How can you rewrite a perfect square as a exponent?

Answer 1

Find its square root and raise it in the second power.

64 = 82

81 = 92

841 = ?2 you need to work a little bit here

1. base would be in two digits (odd number with 3 digits, (3+1)/2)

2. the first digit would be 2 (the square of 2 is less than 8), so the base would be between 20 and 30.

3. the second digit would be 9 (the square of 9 ends in 1)

4. the answer is 29, 841 = 292

529 = ?2

1. base would be in two digits (odd number with 3 digits, (3+1)/2)

2. the first digit would be 2 (the square of 2 is less than 5), so the base would be between 20 and 30.

3. the second digit would be 3 or 7 (the square of 3 and 7 end in 9); is it 23 or 27?

4. 529 - 2232 = 529 - 409 = 120; it is 3, because 2(2*3)=12. So 529 = 232

2809 = ?2

1. base would be in two digits (even number with 4 digits, 4/2)

2. the first digit would be 5 (the square of 5 is less than 28,), so the base would be between 50 and 60.

3. the second digit would be 3 or 7 (the square of 3 and 7 end in 9); is it 53 or 57?

4. 2809 - 5232 = 2809 - 2509 = 300; it is 3, because 2(5*3)=30. So 2809 = 532

7056 = ?2

1. base would be in two digits (even number with 4 digits, 4/2)

2. the first digit would be 8 (the square of 8 is less than 70,), so the base would be between 80 and 90.

3. the second digit would be 4 or 6 (the square of 4 and 6 end in 6); is it 84 or 86?

4. 7056 - 8242 = 7056 - 6416 = 640; it is 4, because 2(8*4)=64. So 7056 = 842

7396 = ?2

1. base would be in two digits (even number with 4 digits, 4/2)

2. the first digit would be 8 (the square of 8 is less than 73), so the base would be between 80 and 90.

3. the second digit would be 4 or 6 (the square of 4 and 6 end in 6); is it 84 or 86?

4. 7396 - 8262 = 7396 - 6436 = 960; it is 6, because 2(8*6)=96. So 7396

(if you test 4 first, you'll see that it doesn't work, so it will be 6)

55696 = ?2

1. base would be in three digits (odd number with 5 digits, (5+1)/2)

2. the first digit would be 2 (the square of 2 is less than 5), so the base would be between 200 and 300.

3. the last digit would be 4 or 6 (the square of 4 and 6 end in 6); is it 2_4 or 2_6?

4. let's work a little bit here:

2_4; 22_42 = 40016

40016+ ... = 45696

00080 (101*2*4*_, we need to find the middle digit)

01600 (102*2*2*4)

04000 (103*2*2*_)

55696 - 45696 = 10000

40016

00080

01600

04000

10000

00080 doesn't change, this tell us the middle digit can be 1 or 6. But one is too small and 6 is too big (2*2*1= 4, 2*2*6=32). So the last digit is not 4, it is 6. Let's find the middle digit.

2_6; 22_62 = 40036

40036+ ... = 46556

00120 (101*2*6*_, we need to find the middle digit)

02400 (102*2*2*6)

04000 (103*2*2*_)

55696 - 46556 = 9140

40036

00120

02400

04000

09140

00120 becomes 00160, and tell us the middle digit can be 3 or 8, but 8 is too big (2*2*8=32), so it is 3

Thus, 55696 = 2362.

263169 = ?2

1. base would be in three digits (even number with 6 digits, 6/2)

2. the first digit would be 5 (the square of 5 is less than 26), so the base would be between 500 and 600.

3. the second digit would be 3 or 7 (the square of 3 and 7 end in 9); is it 5_3 or 5_7?

4. let's work a little bit here:

5_3; 52_32 = 250009

250009+ ... = 263069

000060 (101*2*3*_, we need to find the middle digit)

003000 (102*2*5*3)

010000 (103*2*5*_)

263169 - 263069 = 100

250009

000060

003000

010000

000100

00060 doesn't change so the only possibility is when the middle digit is 1.

Thus, 2631169 = 5132

Answer 2

28