twobagels wrote:
Let n be a 5-digit number, and let q and r be the quotient and 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
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
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.
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.
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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 [ 82.4 KiB | Viewed 2004 times ]
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 BP.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
_________________
Want to discuss quant questions and strategies : Join the quant chat group todayPower of Tiny Gains1.01^(365) = 37.780.99^(365) = 0.03