position, "b+d" in 10's postion & "a" in 100's position: ex: 123+12 = "3+2" in 1's position, "2+1" in 10's position and "1" in 100's position
therfore,
sum of digits at odd position in "
abc+de" =
a+c+e sum of digits at even position in "
abc+de"=
b+dwe need to see if
a+c+e either equal to
b+d or b+d+11this is as good as checking the 11 divisibility of number "
abcde" which is also checking a+c+e = b+d or a+c+e = b+d+11.
hence, if we find the numbers in between 10000 to 99999 that are divisible by 11, it will be the answer.
the numbers divisible by 11 will be in arithematic progression with difference 11. hence we need to find the total such numbers in between 10000 & 99999. The minimum number that is divisible by 11 shall be 10010 (simple you can write this number from 11 multiplication rule also; even and odd positions) & the maximum number will be 99990.
so the first (say x
1) & last number (say x
n) in that arithmatic progression by 11 (numbers divisible by 11) shall be 10010 & 99990
we know in arithmatic progression x
n = x
1 + (total numbers in series -1) * (difference between two consecutive numbers)
= x
1 + (total numbers in series -1) * 11
99990 = 10010 + (total numbers - 1)*11
total numbers -1 = 99990-10010 / 11
= 89980 / 11
= 8180
hence total numbers in that series of numbers that are divisible by 11 = 8180+1 = 8181. Option (B) is correct.