Box Problem Redux
Find largest 9 consecutive integers that can be arranged such that they satisfy the equation above.
What is the value of ?
That is, sum of the volumes of two boxes equals that of the third box. What is the volume of the third box?
Note : The integers are not necessarily in order. Only that 9 consecutive integers are used, each once.
Full credit for this wonderful problem goes to Vikram Pandya.
The answer is 1092.
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OK,maybe there's a more rigorous proof here.
Clearly,in 9 consecutive natural numbers,at least 4 of them are even.
According to the principle of drawer,the three box must have a even number in it at least.i.e.the volume of the three boxes all are even.
Let the Median = n
the all 9 integers can be expressed as (n-4),(n-3),(n-2),(n-1),n,(n+1),(n+2),(n+3),(n+4),n≥5
Meanwhile, these nine integers can not have any one of the more than 13 prime factor(n≤9), or according to the prime decomposition theorem, can not meet the conditions set.
1° n is odd.
there 5 odd numbers and 4 even numbers.
Def.○ is odd,● is even.
Case 1 ●○○ +●○○=●●○
Both sides mod 4,we have ●○○+●○○≡0(mod 4)
∴abc≡def≡0(mod 4) or abc≡1(mod 4),def≡3(mod 4) or abc≡def≡2(mod 4)
if x≡y≡0(mod 4),n=4k+1 or 4k+3
when n=4k+1,we heve
(n+1)(n-4)(n-3)+(n+3)(n+2)(n-2)=n(n+4)(n-1)
here n = 5.abc=12,def=168,ghi=180
if abc≡1(mod 4),def≡3(mod 4),No solution
if abc≡def≡2(mod 4)
(n-4)(n+1)n+(n-1)(n+3)(n-2)=(n+4)(n+2)(n-3)
Only n=5 satisfied.abc=30,def=96,ghi=126
Case 2 ●●○+●○○=●○○
Similarly,we have
(n-3)(n+1)(n+2)+(n-1)(n+4)(n-4)=(n+3)n(n-2)
n=5,abc=84,def=36,ghi=120
(n-4)(n-3)(n+1)+(n-1)n(n+2)=(n-2)(n+3)(n+4)
n=9,abc=300,def=792,ghi=1092
Similarly,discussion n is an even number of cases.
Only n=6,then,abc=48,def=152,ghi=300
SO THE LARGEST VALUE IS 1092