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25 Jan 2020, 00:25
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C
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Difficulty:
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Question Stats:
29% (02:25) correct
71%
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N is an 80-digit positive integer in the decimal system. All digits except for the 44th digit from the left are 2. If N is divisible by 13, find the 44th digit.
A. 1
B. 5
C. 6
D. 8
E. 9
A. 1
B. 5
C. 6
D. 8
E. 9
25 Jan 2020, 03:18
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Question:
N is an 80-digit positive integer in the decimal system. All digits except for the 44th digit from the left are 2. If N is divisible by 13, find the 44th digit.
Solution:
Idea to be used: If we have a 6-digit number with all digits identical, say dddddd, where d is a digit, this number is divisible by 7, 11 and 13. Why?
Note: dddddd = ddd x 1000 + ddd = ddd x (1000 + 1)
= ddd x 1001
= d x (111) x (1001)
= d x (3 x 37) x (7 x 11 x 13)
Now, we have 80 digit number where all digits are 2 except the 44th digit from the left.
Let us group six 2s at a time starting from the left (shown below):
222222222222 ...
---6---|---6---|
We will get 7 groups and reach the 42nd digit
Next, we have the 43rd digit which is also 2; and the 44th digit - say d.
After this, there are 36 more digits (all 2s), which can also be grouped six at a time to form 6 groups (see image)
11.JPG [ 16.73 KiB | Viewed 9234 times ]
The 7 groups initially and the 6 groups at the end are all divisible by 13
Thus, we just need the number formed by the 42nd and the 43rd digits, i.e. "2x" to be divisible by 13
Thus, x = 6 (since 26 is divisible by 13)
Answer C
But I doubt this will be asked in the GMAT
N is an 80-digit positive integer in the decimal system. All digits except for the 44th digit from the left are 2. If N is divisible by 13, find the 44th digit.
Solution:
Idea to be used: If we have a 6-digit number with all digits identical, say dddddd, where d is a digit, this number is divisible by 7, 11 and 13. Why?
Note: dddddd = ddd x 1000 + ddd = ddd x (1000 + 1)
= ddd x 1001
= d x (111) x (1001)
= d x (3 x 37) x (7 x 11 x 13)
Now, we have 80 digit number where all digits are 2 except the 44th digit from the left.
Let us group six 2s at a time starting from the left (shown below):
222222222222 ...
---6---|---6---|
We will get 7 groups and reach the 42nd digit
Next, we have the 43rd digit which is also 2; and the 44th digit - say d.
After this, there are 36 more digits (all 2s), which can also be grouped six at a time to form 6 groups (see image)
Attachment:
11.JPG [ 16.73 KiB | Viewed 9234 times ]
The 7 groups initially and the 6 groups at the end are all divisible by 13
Thus, we just need the number formed by the 42nd and the 43rd digits, i.e. "2x" to be divisible by 13
Thus, x = 6 (since 26 is divisible by 13)
Answer C
But I doubt this will be asked in the GMAT
General Discussion
ConfusedSEA
Joined: 20 Aug 2019
Last visit: 25 Jun 2021
Posts: 30
Own Kudos:
Given Kudos: 66
Location: Malaysia
Concentration: Strategy, General Management
Schools: Mannheim "21 (A$)
GMAT 1: 630 Q40 V36
GPA: 4
25 Jan 2020, 23:21
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sujoykrdatta
Thanks sujoykrdatta for the debrief. May i know what is the range of divisibility rule that is under the GMAT scope?
