Just Think Part#2
3 6 + 6 × 2 4 3 × 2 + 1 5 × 8 1 × 4 + 2 0 × 2 7 × 8 + 1 5 × 9 × 1 6 + 6 × 3 × 3 2 + 6 4 1 8 3 + 7 3 + 3 × 1 8 × 7 × 2 5 = a
Find the value of [ a 1 4 3 9 ]
Note- Don't use calculator
The answer is 1439.
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1 solution
You can't just say that the denominator will be in that form just because 2 6 and 3 6 are there.
The denominator comes in that form because it is in the form of the expansion of ( 2 + 3 ) 6 using binomial theorem according to which, ( x + y ) n = k = 0 ∑ n ( ( k n ) x n − k y k )
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Yes, I agree with you. It does need this full examination..
We Observe that - 18 ^ 3 + 7^3 + 3 * 18* 7*25 is in the form of (a+b)^3 that is (25)^3 in the denominator we see the term 3 ^ 6 and 2 ^ 6 ie. 64 hence the denominator is in the form( 2+3)^6 = 5^6
(25)^3/(5)^6 = 5 ^2^3 / 5 ^6 = 5^6/5^6 which implies a = 1
therefore 1439/1 = 1439