(33076161)^1/3 + (279841)^1/4 + (243)^1/5 = ?
Given options are
- 344
- 345
- 346
- 347
- None of these
The solution involves in four steps
- Finding the cube root of the number 33076161
- Finding the fourth root of 279841
- Find the fifth root of 243
- Adding all the three numbers
- Finding the cube root of the number 33076161
TP SP FP - We will be finding the numbers in these three places as below.
- Divide the number in three parts
- First number (FN) : 33
- Second number (SN) : 076
- Third number (TN) : 161
To find the cube root, we will use TN, then FN and finally SN.
- TN ends with 1, so the number in the ones place (FP) is 1. (Refer Table 1 below)
TP SP 1
- FN is 33, which in between the cube of 3 and 4, so we have to choose the small number which is 3 will be in third place (TP). (Refer table 1 below)
3 SP 1
- To find the number in second place, we can use the formula - 3*(TP^2)*SP= second digit of (TN-FP)
- 3*(1^2)*SP= Second digit of (161-1) = Second digit of 160 = 6
- 3*(1^2)*SP=6
- SP=6/3=2
3 2 1
- Finally, cube root of (33076161)⅓ is 321
2. Finding the fourth root of 279841
- First find the square root of the number 279841
- Square root of 279841 is 529
- Again find the square root of 529
- Square root of 529 is 23
- (23^4)¼=23
3. Find the fifth root of 243
- Rewrite 243 as 3^5
- Cancel the common factor of 5
- And the answer is 3
4. Then, the final step is adding all the three numbers 321+23+3=347
Table 1: Square, Cube, Numbers Ranging 0 - 10
Number x
|
Square x2
|
Cube x3
|
1
|
1
|
1
|
2
|
4
|
8
|
3
|
9
|
27
|
4
|
16
|
64
|
5
|
25
|
125
|
6
|
36
|
216
|
7
|
49
|
343
|
8
|
64
|
512
|
9
|
81
|
729
|
10
|
100
|
1000
|
