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Find three perfect squares whose product is 45 less than the largest 4-digit perfect cube?
There are quite a few solutions to this problem, and here's how to find them: 99991/3 = 21.543... Hence 213 = 9261 is the largest four digit cube number. 45 less than this is 9216. ---- We want to expess 9216 in the form A2B2C2 9216 = A2B2C2 Square root both sides: 96 = ABC And from here we can deduce possible values for A, B and C. The prime factorization of 96 is: 96 = 25 x 3 = 2 x 2 x 2 x 2 x 2 x 3 By separating this string of prime factors into three groups we can find our possible solutions. e.g. 96 = (2 x 2) x (2 x 2 x 2) x (3) so 96 = 4 x 8 x 3 and 9216 = 42 x 82 x 32 Find more solutions by splitting the prime factors differently.
What is ABC?
The alphabetAmerican Broadcasting corporationAustralian Broadcasting commissionThe first 3 letters of the alphabet, the alphabet is a group of 26 letters UPPERCASE and lowercase here is the alphabet:A a B b C c D d E e F f G g H h I i J j K k L l M m N n O o P p Q q R r S s T t U u V v W w X x Y y Z z
