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?
Answers
Answer:
8181
Step-by-step explanation:
Given 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?
When a five digit number is divided by 100, we get first three digits as quotient and last two digits as remainder. Therefore q can be any integer from 100 to 999 and r can be from 0 to 99
So for each value of 9 x 100 = 900 , possible value will be q = 100/11 = 9
Now q + r ≡ 0 (mod 11) will be possible values of r
Since there is one extra , 900 / 11 = = 81
hence q + r ≡ 0 (mod 11)
So possible values of n such that q + r = 0(mod 11) is 9 x 100 + 1 x 81
So it will be 8100 + 81 = 8181
Answer:
8181 numbers
Step-by-step explanation:
n = 100q + r
n = 10000 to 99999
q = 100 to 999
r = 0 to 99
q = 11a + b
q = 100 to 999
a = 9 to 90
b = 0 to 10
if q + r is divisible by 11 then b + r is also divisible by 11
b from 0 to 10
r from 0 to 99
b+r ranges from 0 to 109
b +r divisible by 11 = 0, 11 , 22 , 33 , 44 , 55 , 66 , 77 , 88 , 99 ,
b = 0 r = 0-99 - 10 numbers
b = 1 r = 0-99 - 9 numbers
b = 2 r = 0-99 - 9 numbers
b = 3 r = 0-99 - 9 numbers
b = 4 r = 0-99 - 9 numbers
b = 5 r = 0-99 - 9 numbers
b = 6 r = 0-99 - 9 numbers
b = 7 r = 0-99 - 9 numbers
b = 8 r = 0-99 - 9 numbers
b = 9 r = 0-99 - 9 numbers
b = 10 r = 0-99 - 9 numbers
for a = 9 b starts with 1 ( 1-10) = 90 numbers
for a = 10 to 89 b (0 - 10) = 80*100 = 8000
for a = 90 b (0-9) = 91 numbers
Total numbers = 90 + 8000 + 91 = 8181 Numbers