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.
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The side lengths could be 4 in. x 16 in. because that equals 64 inches^2. It could also be 8 in. x 8 in. because that also equals 64 inches^2.
You can then write really big numbers without running out of space
for example instead of 2300000000000000000000000000
you can just write 2.3 x 10^27
Answer:
Two extraneous solutions
Step-by-step explanation:
Answer:
Step-by-step explanation:
X is 1 y is -4
Which gets the following:
X(1) + y(-4) = -3
x(1) + 2(-4) = -7