Answer: 
<u>Simplify both sides of the equation</u>
<u>Subtract 5n from both sides</u>
<u>Subtract 15 from both sides</u>
Assume a is not divisible by 10. (otherwise the problem is trivial).
<span>Define R(m) to be the remainder of a^m when divided by 10. </span>
<span>R can take on one of 9 possible values, namely, 1,2,...,9. </span>
<span>Now, consider R(1),R(2),......R(10). At least 2 of them must have the sames value (by the Pigeonhole Principle), say R(i) = R(j) ( j>i ) </span>
<span>Then, a^j - a^i is divisible by 10.</span>
Answer:
Step-by-step explanation:
Answer:
1)-
How to solve your question
Your question is
4(4−72)−9(5+2)
4(4y-7y^{2})-9(5y+2)4(4y−7y2)−9(5y+2)
Simplify
1
Rearrange terms
4(4−72)−9(5+2)
4({\color{#c92786}{4y-7y^{2}}})-9(5y+2)4(4y−7y2)−9(5y+2)
4(−72+4)−9(5+2)
4({\color{#c92786}{-7y^{2}+4y}})-9(5y+2)4(−7y2+4y)−9(5y+2)
2
Distribute
4(−72+4)−9(5+2)
{\color{#c92786}{4(-7y^{2}+4y)}}-9(5y+2)4(−7y2+4y)−9(5y+2)
−282+16−9(5+2)
{\color{#c92786}{-28y^{2}+16y}}-9(5y+2)−28y2+16y−9(5y+2)
3
Distribute
−282+16−9(5+2)
-28y^{2}+16y{\color{#c92786}{-9(5y+2)}}−28y2+16y−9(5y+2)
−282+16−45−18
-28y^{2}+16y{\color{#c92786}{-45y-18}}−28y2+16y−45y−18
4
Combine like terms
2)
−17y+17z+24
See steps
Step by Step Solution:

STEP1:Equation at the end of step 1
((24 - 4 • (5y - 6z)) + 3y) - 7z
STEP2:
Final result :
-17y + 17z + 24
−282+16−45−18
-28y^{2}+{\color{#c92786}{16y}}{\color{#c92786}{-45y}}-18−28y2+16y−45y−18
−282−29−18
-28y^{2}{\color{#c92786}{-29y}}-18−28y2−29y−18
Solution
−282−29−18
