let a, b, c be positive integers less than 10 such that (100a + 10b + c)^2 = (a+b+c)^5 what is the value of (a*b-c)
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lets assume
A = 9
B = 9
99+99 = 198
as you can see AAC cannot be bigger then 198 so A has to be 1.
A=1
11C is a three digit number then B has to be 9 !!
since 18+81 = 99 --> two digits
19+91 = 100+10+C
110 = 110+C
C=0
the answer is (E)
A = 9
B = 9
99+99 = 198
as you can see AAC cannot be bigger then 198 so A has to be 1.
A=1
11C is a three digit number then B has to be 9 !!
since 18+81 = 99 --> two digits
19+91 = 100+10+C
110 = 110+C
C=0
the answer is (E)
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