If a; b; c and d are consecutive natural numbers and a^3=b^3+c^3+d^3,what is the least value of 'a'?
Answers
Answer:
6
Step-by-step explanation:
a
3
=b
3
+c
3
+d
3
is possible for consecutive natural numbers only if a>b>c>d
Therefore, b=(a−1),c=(a−2),d=(a−3)
⇒a
3
=(a−1)
3
+(a−2)
3
+(a−3)
3
a cannot be 3 as this will make d=0 and it is given in the question that all a,b,c,d are natural numbers
Substituting for a from the answer option, starting with the least number
If a=6, then LHS=a
3
=3
3
=216, and
RHS= (a−1)
3
+(a−2)
3
+(a−3)
3
=5
3
+4
3
+3
3
=125+64+27=216
We can see LHS=RHS
We can stop here as we want the least value for a.
Hence, 6 is the correct answer.
Answer:
6
Step-by-step explanation:
The correct answer is 6
a
3
=b
3
+c
3
+d
3
is possible for consecutive natural numbers only if a>b>c>d
Therefore, b=(a−1),c=(a−2),d=(a−3)
⇒a
3
=(a−1)
3
+(a−2)
3
+(a−3)
3
a cannot be 3 as this will make d=0 and it is given in the question that all a,b,c,d are natural numbers
Substituting for a from the answer option, starting with the least number
If a=6, then LHS=a
3
=3
3
=216, and
RHS= (a−1)
3
+(a−2)
3
+(a−3)
3
=5
3
+4
3
+3
3
=125+64+27=216
We can see LHS=RHS
We can stop here as we want the least value for a.
Hence, 6 is the correct answer.