\(N\equiv 2 \pmod{LCM[6,7,8]}\\ N\equiv 2 \pmod{168}\)
set up equation
\(168N+2=1000\\ 168N=998\\ N=\frac{499}{84} \approx 5.940\)
round down N = 6
answer is 6.
\(a \equiv 5 \pmod{7}\\ a+1\equiv 5+1 \pmod7\\ a+1 \equiv 6 \pmod7\)
6 is the answer
convert 2/7 to base 16
2/7_10 = 2/7_16 (nothing changes)
do base division
\(\frac{2}{7}_{16} \approx 0.4924924924924925\)
The answer is 49249
The answer is \(3*7=21\)
so 1 is the answer
The answer is \(1*2*6 \equiv 5 \pmod7\)
so 5 is the answer
take the first and last number. sum them
\(140+1=141\)
multiply by half the number of total numbers. \(140/2=70\)
we thus have \(141*70=9870 \)
prime factorize 9870. \(9870 = 2 * 3 * 5 * 7 * 47\)
take largest prime number --> 47
47 is the answer.
Hi EP! :)
\(1\% * 200 = 2\) this means that only 2 kg were not water.
let x be the new amount
\(2 = 10\%x\)
solve for x
\(2(10)=x\\ x=20\)
20kg is the answer.
The answer is \(0\)
