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:
20 square units
Step-by-step explanation:
count the face of the figure one by one
All triangles have angles that add up to 180, so we can take the top triangle, which has angles 14, 45, and x, and say they add up to 180, so 180-14-45=x and x = 121. For E, all angles that look like that and have opposite angles, then these opposites are equal, so both sides of E are 121. then we use the bottom triangle to solve for D with the fact that all triangles' angles add to 180 again. 180-27-121 = D, so D=32
Answer:
132 and 88
Step-by-step explanation:
If there are 20% more boys than girls, then this means that the total percentage of boys us 60% and girls is 40%.
0.6 * 220 = 132
0.4 * 220 = 88
There are 132 boys and 88 girls in year 9
Answer:
Jorge needs 1/2 apple to make the snack.
