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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Answer:
Three quarters of the members of a club are juniors, the rest are seniors. Express this as a ratio of juniors:seniors.
I think we can all agree that 1 centimeter=10millimeters.
And if x is the side in centimeters then 10x will be the side in millimeters.
We just need to rewrite x^3 into (10x)^3.
Because you need to do the exponent to all part of the ( )...
(10x)^3=1000x^3
V(x)=1000x^3 cubic millimeters is the answer
Answer:
✎There will be 30 Chocolate-Covered raisins in each bag.
✎ And 5 Remaining.
Step-by-step explanation:
Take 185 and divide it by 6 and you should get 30 per bag with a remainder of 5 :)
