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N is an 80-digit positive integer in the decimal system. All digits

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N is an 80-digit positive integer in the decimal system. All digits  [#permalink]

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New post 24 Jan 2020, 23:25
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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

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N is an 80-digit positive integer in the decimal system. All digits  [#permalink]

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New post 25 Jan 2020, 02:18
2
1
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)

Attachment:
11.JPG
11.JPG [ 16.73 KiB | Viewed 428 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
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Re: N is an 80-digit positive integer in the decimal system. All digits  [#permalink]

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New post 25 Jan 2020, 22:21
sujoykrdatta wrote:
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)

Attachment:
11.JPG


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





Thanks sujoykrdatta for the debrief. May i know what is the range of divisibility rule that is under the GMAT scope?
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Re: N is an 80-digit positive integer in the decimal system. All digits  [#permalink]

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New post 25 Jan 2020, 22:40
Hi.

You should be knowing the divisibility rules for
2, 4, 8 (powers of 2)
5
3 & 9

And also how to tackle non-primes (composites) like 6, 12, 24 etc.

You may check the rule for 11 (though I don't think that will be asked either).

I feel you won't get questions asking about divisibility of 7, 13, etc.

However, you should try to understand the concept of factors of a number like 555555 (the concept I used).

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N is an 80-digit positive integer in the decimal system. All digits  [#permalink]

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New post 04 Feb 2020, 11:09
sujoykrdatta wrote:
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)

Attachment:
11.JPG


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



Mr sujoykrdatta,
Awesome solution, however there is a small typo, you may wish to correct to prevent confusion. In the figure 2 should be the 43rd digit and the 44th digit should be x. Correspondingly the typo in the last line should be corrected. However the solution is awesome+1.
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N is an 80-digit positive integer in the decimal system. All digits   [#permalink] 04 Feb 2020, 11:09
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