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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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Answer:

The correct answer is:   " x  >  2 "  .

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Step-by-step explanation:

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Given the inequality:

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Divide each side of the inequality by "6" ;

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4 0
3 years ago
The some of two consecutive even integers is 170. find the two integers.
Tatiana [17]
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x + x + 2 = 170
2x + 2 = 170
2x = 170 - 2
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x = 168/2
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x + 2 = 84 + 2 = 86

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8 0
3 years ago
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Step-by-step explanation:

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Dmitrij [34]

Answer:

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Step-by-step explanation:

Our understanding of your figure is shown below.

The question says the "shortest side" and the "width" have the same dimension. If the "width" is a reference to "height 6 yards", then it seems the "shortest side" is 6 yards. Since the slant sides are longer than the height, the "shortest side" is also the "short base."

The short base is 6 yards.

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The long base overhangs the short base by 1 yard on either end, so it is a total of 2 yards longer than the short base. It it 6+2 = 8 yards long.

The long base is 8 yards.

6 0
3 years ago
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