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13 May 2019, 23:08
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
39% (02:17) correct
61%
(02:21)
wrong
based on 119
sessions
History
Date
Time
Result
Not Attempted Yet
How many of the first 2018 numbers in the sequence 101, 1001, 10001, 100001, ... are divisible by 101?
A. 253
B. 504
C. 505
D. 506
E. 1009
A. 253
B. 504
C. 505
D. 506
E. 1009
17 May 2019, 04:06
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This is a question which requires you to know some advanced divisibility rules. It also requires you to be able to interpret the given sequence and develop a pattern, to solve the question.
Firstly, 101 is a number which can be written in the form of \(10^2\) + 1. Therefore, the divisibility rule for 101 is:
1. Starting from the units digit, divide the given number into groups of 2 digits each (because of \(10^2\) + 1) and number the groups as 1,2 ,3 and so on, starting from the units digit side again.
2. Add the numbers in the odd numbered groups and add the numbers in the even numbered groups.
3. Calculate the difference between the two sums.
4. If this difference is ZERO or a multiple of 101, then the given number is divisible by 101; otherwise, it is not.
101 is clearly divisible by 101.
The next number is 1001. If we apply the divisibility rule on 1001, we see that the difference of the numbers in the groups (there are only 2 groups) is 9 (10 – 01). 9 is not divisible by 101.
The next number 10001 is also not divisible, and so is 100001.
1000001 is divisible by 101. Working out in the same way, the next number divisible is 10000000001.
So, the pattern is there should be 1 zero or 5 zeroes or 9 zeroes or 13 zeroes in the numbers to be divisible by 101. Therefore, we are looking at a 3-digit number or a 7-digit number or a 11-digit number and so on.
There is a uniform spacing between the values – 3, 7, 11, 15 and so on. This spacing is 4.
The biggest number in our sequence will have 2020 digits, since our first number has 3 digits and there are 2018 numbers.
The closest multiple of 4 ( spacing of the sequence) to 2018 is 2016. Dividing 2016 by 4, we get 504. Since we have to include both the values at the extremes, we will have to add 1 to 504. This means there will be 504 + 1 i.e. 505 values in this sequence. And this sequence consists of numbers which are divisible by 101.
So, the correct answer option is C.
As mentioned earlier, unless you develop a pattern from the sequence given, it will be hard to calculate how many terms have to be considered.
Hope this helps!
Firstly, 101 is a number which can be written in the form of \(10^2\) + 1. Therefore, the divisibility rule for 101 is:
1. Starting from the units digit, divide the given number into groups of 2 digits each (because of \(10^2\) + 1) and number the groups as 1,2 ,3 and so on, starting from the units digit side again.
2. Add the numbers in the odd numbered groups and add the numbers in the even numbered groups.
3. Calculate the difference between the two sums.
4. If this difference is ZERO or a multiple of 101, then the given number is divisible by 101; otherwise, it is not.
101 is clearly divisible by 101.
The next number is 1001. If we apply the divisibility rule on 1001, we see that the difference of the numbers in the groups (there are only 2 groups) is 9 (10 – 01). 9 is not divisible by 101.
The next number 10001 is also not divisible, and so is 100001.
1000001 is divisible by 101. Working out in the same way, the next number divisible is 10000000001.
So, the pattern is there should be 1 zero or 5 zeroes or 9 zeroes or 13 zeroes in the numbers to be divisible by 101. Therefore, we are looking at a 3-digit number or a 7-digit number or a 11-digit number and so on.
There is a uniform spacing between the values – 3, 7, 11, 15 and so on. This spacing is 4.
The biggest number in our sequence will have 2020 digits, since our first number has 3 digits and there are 2018 numbers.
The closest multiple of 4 ( spacing of the sequence) to 2018 is 2016. Dividing 2016 by 4, we get 504. Since we have to include both the values at the extremes, we will have to add 1 to 504. This means there will be 504 + 1 i.e. 505 values in this sequence. And this sequence consists of numbers which are divisible by 101.
So, the correct answer option is C.
As mentioned earlier, unless you develop a pattern from the sequence given, it will be hard to calculate how many terms have to be considered.
Hope this helps!
17 May 2019, 07:09
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Firstly, 101 is a number which can be written in the form of \(10^2\) + 1. Therefore, the divisibility rule for 101 is:
1. Starting from the units digit, divide the given number into groups of 2 digits each (because of \(10^2\) + 1) and number the groups as 1,2 ,3 and so on, starting from the units digit side again.
2. Add the numbers in the odd numbered groups and add the numbers in the even numbered groups.
3. Calculate the difference between the two sums.
4. If this difference is ZERO or a multiple of 101, then the given number is divisible by 101; otherwise, it is not.
101 is clearly divisible by 101.
The next number is 1001. If we apply the divisibility rule on 1001, we see that the difference of the numbers in the groups (there are only 2 groups) is 9 (10 – 01). 9 is not divisible by 101.
The next number 10001 is also not divisible, and so is 100001.
1000001 is divisible by 101. Working out in the same way, the next number divisible is 10000000001.
So, the pattern is there should be 1 zero or 5 zeroes or 9 zeroes or 13 zeroes in the numbers to be divisible by 101. Therefore, we are looking at a 3-digit number or a 7-digit number or a 11-digit number and so on.
Hello Arvind!
Can you please explain the grouping for divisblity rule of 101?
Posted from my mobile device
1. Starting from the units digit, divide the given number into groups of 2 digits each (because of \(10^2\) + 1) and number the groups as 1,2 ,3 and so on, starting from the units digit side again.
2. Add the numbers in the odd numbered groups and add the numbers in the even numbered groups.
3. Calculate the difference between the two sums.
4. If this difference is ZERO or a multiple of 101, then the given number is divisible by 101; otherwise, it is not.
101 is clearly divisible by 101.
The next number is 1001. If we apply the divisibility rule on 1001, we see that the difference of the numbers in the groups (there are only 2 groups) is 9 (10 – 01). 9 is not divisible by 101.
The next number 10001 is also not divisible, and so is 100001.
1000001 is divisible by 101. Working out in the same way, the next number divisible is 10000000001.
So, the pattern is there should be 1 zero or 5 zeroes or 9 zeroes or 13 zeroes in the numbers to be divisible by 101. Therefore, we are looking at a 3-digit number or a 7-digit number or a 11-digit number and so on.
Hello Arvind!
Can you please explain the grouping for divisblity rule of 101?
Posted from my mobile device
