in a five digit no 1b6a3,a is the greatest single digit no. and twice of it exceeds b by 7. then the sum of the number and its cube root is?with solution
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Greatest single digit perfect cube = 8
a=8
b=2a-7=9
No. : 19683
Ans : 19683+(19683)^1/3 = 19683+27
= 19710
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a=8
b=2a-7=9
No. : 19683
Ans : 19683+(19683)^1/3 = 19683+27
= 19710
hope it helps you. . . . mark as a brainlist. . . . follow me. . .
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