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Vlad1618 [11]
2 years ago
7

The first digit of a string of 2002 digits is a 1. Any two-digit number formed by consecutive digits within this string is divis

ible by 19 or 31. What is the largest possible last digit in this string
Mathematics
1 answer:
Umnica [9.8K]2 years ago
4 0

The largest possible last digit in the string of 2002 digits and number divisible by 19 or 31 is 9.

Given the first digit of a string of 2002 digits is 1 and the two digit number formed by consecutive digits within the string is divisible by 19 or 31.

We have to tell the last largest digit of such number.

Two digit numbers divisible by 19=19,38,57,76,95.

Two digit numbers divisible by 31=31,62,93,124

Number started with 1 =19

Last digit is 9

We have said that the number should be divisible by 19 or 31 not from both and started with 1.

Hence the largest possible last digit and number divisible by 19 or 31 in this string is 9.

Learn more about digits at brainly.com/question/26856218

#SPJ4

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Step-by-step explanation:

<u><em>Explanation:-</em></u>

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Step-by-step explanation:

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