Let n be a 5-digit number, and let q and r be the quotient and the rem

Question

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

chetan2u · Accepted Answer

There are 99999-10000+1=90000 possible values of n...

q+r divisible by 11 means the same as the number divisible by 11..
example - 11000 is divisible by 11... 11000=100*110+0 so q+r=110+0=110, divisible by 11
10000 is not divisible by 11, so is not q+r
  Thus, we are looking for 5-digit numbers that are the multiples of 11   
10010 is first multiple => 11*910, and 99990 is the last 5-digit multiple of 11=>11*9090
Total multiples = 9090-910+1=8180+1=8181

B

KarishmaB · Answer

The main point in this question is how do you figure out that if upon division by 100, q + r is divisible by 11, then n is also divisible by 11?

Assume a 5 digit number abcde.
When divided by 100, we get 
abcde = 100abc + de 
where de is the 2 digit remainder and abc is the 3 digit quotient.

q + r = abc + de = 100a + 10b + c + 10d + e = 100a + c + e + 10 (b + d) 
To find whether it is divisible by 11, let's re-write it as

99a + a + c + e + 11(b + d) - b - d

Now, 99a + 11 (b + d) is divisible by 11. 
If (a + c + e) - (b + d) is divisible by 11, then so is the entire number. 
But this is the same rule as the divisibility rule for division by 11 for a 5 digit number abcde.
Hence, if q + r is divisible by 11, the five digit number should also be divisible by 11. So we just need the multiples of 11 in all 5 digit numbers from 10000 to 99999

First such number will be 10010 (11 * 910) and last will be 99990 (11 + 9090)
Total such numbers = 9090 - 910 + 1 = 8181

Answer (B)

gmatophobia · Answer

The question can be solved using counting techniques.

Let the five-digit number be \(abcde\). In this representation, 'e' represents the units digit of the number, and 'a' represents the ten-thousands digit of the number.

The quotient of when \(abcde\) is divided by 100, is '\(abc\)', the remainder is '\(de\)'.

q = abc

r = de

The value of '\(abc\)' will be between '100' and '999'; the value of 'de' will be between 00 and 99.

We want to find the remainder when q + r is divided by 11.

As there are two parts to the sum, we can have multiple possibilities such as

1) Remainder(\(\frac{q}{11}\)) = 0 & Remainder(\(\frac{r}{11}\)) = 00
2) Remainder(\(\frac{q}{11}\)) = 1 & Remainder(\(\frac{r}{11}\)) = 10
3) Remainder(\(\frac{q}{11}\)) = 2 & Remainder(\(\frac{r}{11}\)) = 09
4) Remainder(\(\frac{q}{11}\)) = 3 & Remainder(\(\frac{r}{11}\)) = 08
.
.
.
.
.
10) Remainder(\(\frac{q}{11}\)) = 09 & Remainder(\(\frac{r}{11}\)) = 02
11) Remainder(\(\frac{q}{11}\)) = 10 & Remainder(\(\frac{r}{11}\)) = 01

Let's try to find the the possible values each of '\(abc\)' and '\(de\)' can take for the cases listed above -

Case 1: Remainder(\(\frac{q}{11}\)) = 0 & Remainder(\(\frac{r}{11}\)) = 00

abc - Remainder of (\(\frac{abc}{11}\)) = 0

Lowest Value: \(110\)
Highest Value: \(990\)
Number of terms: \(\frac{990-110}{11} + 1 = 81\)

de - Remainder of (\(\frac{de}{11}\)) = 0

Lowest Value: \(00\)
Highest Value: \(99\)
Number of terms: \(\frac{99-11}{11} + 1= 10\)

Total Possible Values = \(81 * 10 = 810\)

Case 2: Remainder(\(\frac{q}{11}\)) = 1 & Remainder(\(\frac{r}{11}\)) = 10

abc - Remainder of (\(\frac{abc}{11}\)) = 1

Lowest Value: \(100\)
Highest Value: \(991\)
Number of terms: \(\frac{991-100}{11} + 1 = 82\)

de - Remainder of (\(\frac{de}{11}\)) = 10

Lowest Value: \(10\)
Highest Value: \(98\)
Number of terms: \(\frac{98-10}{11} + 1= 9\)

Total Possible Values = \(82 * 09 = 738\)

We can find the values of the remaining cases as shown below.

Attachment:

Screenshot 2023-04-03 114026.jpg

Note: Once we have identified a pattern, we don't have to calculate for each case. For example, the number of possible values that '\(abc\)' can take remains the same for the remainder of 1 to 9. In a similar manner, the number of possible values '\(de\)' can take also remains constant for each possible remainder. This will help speed up the calculation process.

Total number of terms = (81*10) + (82 * 9 * 9) + (81 * 9) = 810 + 6642 + 729 = 8181

Option B

P.S. While we can optimize the process by not actually calculating the number of terms for each possible combination of the remainder, I still feel this is a bit tedious process (the working took me more than 2 mins

). Hope there is a shorter and more elegant version to solve this question

). Hope there is a shorter and more elegant version to solve this question

samarpan.g28 · Answer

We are dividing n by 100 -> q will be our quotient and r will be our reminder. So we can write n=100q+r.
Now, q is always greater than r. Also, we need to ensure n from above eqn has exactly five digits. This is possible when q has a minimum of three digits and n has a minimum of two digits.

Considering the above, the minimum three-digit number is 100 (value of q) and the minimum two-digit number is 10 (value of r). If I add them, 100+10=110 -> this is divisible by 11 (11*10 = 110).
Also, 100*100+10=10010 (value of n) is a five-digit number.
Do we notice one thing here? 10010 i.e. n is also divisible by 11 (11*910=10010).

What will be the next q+r? 121 since 11*11=121. So q=100 and r=21. Also, 100*100+21=10021 (11*911=10021). This is enough to understand that  we only have to find five-digit numbers divisible by 11,  considering q and r satisfying the above condition.
10010,10021,.....99990.

Apply AP: 99990=10010+(n-1)*11 where n is the number of elements (not to confuse with the constant given in the problem). n=8181. Option (B) is correct.

ANVESHCH · Answer

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+d 
we need to see if  a+c+e  either equal to  b+d or b+d+11 
this 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.

MehakKaushal · Answer

When n is divided by 100,
n=100q+r,0≤r<100.
Given q+r is divisible by 11,
q+r=11k for some integer k.
So, q=11k−r.
Substitute into n=100q+r:
n =100(11k−r)+r
=1100k−100r+r
=1100k−99r
=11(100k−9r).​
Hence, n must be divisible by 11.
The 5-digit multiples of 11 are:
 Smalles t: Divide 10,000 by 11, which gives 909 with a remainder of 1. 11 times 909 = 9,999 (too small, only 4 digits). Add 11 to get the first 5-digit one: 9,999 + 11 = 10,010
10010=11×910
 Largest : Divide 99,999 by 11, which gives 9,090 with a remainder of 9.Subtract the extra 9 leftover: 99,999 - 9 =99,990
99990=11×9090
These form an arithmetic progression with common difference 11.
99990= 10010 +(n-1)11
n=8181​
Answer: 
(B) 8181
 ​

Adit_ · Answer

It is actually really simple if we note it down algebraically:

n = 100q+r.
n = 99q+ 
We know q+r = Multiple of 11, lets say 11y.

n = 99q+11y.
n = 11(9q+y)
q and y are independent of each other and n>10,000 so this covers all multiples of 11.
All multiples of 11 greater than 10,000 =  8181 .

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