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:
X =8
Step-by-step explanation:
DF = DE + EF
15 = X+1 + X-2
16 = 2X
X = 8
Y=t+12 this is one the answers
Answer:
x<13
Step-by-step explanation:
1. Move variable [x] to left-hand side and change its sign
Move constant [10] to the right-hand side and change its sign
10>x-3
-x+10>-3-10
2. Calculate the difference [-3-10]
-x>-3-10
-x>-13
3. Change the signs [-x>-13] on both sides of the inequality and flip the inequality sign
x<13
4. Solution
x<13
Answer:
Step-by-step explanation:
