If an integer is both a square and a cube, it can be of the form:
<span>(<span>a3</span><span>)^2</span></span>
Now,
since a cube can be of the form 7k or 7k+-1(thanks to FoolForMath),
we write
<span><span>a^3</span>=7k</span>
and get the no to be
49k^2
, which is in the form of 7 times something
<span>49<span>k^2</span>=7×(7<span>k^2</span>)</span>
Now put
<span><span>a^3</span>=7k+−1</span>
Square it
and you'll get a number in the form of (7times something +1)
7x. Try to ignore the X's for a second and think of the problem as -12+5 (aka 5-12). The answer to that is 7. now add the x back again to get: 7x
To check it we can use Pythagorean theorem.
, where c is the longest side.
Lets check it
Assume that a=10 and b= 24
Now lets compare with c^2=676
Its equal, so its right triangle.
*I hope I read these problems correctly. The photo is very blurry so I tried my best.*
5. 3 + (16 ÷ 4)
3 + 4
= 7
6. 12 ÷ 4 + 5
3 + 5
= 8
7. 6 + 4 x 2
6 + 8
= 14
8. (9 + 7) - 8
16 - 8
= 8
9. 6 x 3 - 4
18 - 4
= 14
10. 6 + (10 ÷ 5)
6 + 5
= 11
11. (7 x 0) + 7
0 + 7
= 7
12. 8 ÷ (3 + 1)
8 ÷ 4
= 2
13. (7 x 6) + 7
42 + 7
= 49
14. 10 - 2 x 4
8 x 4
= 32
15. 6 - (4 - 4)
6 - 0
= 6
Hope I helped. >w< ❤