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Consider a 50 digit number such that its all digits, except the 26th

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Bunuel
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Consider a 50 digit number such that its all digits, except the 26th [#permalink] Consider a 50 digit number such that its all digits, except the 26th   09 Dec 2021, 05:00
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Consider a 50 digit number such that its all digits, except the 26th digit from left are 1. If the number is divisible by 13, find the 26th digit from the left.

A. 1
B. 9
C. 6
D. 5
E. 3


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E
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Aeyvei
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Consider a 50 digit number such that its all digits, except the 26th [#permalink] Consider a 50 digit number such that its all digits, except the 26th   17 Dec 2021, 00:00
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Bunuel wrote:
Consider a 50 digit number such that its all digits, except the 26th digit from left are 1. If the number is divisible by 13, find the 26th digit from the left.

A. 1
B. 9
C. 6
D. 5
E. 3


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Because the number is too big, there must be a pattern.

The number 111111 (6 times 1) is completely divisible by 13, after which the quotient starts repeating itself.

So, till 24th digit (a multiple of 6), the number will be completely divisible by 13. And after 26th digit as well, there will 24 such digits.

We then basically need to deal with 25th and 26th digit. Since 25th digit is given as 1, 26th digit will have to be 3 for these 2 digits to be divisible by 13 as well to make the whole number divisible by 13.

OA: E. 3
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Consider a 50 digit number such that its all digits, except the 26th [#permalink] Consider a 50 digit number such that its all digits, except the 26th   25 May 2024, 10:52
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Bunuel KarishmaB ScottTargetTestPrep MartyMurray

Hello Experts!! Could you please present the answer for this question? Also, what is the general divisibility rule for 13, if any?

Thanks in advance!­
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Consider a 50 digit number such that its all digits, except the 26th [#permalink]
25 May 2024, 10:52
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Bunuel
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